Iran’s Geometry Problems
Problems and Solutions from Contests
2014-2015
This booklet is prepared by Hirad Aalipanah, Iman Maghsoudi.
With special thanks to Morteza Saghafian, Mahdi Etesami Fard,
Davood Vakili, Erfan Salavati.
Copyright c Young Scholars Club 2014-2015. All rights reserved.
Ministry of education, Islamic Republic of Iran.
www.ysc.ac.ir - www.igo-official.ir
The first Iranian Geometry Olympiad was held simultaneously in Tehran and Isfahan on September 4th, 2014 with over 300 participants. This competition had two
levels, junior and senior which each level had 5 problems. The contestants solved
problems in 4 hours and 30 minutes.
In the end, the highest ranked participants in each level awarded with gold ruler,
silver ruler or bronze ruler respectively.
This booklet have the problems of this competition plus other geometry problems
used in other Iranian mathematical competition since summer of 2014 till spring of
2015.
This year the second Iranian Geometry Olympiad will be held in Tehran on September 3th, 2015. We tend to provide online presence for those who are interested from
other countries. Those who wish to participate can contact Mr. Salavati for more
information at erfan.salavati@gmail.com
Iranian Geometry Olympiads website: www.igo-official.ir
Problems
Problems
4
1.(Geometry Olympiad(Junior and Senior level)) In a right triangle ABC
we have ∠A = 90◦ , ∠C = 30◦ . Denot by C the circle passing through A which is
tangent to BC at the midpoint.Assume that C intersects AC and the circumcircle of
ABC at N and M respectively. Prove that M N ⊥BC.
Proposed by Mahdi Etesami Fard
2.(Geometry Olympiad(Junior Level)) The inscribed circle of 4ABC touches
BC, AC and AB at D, E and F respectively. Denote the perpendicular foots from
F , E to BC by K, L respectively. Let the second intersection of these perpendiculars
S
D
with the incircle be M , N respectively. Show that S4BM
= DK
DL
4CN D
Proposed by Mahdi Etesami Fard
3.(Geometry Olympiad (Junior Level)) Each of Mahdi and Morteza has
drawn an inscribed 93-gon. Denote the first one by A1 A2 ...A93 and the second by
B1 B2 ...B93 . It is known that Ai Ai+1 k Bi Bi+1 for 1 6 i 6 93 (A93 = A1 , B93 = B1 ).
Ai Ai+1
is a constant number independent of i.
Show that B
i Bi+1
Proposed by Morteza Saghafian
4.(Geometry Olympiad (Junior Level)) In a triangle ABC we have ∠C =
∠A + 90◦ . The point D on the continuation of BC is given such that AC = AD. A
point E in the side of BC in which A doesnt lie is chosen such that
1
∠EBC = ∠A, ∠EDC = ∠A
2
Prove that ∠CED = ∠ABC.
Proposed by Morteza Saghafian
5.(Geometry Olympiad (Junior Level)) Two points X, Y lie on the arc BC of
the circumcircle of 4ABC (this arc does not contain A) such that ∠BAX = ∠CAY .
Let M denotes the midpoint of the chord AX . Show that BM + CM > AY
Proposed by Mahan Tajrobekar
Problems
5
6.(Geometry Olympiad(Senior level)) In a quadrilateral ABCD we have
∠B = ∠D = 60◦ . Consider the line whice is drawn from M , the midpoint of AD,
parallel to CD. Assume this line intersects BC at P . A point X lies on CD such
that BX = CX. Prove that AB = BP ⇔ ∠M XB = 60◦
Proposed by Davood Vakili
7.(Geometry Olympiad(Senior level)) An acute-angled triangle ABC is given.
The circle with diameter BC intersects AB, AC at E, F respectively. Let M be the
midpoint of BC and P the intersection point of AM and EF . X is a point on the arc
EF and Y the second intersection point of XP with circle mentioned above. Show
that ∠XAY = ∠XY M .
Proposed by Ali Zooelm
8.(Geometry Olympiad(Senior level)) The tangent line to circumcircle of the
acute-angled triangle ABC (AC > AB) at A intersects the continuation of BC at P .
We denote by O the circumcenter of ABC. X is a point OP such that ∠AXP = 90◦ .
Two points E, F respectively on AB, AC at the same side of OP are chosen such
that
∠EXP = ∠ACX, ∠F XO = ∠ABX
If K, L denote the intersection points of EF with the circumcircle of 4ABC, show
that OP is tangent to the circumcircle of 4KLX.
Proposed by Mahdi Etesami Fard
9.(Geometry Olympiad(Senior level)) Two points P , Q lie on the side BC
of triangle ABC and have the same distance to the midpoint. The pependiculars
fromP , Q tp BC intesects AC, AB at E, F respectively. LEt M be the intersection
point of P F and EQ. If H1 and H2 denote the orthocenter of 4BF P and 4CEQ
recpectively, show that AM ⊥ H1 H2 .
Proposed by Mahdi Etesami Fard
10.(IGO Short list)Suppose that I is incenter of 4ABC and CI inresects AB
at D.In circumcircle of 4ABC, T is midpoint of arc BAC and BI intersect this circle
at M . If M D intersects AT at N , prove that: BM k CN .
Proposed by Ali Zooelm
Solutions
Solutions
11
1.(Geometry Olympiad(Junior and Senior Level)) In a right triangle ABC
we have ∠A = 90◦ , ∠C = 30◦ . Denot by C the circle passing through A which is
tangent to BC at the midpoint.Assume that C intersects AC and the circumcircle of
ABC at N and M respectively. Prove that M N ⊥BC.
Proposed by Mahdi Etesami Fard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Let K midpoint of side BC. Therefore:
AK = KC ⇒ ∠KAC = ∠N KC = 30◦
∠AN K = ∠N KC + ∠ACB = 60◦
A, K, N, M lie on circle (C). Therefore:
∠KAN = ∠KM N = 30◦ , ∠AM K = 60◦
We know that K is the circumcenter of 4ABC. So we can say KM = KC = AK.
Therefore 4AKM is equilateral.( because of ∠AM K = 60◦ ). So ∠AKM = 60◦ . We
know that ∠AKB = 60◦ , so we have ∠M KC = 60◦ . On the other hand:
∠KM N = 30◦ ⇒ M N ⊥BC
Solutions
12
2.(Geometry Olympiad(Junior Level)) The inscribed circle of 4ABC touches
BC, AC and AB at D, E and F respectively. Denote the perpendicular foots from
F , E to BC by K, L respectively. Let the second intersection of these perpendiculars
S
D
= DK
with the incircle be M , N respectively. Show that S4BM
DL
4CN D
Proposed by Mahdi Etesami Fard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Let I be the incenter of 4ABC. We know that
)
∠BF K = 90◦ − ∠B
1
⇒ ∠DF M = ∠B
1
◦
2
∠BF D = 90 − 2 ∠B
But ∠DF M = ∠M DK. Therefore
1
∠M DK = ∠B
2
Hense 4M DK and 4BID are similar (same angles) and
L
r
way we have N
= CD
. Therefore
DL
r=
MK
DK
=
r
.
BD
In the same
N L · CD
area of 4BM D
M K · BD
DK
M K · BD
=
⇒
=
=
DK
DL
area of 4CN D
N L · CD
DL
Solutions
13
3.(Geometry Olympiad (Junior Level)) Each of Mahdi and Morteza has
drawn an inscribed 93-gon. Denote the first one by A1 A2 ...A93 and the second by
B1 B2 ...B93 . It is known that Ai Ai+1 k Bi Bi+1 for 1 6 i 6 93 (A93 = A1 , B93 = B1 ).
Ai Ai+1
Show that B
is a constant number independent of i.
i Bi+1
Proposed by Morteza Saghafian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
We draw a 93-gon similar with the second 93-gon in the circumcircle of the first
93-gon (so the sides of the second 93-gon would be multiplying by a constant number
c). Now we have two 93-gons witch are inscribed in the same circle and apply the
problem’s conditions. We name this 93-gons A1 A2 ...A93 and C1 C2 ...C93 .
_
_
We know that A1 A2 k C1 C2 . Therefore A1 C1 =A2 C2 but they lie on the opposite
_
_
_
_
_
_
other for all 1 6 i 6 93 (A94 C94 =A1 C1 ). Therefore A1 C1 and A1 C1 lie on the opposite
_
◦
◦
side of each other. In fact, Ai Ci =Ai+1 Ci+1 and they lie on the opposite side of each
side of each other. So A1 C1 = 0 or 180 . This means that the 93-gons are coincident
or reflections of each other across the center. So Ai Ai+1 = Ci Ci+1 for 1 6 i 6 93.
Ai Ai+1
= c.
Therefore, B
i Bi+1
Solutions
14
4.(Geometry Olympiad (Junior Level)) In a triangle ABC we have ∠C =
∠A + 90◦ . The point D on the continuation of BC is given such that AC = AD. A
point E in the side of BC in which A doesnt lie is chosen such that
1
∠EBC = ∠A, ∠EDC = ∠A
2
Prove that ∠CED = ∠ABC.
Proposed by Morteza Saghafian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Suppose M is the midpoint of CD. Hense AM is the perpendicular bisector of
CD. AM intersects DE and BE at P, Q respectively. Therefore, P C = P D. We
have
∠EBA + ∠CAB = ∠A + ∠B + ∠A = 180◦ − ∠C + ∠A = 90◦
Hense AC ⊥ BE. Thus in 4ABQ, BC, AC are altitudes. This means C is the
orthocenter of this triangle and
1
1
∠CQE = ∠CQB = ∠A = ∠A + ∠A = ∠P DC + ∠P CD = ∠CP E
2
2
Hense CP QE is cyclic. Therefore
∠CED = ∠CEP = ∠CQP = ∠CQA = ∠CBA = ∠B.
Solutions
15
5.(Geometry Olympiad (Junior Level)) Two points X, Y lie on the arc BC of
the circumcircle of 4ABC (this arc does not contain A) such that ∠BAX = ∠CAY .
Let M denotes the midpoint of the chord AX . Show that
BM + CM > AY
Proposed by Mahan Tajrobekar
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
O is the circumcenter of 4ABC, so OM ⊥ AX. We draw a perpendicular line
from B to OM . This line intersects with the circumcircle at Z. Since OM ⊥ BZ,
OM is the perpendicular bisector of BZ. This means M Z = M B. By using triangle
inequality we have
BM + M C = ZM + M C > CZ
But BZ k AX, thus
_
_
_
_
_
AZ = BX = CY ⇒ ZAC = Y CA ⇒ CZ = AY
Hense BM + CM > AY.
Solutions
16
6.(Geometry Olympiad(Senior level)) In a quadrilateral ABCD we have
∠B = ∠D = 60◦ . Consider the line whice is drawn from M , the midpoint of AD,
parallel to CD. Assume this line intersects BC at P . A point X lies on CD such
that BX = CX. Prove that:
AB = BP ⇔ ∠M XB = 60◦
Proposed by Davood Vakili
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Suppose X 0 is a point such that 4M BX 0 is equilateral.(X 0 and X lie on the same
side of M B) It’s enough to show that:
AB = BP ⇔ X 0 ≡ X
We want to prove that if AB = BP then ∠M XB = 60◦ .
AB = BP therefore 4ABP is equilateral. We know that ∠ABP = ∠M BX 0 = 60◦ ,
Therefore ∠ABM = ∠P BX 0 . On the other hand AB = BP, BM = BX 0 therefore
4BAM and 4BP X 0 are equal.
∠X 0 P M = 360◦ − ∠M P B − ∠BP X 0 = 360◦ − ∠DCB − ∠BAM 0 = 120◦
Solutions
17
M P k DC, so we can say ∠P M D = 120◦ . If we draw the line passing through
X 0 such that be parallel with CD and this line intersects AD in D0 , then quadrilateral M P X 0 D0 is isosceles trapezoid. Therefore P X 0 = M D0 . In the other hand
P X 0 = AM = M D ( becauese 4BAM and 4BP X 0 are equal.) According to the
statements we can say M D0 = M D. In other words, D0 ≡ D and X 0 lie on CD.
Therefore both of X and X 0 lie on intersection of DC and perpendicular bisector of
M B, so X 0 ≡ X.
Now we prove if ∠M XB = 60◦ then AB = BP .
Let P 0 such that 4M P 0 X be equilateral.(P 0 and X be on the same side of AB) It’s
enough to show that P 0 ≡ P .
Draw the line passing through P 0 such that be parallel with CD. Suppose that this
line intersects AD in M 0 .
∠XP 0 M 0 = 360◦ − ∠M 0 P 0 B − ∠BP 0 X = 360◦ − ∠DCA − ∠BAM = 120◦
Also ∠P 0 M 0 D = 120◦ . Therefore quadrilateral XP 0 M 0 D is isosceles trapezoid and
DM 0 = P 0 X = AM = DM . So we can say M 0 ≡ M ⇒ P 0 ≡ P .
Solutions
18
7.(Geometry Olympiad(Senior level)) An acute-angled triangle ABC is given.
The circle with diameter BC intersects AB, AC at E, F respectively. Let M be the
midpoint of BC and P the intersection point of AM and EF . X is a point on the arc
EF and Y the second intersection point of XP with circle mentioned above. Show
that ∠XAY = ∠XY M .
Proposed by Ali Zooelm
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Suppose point K is intersection AM and circumcircle of 4AEF . M F tangent to
circumcircle of 4AEF at F .
( because of ∠M F C = ∠M CF = ∠AEF ). Therefore M F 2 = M K.M A . In the
other hand, M Y = M F so M Y 2 = M K.M A. It means
∠M Y K = ∠Y AM
(1)
Also AP.P K = P E.P F = P X.P Y therefore AXKY is(...??) .Therefore
∠XAY = ∠XY K
According to equation 1 and 2 we can say ∠XAY = ∠XY M .
(2)
Solutions
19
8.(Geometry Olympiad(Senior level)) The tangent line to circumcircle of the
acute-angled triangle ABC (AC > AB) at A intersects the continuation of BC at P .
We denote by O the circumcenter of ABC. X is a point OP such that ∠AXP = 90◦ .
Two points E, F respectively on AB, AC at the same side of OP are chosen such
that
∠EXP = ∠ACX, ∠F XO = ∠ABX
If K, L denote the intersection points of EF with the circumcircle of 4ABC, show
that OP is tangent to the circumcircle of 4KLX.
Proposed by Mahdi Etesami Fard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
Let M and N on continuation of XF and XE such that M, L, X, N, K lie on
same circle. We have to prove ∠AM X = ∠ACX. In other hand, ∠ACX = ∠N XP
so we have to prove ∠ACX = ∠N M X.
We know that XF.F M = F L.F K = AF.F C. Therefore AM CX is cyclic and
∠AM X = ∠ACX. similarly we can say AN BX is cyclic. Now it’s enough to show
that ∠AM X = ∠N M X. In other words, we have to show that A, N , M lie on same
line. we know that AN BX is cyclic therefore:
∠N AM = ∠N AE + ∠A + ∠F AM = ∠EXB + ∠A + ∠CXF
= ∠A + 180◦ − ∠BXC + ∠ABX + ∠ACX
= ∠A + 180◦ − ∠BXC + ∠BXC − ∠A = 180◦
Solutions
20
9.(Geometry Olympiad(Senior level)) Two points P , Q lie on the side BC
of triangle ABC and have the same distance to the midpoint. The pependiculars
fromP , Q tp BC intesects AC, AB at E, F respectively. LEt M be the intersection
point of P F and EQ. If H1 and H2 denote the orthocenter of 4BF P and 4CEQ
recpectively, show that AM ⊥ H1 H2 .
Proposed by Mahdi Etesami Fard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
solution.
First we show that if we move P and Q, the line AM doesn’t move. To show that
sin ∠A1
. By the law of sines in 4AF M and 4AEM we have
we calculate sin
∠A2
sin ∠A1
sin ∠F1 F M
=
·
sin ∠A2
sin ∠E1 EM
(3)
also, for 4F BP and 4CEQ we have
sin ∠F1 =
sin ∠E1 =
BP
PF
CQ
EQ
· sin ∠B
· sin ∠C
)
⇒
sin ∠F1
sin ∠B EQ
=
·
sin ∠E1
sin ∠C F P
(4)
from (3) and (4) we have
sin ∠A1
sin ∠B EQ F M
=
·
·
sin ∠A2
sin ∠C F P EM
4F M Q and 4EM P are similar, thus
FM
FQ
EQ
F Q + EP
=
,
=
FP
F Q + EP EM
EP
with putting this into (5) we have
(5)
Solutions
21
sin ∠A1
sin ∠B F Q
=
·
sin ∠A2
sin ∠C EP
on the other hand
tan ∠B =
tan ∠C =
BQ = CP
FQ
BQ 

EP
CP 
(6)


⇒
FQ
tan ∠B
=
EP
tan ∠C
if we put this in (6) we have
sin ∠B tan ∠B
sin ∠A1
=
·
sin ∠A2
sin ∠C tan ∠C
wich is constant.
now we show that H1 H2 s are parallel. consider α the angle between H1 H2 and
BC. Hense we have
tan α =
H2 P − H1 Q
QP
(7)
H1 and H2 are the orthometers of 4BF P and 4CQE respectively. Thus we have
QF · H1 Q = BQ · QP ⇒ H1 Q =
BQ · QP
FQ
EP · H2 P = CP · P Q ⇒ H2 P =
CP · P Q
EP
but CP = BQ. Thus
H2 P − H1 Q =
P Q · BQ · (F Q − EP )
EP · F Q
by putting this in (7) :
tan α =
BQ · (F Q − EP )
BQ BQ
CP
BQ
=
−
=
−
EP · F Q
EP
FQ
EP
FQ
⇒ tan α = cot ∠B − cot ∠C
(8)
Solutions
22
hense tan α is constant, thus H1 H2 s are parallel.
Soppuse θ is the angle between AM and BC. we have to show
tan α · tan θ = 1
let AM intersects with BC at X. We have
sin ∠A1 sin ∠C
BX
tan ∠B
BX
=
·
⇒
=
CX
sin ∠A2 sin ∠B
CX
tan ∠C
let D be the foot of the altitude drawn from A. We have
BX
tan ∠B
=
=
CX
tan ∠C
tan θ =
AD
BD
AD
CD
=
CD
⇒ BD = CX
BD
AD
AD
AD
=
=
=
DX
CD − CX
CD − BD
this equality and (8) implies that AM ⊥ H1 H2 .
CD
AD
1
−
BD
AD
=
1
cot ∠B − cot ∠C
Problems of 2nd Iranian Geometry Olympiad 2015 (Elementary)
1. We have four wooden triangles with sides 3, 4, 5 centimeters. How many convex
polygons can we make by all of these triangles?(Just draw the polygons without any
proof)
A convex polygon is a polygon which all of it’s angles are less than 180◦ and there
isn’t any hole in it. For example:
This polygon isn’t convex
This polygon is convex
P roposed by M ahdi Etesami F ard
2. Let ABC be a triangle with ∠A = 60◦ . The points M, N, K lie on BC, AC, AB
respectively such that BK = KM = M N = N C. If AN = 2AK, find the values of
∠B and ∠C.
P roposed by M ahdi Etesami F ard
3. In the figure below, we know that AB = CD and BC = 2AD. Prove that
∠BAD = 30◦ .
P roposed by M orteza Saghaf ian
1
4. In rectangle ABCD, the points M, N, P, Q lie on AB, BC, CD, DA respectively
such that the area of triangles AQM, BM N, CN P, DP Q are equal. Prove that the
quadrilateral M N P Q is parallelogram.
P roposed by M ahdi Etesami F ard
5. Do there exist 6 circles in the plane such that every circle passes through centers
of exactly 3 other circles?
P roposed by M orteza Saghaf ian
2
Problems of 2nd Iranian Geometry Olympiad 2015 (Medium)
1. In the figure below, the points P, A, B lie on a circle. The point Q lies inside the
circle such that ∠P AQ = 90◦ and P Q = BQ. Prove that the value of ∠AQB−∠P QA
is equal to the arc AB.
P roposed by Davood V akili
2. In acute-angled triangle ABC, BH is the altitude of the vertex B. The points D
and E are midpoints of AB and AC respectively. Suppose that F be the reflection of
H with respect to ED. Prove that the line BF passes through circumcenter of ABC.
P roposed by Davood V akili
3. In triangle ABC, the points M, N, K are the midpoints of BC, CA, AB respectively. Let ωB and ωC be two semicircles with diameter AC and AB respectively,
outside the triangle. Suppose that M K and M N intersect ωC and ωB at X and Y
respectively. Let the tangents at X and Y to ωC and ωB respectively, intersect at Z.
prove that AZ⊥BC.
P roposed by M ahdi Etesami F ard
3
4. Let ABC be an equilateral triangle with circumcircle ω and circumcenter O. Let
P be the point on the arc BC( the arc which A doesn’t lie ). Tangent to ω at P intersects extensions of AB and AC at K and L respectively. Show that ∠KOL > 90◦ .
P roposed by Iman M aghsoudi
5. a) Do there exist 5 circles in the plane such that every circle passes through centers
of exactly 3 circles?
b) Do there exist 6 circles in the plane such that every circle passes through centers
of exactly 3 circles?
P roposed by M orteza Saghaf ian
4
Problems of 2nd Iranian Geometry Olympiad 2015 (Advanced)
1. Two circles ω1 and ω2 (with centers O1 and O2 respectively) intersect at A and B.
The point X lies on ω2 . Let point Y be a point on ω1 such that ∠XBY = 90◦ . Let
X 0 be the second point of intersection of the line O1 X and ω2 and K be the second
point of intersection of X 0 Y and ω2 . Prove that X is the midpoint of arc AK.
P roposed by Davood V akili
2. Let ABC be an equilateral triangle with circumcircle ω and circumcenter O. Let
P be the point on the arc BC( the arc which A doesn’t lie ). Tangent to ω at P
intersects extensions of AB and AC at K and L respectively. Show that ∠KOL > 90◦ .
P roposed by Iman M aghsoudi
3. Let H be the orthocenter of the triangle ABC. Let l1 and l2 be two lines passing
through H and perpendicular to each other. l1 intersects BC and extension of AB
at D and Z respectively, and l2 intersects BC and extension of AC at E and X respectively. Let Y be a point such that Y D k AC and Y E k AB. Prove that X, Y, Z
are collinear.
P roposed by Ali Golmakani
4. In triangle ABC, we draw the circle with center A and radius AB. This circle
intersects AC at two points. Also we draw the circle with center A and radius AC and
this circle intersects AB at two points. Denote these four points by A1 , A2 , A3 , A4 .
Find the points B1 , B2 , B3 , B4 and C1 , C2 , C3 , C4 similarly. Suppose that these 12
points lie on two circles. Prove that the triangle ABC is isosceles.
P roposed by M orteza Saghaf ian
5. Rectangles ABA1 B2 , BCB1 C2 , CAC1 A2 lie otside triangle ABC. Let C 0 be a
point such that C 0 A1 ⊥ A1 C2 and C 0 B2 ⊥ B2 C1 . Points A0 and B 0 are defined similarly. Prove that lines AA0 , BB 0 , CC 0 concur.
P roposed by Alexey Zaslavsky (Russia)
5
Solutions of 2nd Iranian Geometry Olympiad 2015 (Elementary)
1. We have four wooden triangles with sides 3, 4, 5 centimeters. How many convex
polygons can we make by all of these triangles?(Just draw the polygons without any
proof)
A convex polygon is a polygon which all of it’s angles are less than 180◦ and there
isn’t any hole in it. For example:
This polygon isn’t convex
This polygon is convex
P roposed by M ahdi Etesami F ard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
6
7
2. Let ABC be a triangle with ∠A = 60◦ . The points M, N, K lie on BC, AC, AB
respectively such that BK = KM = M N = N C. If AN = 2AK, find the values of
∠B and ∠C.
P roposed by M ahdi Etesami F ard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Suppose the point P be the midpoint of AN . Therefore AK = AP = AN and
A
so we can say 4AP K is the equilateral triangle. So ∠AN K = ∠KP
= 30◦ Let
2
∠ACB = ∠N M C = α. Therfore ∠ABC = ∠KM B = 120◦ − α. So ∠KM N = 60◦ .
Therefore 4KM N is the equilateral triangle. Now we know that ∠M N A = 90◦ .
Therefore α = 45◦ . So we have ∠C = 45◦ and ∠B = 75◦ .
8
3. In the figure below, we know that AB = CD and BC = 2AD. Prove that
∠BAD = 30◦ .
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Let two points E and F on BC and AB respectively such that DF ⊥BC and
= AB
.(because of ∠BCD = 30◦ and ∠DF C = 90◦ )
DE⊥AB. We can say DF = DC
2
2
Also we know that DF = BE, therfore DE is the perpendicular bisector of AB. So
BD = AD.
Let H be a point on CD such that BH⊥CD. therefore BH = BC
= BD, so we
2
◦
◦
can say D ≡ H and ∠BDC = 90 . Therefore ∠ABD = ∠BAD = 30 .
9
Solution 2.
Suppose that P is the point such that triangle DCP is Equilateral. We know that
P C⊥BC and P C = CD = AB, therfore quadrilateral ABCP is Rectangular.
⇒
∠AP D = ∠AP C − ∠DP C = 90◦ − 60◦ = 30◦
In other hand, DP = DC and AP = BC. So 4ADP and 4BDC are congruent.
Therfore AD = BD.
Let the point H on CD such that BH⊥CD. therefore BH = BC
= BD, so we
2
◦
◦
can say D ≡ H and ∠BDC = 90 . Therefore ∠ABD = ∠BAD = 30 .
10
4. In rectangle ABCD, the points M, N, P, Q lie on AB, BC, CD, DA respectively
such that the area of triangles AQM, BM N, CN P, DP Q are equal. Prove that the
quadrilateral M N P Q is parallelogram.
P roposed by M ahdi Etesami F ard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Let AB = CD = a, AD = BC = b and AM = x, AQ = z, P C = y, N C = t. If
x 6= y, we can assume that x > y. We know that:
y<x ⇒
SAQM = SCN P
⇒
a−x<a−y
zx = yt ⇒
(1)
z<t ⇒
b−t<b−z
(2)
According to inequality 1, 2:
(a − x)(b − t) < (a − y)(b − z) ⇒
SBM N < SDP Q
it’s a contradiction. Therfore x = y, so z = t Now we can say two triangles AM Q
and CP N are congruent. Therefore M Q = N P and similarly M N = P Q. So the
quadrilateral M N P Q is parallelogram.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Comment.
If quadrilateral ABCD be the parallelogram, similarly we can show that quadrilateral M N P Q is parallelogram.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
11
5. Do there exist 6 circles in the plane such that every circle passes through centers
of exactly 3 other circles?
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
In the picture below, we have 6 points in the plane such that for every point there
exists exactly 3 other points on a circle with radius 1 centimeter.
12
Solutions of 2nd Iranian Geometry Olympiad 2015 (Medium)
1. In the figure below, the points P, A, B lie on a circle. The point Q lies inside the
circle such that ∠P AQ = 90◦ and P Q = BQ. Prove that the value of ∠AQB−∠P QA
is equal to the arc AB.
P roposed by Davood V akili
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Let point M be the midpoint of P B. So we can say ∠P M Q = 90◦ and we know
that ∠P AQ = 90◦ , therefore quadrilateral P AM Q is cyclic. Therefore:
∠AP M = ∠AQM
In the other hand:
∠AQB − ∠AQP = ∠P QM + ∠AQM − ∠AQP = 2∠AQM
So we can say that the subtract ∠AQB from ∠P QA is equal to arc AB.
13
Solution 2.
Let the point K be the reflection of P to AQ. We have to show:
2∠AP B = ∠AQB − ∠AQP
Now we know that AQ is the perpendicular bisector of P K. So ∠AQP = ∠AQK
and P Q = KQ = BQ, therefore the point Q is the circumcenter of triangle P KB.
We know that:
2∠AP B = ∠KQB = ∠AQB − ∠AQK = ∠AQB − ∠AQP
Therefore the subtract ∠AQB from ∠P QA is equal to arc AB.
14
2. In acute-angled triangle ABC, BH is the altitude of the vertex B. The points D
and E are midpoints of AB and AC respectively. Suppose that F be the reflection of
H with respect to ED. Prove that the line BF passes through circumcenter of ABC.
P roposed by Davood V akili
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
The circumcenter of 4ABC denote by O. We know that ∠OBA = 90◦ − ∠C,
therfore we have to show that ∠F BA = 90◦ − ∠C. We know that AD = BD = DH,
also DH = DF .
Therfore quadrilateral AHF B is cyclic (with circumcenter D)
⇒
∠F BA = ∠F HE = 90◦ − ∠DEH
⇒
,
DE k BC
∠F BA = 90◦ − ∠C
15
⇒
∠DEH = ∠C
Solution 2.
The circumcenter of 4ABC denote by O. We know that quadrilateral ADOE is
cyclic. Also we know that AD = HD = DB, therefore:
∠A = ∠DHA = 180◦ − ∠DHE = 180◦ − ∠DF E ⇒ ADF E : cyclic
So we can say ADF OE is cyclic, therefore quadrilateral DF OE is cyclic.
∠C = ∠DEA = ∠DEF = ∠DOF
In the other hand: ∠C = ∠DOB so ∠DOF = ∠DOB, therefore B, F, O are collinear.
16
3. In triangle ABC, the points M, N, K are the midpoints of BC, CA, AB respectively. Let ωB and ωC be two semicircles with diameter AC and AB respectively,
outside the triangle. Suppose that M K and M N intersect ωC and ωB at X and Y
respectively. Let the tangents at X and Y to ωC and ωB respectively, intersect at Z.
prove that AZ⊥BC.
P roposed by M ahdi Etesami F ard
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Let point H on BC such that AH⊥BC. Therefore quadrilaterals AXBH and
AY CH are cyclic. We know that KM and M N are parallel to AC and AB respecand
tively. So we can say ∠AKX = ∠AN Y = ∠A, therefore ∠ABX = ∠ACY = ∠A
2
∠XAB = ∠Y AC = 90◦ − ∠A
.
So
X,
A,
Y
are
collinear.
2
∠AHX = ∠ABX =
∠A
∠A
, ∠AHY = ∠ACY =
⇒ ∠XHY = ∠XM Y = ∠A
2
2
Therefore quadrilateral XHM Y is cyclic. Also we know that ∠M XZ = ∠M Y Z =
90 , therefore quadrilateral M XZY is cyclic. So we can say ZXHM Y is cyclic. therfore quadrilateral HXZY is cyclic.
◦
17
In the other hand: ∠ZY X = ∠ACY =
∠ZHX = ∠ZY X =
∠A
2
∠A
∠A
, ∠AHX =
2
2
⇒
∠ZHX = ∠AHX
So the points Z, A, H are collinear, therefore AZ⊥BC.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 2.
Let point H on BC such that AH⊥BC. We know that KM and M N are parallel
to AC and AB respectively. So we can say ∠AKX = ∠AN Y = ∠A, therefore
∠ABX = ∠ACY = ∠A
and ∠XAB = ∠Y AC = 90◦ − ∠A
. So X, A, Y are collinear.
2
2
∠A
⇒ ZX = ZY
2
So the point Z lie on the radical axis of two these semicirculars. Also we know that
the line AH is the radical axis of two these semicirculars. Therefore the points Z, A, H
are collinear, therefore AZ⊥BC.
⇒ ∠ZXY = ∠ZY X =
18
4. Let ABC be an equilateral triangle with circumcircle ω and circumcenter O. Let
P be the point on the arc BC( the arc which A doesn’t lie ). Tangent to ω at P intersects extensions of AB and AC at K and L respectively. Show that ∠KOL > 90◦ .
P roposed by Iman M aghsoudi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Suppose that M and N be the midpoints of AB and AC respectively. We know
that quadrilateral BM N C is cyclic. Also ∠BP C = 120◦ > 90◦ , so we can say
the point P is in the circumcircle of quadrilateral BM N C. Therefore: ∠M P N >
∠M BN = 30◦
In the other hand, quadrilaterals KM OP and N OP L are cyclic. Therefore:
∠M KO = ∠M P O , ∠N LO = ∠N P O ⇒ ∠AKO + ∠ALO = ∠M P N > 30◦
⇒ ∠KOL = ∠A + ∠AKO + ∠ALO > 90◦
19
Solution 2.
Suppose that ∠KOL ≤ 90◦ , therfore KL2 ≤ OK 2 + OL2 . Assume that R is the
radius of a circumcircle 4ABC. Let BK = x and LC = y and AB = AC = BC = a.
According to law of cosines in triangle AKL, we have:
KL2 = AK 2 + AL2 − AK.AL.cos(∠A) ⇒ KL2 = (a + x)2 + (a + y)2 − (a + x)(a + y)
In the other hand:
KB.KA = OK 2 − R2
⇒
LC.LA = OL2 − R2
⇒
OK 2 = R2 + x(a + x)
OL2 = R2 + y(a + y)
√
We know that KL2 ≤ OK 2 + OL2 and a = R 3, therfore:
(a + x)2 + (a + y)2 − (a + x)(a + y) ≤ 2R2 + x(a + x) + y(a + y)
⇒ R2 ≤ xy
(1)
KL is tangent to circumcircle of 4ABC at P . So we have:
KP 2 = KB.KA = x(a + x) > x2 ⇒ KP > x
LP 2 = LC.LA = y(a + y) > y 2 ⇒ LP > y
According to inequality 2, 3 we can say:
Now According to inequality 1, 4 we have:
20
xy < KP.LP
R2 < KP.LP
(2)
(3)
(4)
(5)
We know that ∠KOL ≤ 90◦ , therefore KOL is acute-triangle. Suppose that H is
orthocenter of 4KOL. So the point H lies on OP and we can say HP ≤ OP .
In other hand, ∠HKP = ∠P OL and ∠KHP = ∠OLP , therefore two triangles
T HP and OP L are similar. So we have:
OP
KP
=
HP
LP
⇒
KP.LP = HP.OP ≤ OP 2 = R2
But according to inequality 5, we have R2 < KP.LP and it’s a contradiction.
Therfore ∠KOL > 90◦ .
21
5. a) Do there exist 5 circles in the plane such that every circle passes through centers
of exactly 3 circles?
b) Do there exist 6 circles in the plane such that every circle passes through centers
of exactly 3 circles?
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
a)Solution.
There aren’t such 5 circles. Suppose that these circles exists, therefore their centers
are 5 points that each point has same distance from 3 other points and has diffrent
distance from the remaining point. We draw an arrow from each point to it’s diffrent
distance point.
- lemma 1. We don’t have two points such Oi , Oj that each one is the diffrent
distance point of the other one.
proof. If we have such thing then Oi and Oj both have same distance to the
remaining points, therefore both of them are circumcenter of the remaining points,
which is wrong.
- lemma 2. We don’t have 4 points such Oi , Oj , Ok , Ol that Oi , Oj put their arrow
in Ok and OK puts it’s arrow in Ol .
proof. If we name the remaining point Om then the distances of Oi from Oj , Ol ,
Om are equal and the distances of Oj from Oi , Ol , Om are equal. Therefore each of
Ol , Om is the diffrent distance point of another which is wrong (according to lemma
1).
so each point sends an arrow and recives an arrow. Because of lemma 1 we don’t
have 3 or 4 points cycles. Therefore we only have one 5 points cycle. So each pair of
these 5 points should have equal distance. which is impossible.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
b)Solution.
in the picture below, we have 6 points in the plane such that for every point there
exists exactly 3 other points on a circle with radius 1 centimeter.
22
Solutions of 2nd Iranian Geometry Olympiad 2015 (Advanced)
1. Two circles ω1 and ω2 (with centers O1 and O2 respectively) intersect at A and B.
The point X lies on ω2 . Let point Y be a point on ω1 such that ∠XBY = 90◦ . Let
X 0 be the second point of intersection of the line O1 X and ω2 and K be the second
point of intersection of X 0 Y and ω2 . Prove that X is the midpoint of arc AK.
P roposed by Davood V akili
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Suppose that the point Z be the intersection of BX and circle ω1 . We know that
∠Y BZ = 90◦ , therefore the points Y, O1 , Z are collinear.
∠O1 Y A = ∠ABX = ∠AX 0 X
⇒
Y AX 0 O1 : cyclic
In the other hand, we know that AO1 = Y O1 so ∠AX 0 X = ∠Y X 0 O1 = ∠XX 0 K.
Therefore the point X lies on the midpoint of arc AK.
23
2. Let ABC be an equilateral triangle with circumcircle ω and circumcenter O. Let
P be the point on the arc BC( the arc which A doesn’t lie ). Tangent to ω at P intersects extensions of AB and AC at K and L respectively. Show that ∠KOL > 90◦ .
P roposed by Iman M aghsoudi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Suppose that M and N be the midpoints of AB and AC respectively. We know
that quadrilateral BM N C is cyclic. Also ∠BP C = 120◦ > 90◦ , so we can say
the point P is in the circumcircle of quadrilateral BM N C. Therefore: ∠M P N >
∠M BN = 30◦
In the other hand, quadrilaterals KM OP and N OP L are cyclic. Therefore:
∠M KO = ∠M P O , ∠N LO = ∠N P O ⇒ ∠AKO + ∠ALO = ∠M P N > 30◦
⇒ ∠KOL = ∠A + ∠AKO + ∠ALO > 90◦
24
Solution 2.
Suppose that ∠KOL ≤ 90◦ , therfore KL2 ≤ OK 2 + OL2 . Assume that R is the
radius of a circumcircle 4ABC. Let BK = x and LC = y and AB = AC = BC = a.
According to law of cosines in triangle AKL, we have:
KL2 = AK 2 + AL2 − AK.AL.cos(∠A) ⇒ KL2 = (a + x)2 + (a + y)2 − (a + x)(a + y)
In the other hand:
KB.KA = OK 2 − R2
⇒
LC.LA = OL2 − R2
⇒
OK 2 = R2 + x(a + x)
OL2 = R2 + y(a + y)
√
We know that KL2 ≤ OK 2 + OL2 and a = R 3, therfore:
(a + x)2 + (a + y)2 − (a + x)(a + y) ≤ 2R2 + x(a + x) + y(a + y)
⇒ R2 ≤ xy
(1)
KL is tangent to circumcircle of 4ABC at P . So we have:
KP 2 = KB.KA = x(a + x) > x2 ⇒ KP > x
LP 2 = LC.LA = y(a + y) > y 2 ⇒ LP > y
According to inequality 2, 3 we can say:
Now According to inequality 1, 4 we have:
25
xy < KP.LP
R2 < KP.LP
(2)
(3)
(4)
(5)
We know that ∠KOL ≤ 90◦ , therefore KOL is acute-triangle. Suppose that H is
orthocenter of 4KOL. So the point H lies on OP and we can say HP ≤ OP .
In other hand, ∠HKP = ∠P OL and ∠KHP = ∠OLP , therefore two triangles
T HP and OP L are similar. So we have:
OP
KP
=
HP
LP
⇒
KP.LP = HP.OP ≤ OP 2 = R2
But according to inequality 5, we have R2 < KP.LP and it’s a contradiction.
Therfore ∠KOL > 90◦ .
26
3. Let H be the orthocenter of the triangle ABC. Let l1 and l2 be two lines passing
through H and perpendicular to each other. l1 intersects BC and extension of AB
at D and Z respectively, and l2 intersects BC and extension of AC at E and X respectively. Let Y be a point such that Y D k AC and Y E k AB. Prove that X, Y, Z
are collinear.
P roposed by Ali Golmakani
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Suppose that HZ intersects AC at P and HX intersects AB at Q. According to
Menelaus’s theorem in two triangles AQX and AP Z we can say:
CX AB QE
.
.
=1
AC BQ EX
(1)
and
BZ AC P D
.
.
=1
AB P C DZ
(2)
In the other hand, H is the orthocenter of 4ABC. So BH⊥AC and we know that
∠DHE = 90◦ , therefore ∠HXA = ∠BHZ = α. Similarly we can say ∠HZA =
∠CHX = θ.
27
According to law of sines in 4HP C, 4HCX and 4HP X:
sin(90 − θ)
sin(∠HCP )
=
PC
HP
⇒
,
sin(θ)
sin(∠HCX)
=
CX
HX
,
HP
sin(α)
=
HX
sin(90 − α)
PC
tan(α)
=
CX
tan(θ)
Similarly, according to law of sines in 4HBQ, 4HBZ and 4HQZ, we can show:
⇒
tan(α)
BZ
=
BQ
tan(θ)
⇒
BZ
PC
=
BQ
CX
⇒
PC
CX
=
BZ
BQ
(3)
According to equality 1, 2 and 3, we can say:
XE
PD
=
EQ
ZD
(4)
Suppose that the line which passes through E and parallel to AB, intersects ZX at
Y1 and the line which passes through D and parallel to AC, intersects ZX at Y2 .
According to Thales’s theorem we can say:
Y1 X
XE
=
ZY1
EQ
,
PD
Y2 X
=
ZY2
ZD
According to equality 4, we show that Y1 ≡ Y2 , therefore the point Y lies on ZX.
28
4. In triangle ABC, we draw the circle with center A and radius AB. This circle
intersects AC at two points. Also we draw the circle with center A and radius AC and
this circle intersects AB at two points. Denote these four points by A1 , A2 , A3 , A4 .
Find the points B1 , B2 , B3 , B4 and C1 , C2 , C3 , C4 similarly. Suppose that these 12
points lie on two circles. Prove that the triangle ABC is isosceles.
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 1.
Suppose that triangle ABC isn’t isosceles and a > b > c. In this case, there
are four points (from these 12 points) on each side of 4ABC. Suppose that these
12 points lie on two circles ω1 and ω2 . Therefore each one of the circles ω1 and ω2
intersects each side of 4ABC exactly at two points. Suppose that P (A, ω1 ), P (A, ω2 )
are power of the point A with respect to circles ω1 , ω2 respectively. Now we know
that:
P (A, ω1 ).P (A, ω2 ) = b.b.(a − c).(a + c) = c.c.(a − b)(a + b)
⇒
b2 (a2 − c2 ) = c2 (a2 − b2 )
⇒
a2 (b2 − c2 ) = 0
⇒
b=c
But we know that b > c and it’s a contradiction. Therefore the triangle ABC is
isosceles.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution 2.
Suppose that triangle ABC isn’t isosceles. In this case, there are four points
(from these 12 points) on each side of 4ABC. Suppose that these 12 points lie on
two circles ω1 and ω2 . Therefore each one of the circles ω1 and ω2 intersects each side
of 4ABC exactly at two points (and each one of the circles ω1 and ω2 doesn’t pass
through A, B, C). We know that the intersections of ω1 and the sides of 4ABC is
even number. Also the intersections of ω2 and the sides of 4ABC is even number.
But Among the these 12 points, just 3 points lie on the sides of 4ABC and this is
odd number. So it’s a contradiction. Therefore the triangle ABC is isosceles.
29
5. Rectangles ABA1 B2 , BCB1 C2 , CAC1 A2 lie otside triangle ABC. Let C 0 be a
point such that C 0 A1 ⊥ A1 C2 and C 0 B2 ⊥ B2 C1 . Points A0 and B 0 are defined similarly. Prove that lines AA0 , BB 0 , CC 0 concur.
P roposed by Alexey Zaslavsky (Russia)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Suppose that lA is the line which passes through A and perpendicular to B2 C1 .
Let lB and lC similarly. Suppose that CB1 = BC2 = x and BA1 = AB2 = y and
AC1 = CA2 = z. According to angles equality, we can say:
sin(∠A1 )
y
=
sin(∠A2 )
z
,
sin(∠B1 )
x
=
sin(∠B2 )
y
,
sin(∠C1 )
z
=
sin(∠C2 )
x
According to sine form of Ceva’s theorem in 4ABC, lA , lB , lC are concur. Suppose
that lA , lB , lC pass through the point P . We know that 4P BC and 4A0 C2 B1 are
equal. ( because of BP k A0 C2 , CP k A0 B1 , BC k B1 C2 and BC = B1 C2 ). So we
have:
P A0 = x , P C 0 = y , P B 0 = z
P A0 ⊥BC , P B 0 ⊥AC , P C 0 ⊥AB
30
Suppose that P A0 , P B 0 , P C 0 intersects BC, AC, AB at D, E, F respectively and:
P D = m , P E = n , P F = t. According to before figure, we have:
sin(∠A1 )
n
y
= =
sin(∠A2 )
t
z
,
sin(∠B1 )
t
x
=
=
sin(∠B2 )
m
y
If n = ky, then: t = kz , m =
,
sin(∠C1 )
m
z
=
=
sin(∠C2 )
n
x
kyz
.
x
Now draw the line from A0 such that be parallel to BC. The intersection of this
line and extension AB and AC denote by B3 and C3 respectively. Let the point A00
be the intersection of AA0 and BC. According to Thales’s theorem, we have:
BA00
B3 A0
=
CA00
C3 A0
31
Let ∠B3 P A0 = α and ∠C3 P A0 = θ. We know that the quadrilaterals P F B3 A0
and P EC3 A0 are cyclic. Therefore ∠B3 F A0 = α and ∠C3 EA0 = θ.
According to law of sines in 4P B3 A0 and 4P C3 A0 and 4P C3 B3 :
B3 A0
tan(α)
=
0
C3 A
tan(θ)
Also according to law of sines in 4P F A0 :
sin(∠B + α − 90)
cos(∠B + α)
t
=
=
= cos(∠B) − tan(α).sin(∠B)
x
cos(α)
cos(α)
⇒ tan(α) =
cos(∠B) −
sin(∠B)
t
x
Similarly we can say:
cos(∠C) −
tan(θ) =
sin(∠C)
n
x
⇒
B3 A0
BA00
x.cos(∠B) − t sin(∠C)
=
=
.
0
00
C3 A
CA
x.cos(∠C) − n sin(∠B)
Similarly, two other fractions can be calculated.
32
According to Ceva’s theorem in 4ABC, we have to that:
x.cos(∠B) − t sin(∠C) z.cos(∠C) − m sin(∠A) y.cos(∠A) − n sin(∠B)
.
.
.
.
.
=1
x.cos(∠C) − n sin(∠B) z.cos(∠A) − t sin(∠C) y.cos(∠B) − m sin(∠A)
⇐⇒
x.cos(∠B) − t z.cos(∠C) − m y.cos(∠A) − n
.
.
=1
x.cos(∠C) − n z.cos(∠A) − t y.cos(∠B) − m
In other hand, we know that:
n = ky
⇐⇒
,
t = kz
,
m=
kyz
x
x.cos(∠B) − kz x.cos(∠C) − ky x.cos(∠A) − kx
.
.
=1
x.cos(∠C) − ky x.cos(∠A) − kx x.cos(∠B) − kz
Therfore, we show that AA0 , BB 0 , CC 0 are concur.
33
1
Problems of 3rd Iranian Geometry Olympiad 2016 (Elementary)
1. Ali wants to move from point A to point B. He cannot walk inside the
black areas but he is free to move in any direction inside the white areas (not
only the grid lines but the whole plane). Help Ali to find the shortest path
between A and B. Only draw the path and write its length.
P roposed by M orteza Saghaf ian
2. Let ω be the circumcircle of triangle ABC with AC > AB. Let X be a
point on AC and Y be a point on the circle ω, such that CX = CY = AB.
(The points A and Y lie on different sides of the line BC). The line XY intersects ω for the second time in point P . Show that P B = P C.
P roposed by Iman M aghsoudi
3. Suppose that ABCD is a convex quadrilateral with no parallel sides. Make
a parallelogram on each two consecutive sides. Show that among these 4 new
points, there is only one point inside the quadrilateral ABCD.
P roposed by M orteza Saghaf ian
2
4. In a right-angled triangle ABC (∠A = 90◦ ), the perpendicular bisector
of BC intersects the line AC in K and the perpendicular bisector of BK intersects the line AB in L. If the line CL be the internal bisector of angle C, find
all possible values for angles B and C.
P roposed by M ahdi Etesami F ard
5. Let ABCD be a convex quadrilateral with these properties:
√
∠ADC = 135◦ and ∠ADB−∠ABD = 2∠DAB = 4∠CBD. If BC = 2CD
prove that AB = BC + AD.
P roposed by M ahdi Etesami F ard
3
Problems of 3rd Iranian Geometry Olympiad 2016 (Medium)
1. In trapezoid ABCD with AB k CD, ω1 and ω2 are two circles with diameters AD and BC, respectively. Let X and Y be two arbitrary points on ω1
and ω2 , respectively. Show that the length of segment XY is not more than
half of the perimeter of ABCD.
P roposed by M ahdi Etesami F ard
2. Let two circles C1 and C2 intersect in points A and B. The tangent to
C1 at A intersects C2 in P and the line P B intersects C1 for the second time
in Q (suppose that Q is outside C2 ). The tangent to C2 from Q intersects C1
and C2 in C and D, respectively (The points A and D lie on different sides of
the line P Q). Show that AD is bisector of the angle CAP .
P roposed by Iman M aghsoudi
3. Find all positive integers N such that there exists a triangle which can
be dissected into N similar quadrilaterals.
P roposed by N ikolai Beluhov (Bulgaria) and M orteza Saghaf ian
4. Let ω be the circumcircle of right-angled triangle ABC (∠A = 90◦ ). Tangent to ω at point A intersects the line BC in point P . Suppose that M is the
midpoint of (the smaller) arc AB, and P M intersects ω for the second time in
Q. Tangent to ω at point Q intersects AC in K. Prove that ∠P KC = 90◦ .
P roposed by Davood V akili
5. Let the circles ω and ω 0 intersect in points A and B. Tangent to circle
ω at A intersects ω 0 in C and tangent to circle ω 0 at A intersects ω in D.
Suppose that the internal bisector of ∠CAD intersects ω and ω 0 at E and F ,
respectively, and the external bisector of ∠CAD intersects ω and ω 0 in X and
Y , respectively. Prove that the perpendicular bisector of XY is tangent to the
circumcircle of triangle BEF .
P roposed by M ahdi Etesami F ard
4
Problems of 3rd Iranian Geometry Olympiad 2016 (Advanced)
1. Let the circles ω and ω 0 intersect in A and B. Tangent to circle ω at A
intersects ω 0 in C and tangent to circle ω 0 at A intersects ω in D. Suppose that
the segment CD intersects ω and ω 0 in E and F , respectively (assume that E
is between F and C). The perpendicular to AC from E intersects ω 0 in point
P and perpendicular to AD from F intersects ω in point Q (The points A, P
and Q lie on the same side of the line CD). Prove that the points A, P and Q
are collinear.
P roposed by M ahdi Etesami F ard
2. In acute-angled triangle ABC, altitude of A meets BC at D, and M is
midpoint of AC. Suppose that X is a point such that ∠AXB = ∠DXM = 90◦
(assume that X and C lie on opposite sides of the line BM ). Show that
∠XM B = 2∠M BC.
P roposed by Davood V akili
3. Let P be the intersection point of sides AD and BC of a convex qualrilateral ABCD. Suppose that I1 and I2 are the incenters of triangles P AB and
P DC, respectively. Let O be the circumcenter of P AB, and H the orthocenter
of P DC. Show that the circumcircles of triangles AI1 B and DHC are tangent
together if and only if the circumcircles of triangles AOB and DI2 C are tangent
together.
P roposed by Hooman F attahimoghaddam
4. In a convex quadrilateral ABCD, the lines AB and CD meet at point
E and the lines AD and BC meet at point F . Let P be the intersection point
of diagonals AC and BD. Suppose that ω1 is a circle passing through D and
tangent to AC at P . Also suppose that ω2 is a circle passing through C and
tangent to BD at P . Let X be the intersection point of ω1 and AD, and Y
be the intersection point of ω2 and BC. Suppose that the circles ω1 and ω2
intersect each other in Q for the second time. Prove that the perpendicular
from P to the line EF passes through the circumcenter of triangle XQY .
P roposed by Iman M aghsoudi
5. Do there exist six points X1 , X2 , Y1 , Y2 , Z1 , Z2 in the plane such that all
of the triangles Xi Yj Zk are similar for 1 ≤ i, j, k ≤ 2?
P roposed by M orteza Saghaf ian
5
Solutions of 3nd Iranian Geometry Olympiad 2016 (Elementary)
1. Ali wants to move from point A to point B. He cannot walk inside the
black areas but he is free to move in any direction inside the white areas (not
only the grid lines but the whole plane). Help Ali to find the shortest path
between A and B. Only draw the path and write its length.
P roposed by M orteza Saghaf ian
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Solution.
According to Pythagorean theorem, the length of the path AB is equal to:
p
p
p
√
32 + 3 2 + 32 + 4 2 + 1 + 22 + 2 2 + 1 = 7 + 5 2
6
2.Let ω be the circumcircle of triangle ABC with AC > AB. Let X be a point
on AC and Y be a point on the circle ω, such that CX = CY = AB. (The
points A and Y lie on different sides of the line BC). The line XY intersects ω
for the second time in point P . Show that P B = P C.
P roposed by Iman M aghsoudi
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Solution.
We know that CX = CY therefore:
∠Y XC = ∠XY C
_
_
_ _
⇒AP + CY =P C
_ _
_ _
Also we have AB = CY therefore AP + CY =AP + AB=P B, so P B = P C.
7
3. Suppose that ABCD is a convex quadrilateral with no parallel sides. Make
a parallelogram on each two consecutive sides. Show that among these 4 new
points, there is only one point inside the quadrilateral ABCD.
P roposed by M orteza Saghaf ian
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Solution.
It’s clear that the ray from B parallel to AD passes through the quadrilateral
if and only if ∠DAB + ∠ABC > 180◦ .
We have to find a parallelogram such that both of it’s rays pass thorough
ABCD. Among A, B and C, D there is exactly one set with sum of angles
greater than 180◦ . Also among A, D and B, D there is exactly one set with sum
of angles greater than 180◦ . These two good sets have a vertex in common, say
A. So both of the rays from B parallel to AD, and from D parallel to AB, are
inside the quadlirateral.
8
4. In a right-angled triangle ABC (∠A = 90◦ ), the perpendicular bisector of
BC intersects the line AC in K and the perpendicular bisector of BK intersects
the line AB in L. If the line CL be the internal bisector of angle C, find all
possible values for angles B and C.
P roposed by M ahdi Etesami F ard
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Solution.
We have three cases:
Case i. AC > AB. We know that:
∠LBK = ∠LKB = α ⇒ ∠KLA = 2α ⇒ ∠LKA = 90◦ − 2α
α
∠BKA
= 45◦ −
2
2
Let T be a point on BC such that LT ⊥BC. We know that the line CL is the
internal bisector of angle C, so LT = LA also we have LB = LK therefore two
triangles BT L and KAL are congruent.
BK = CK
⇒ ∠KBC = ∠KCB =
α
= 90◦ − 2α ⇒ α = 18◦
2
= 54◦ and ∠C = 36◦
⇒ ∠LBT = ∠LKA ⇒ 45◦ +
Therefore ∠B = 45◦ +
α
2
9
Case ii. AC < AB. We know that:
∠LBK = ∠LKB = α ⇒ ∠KLA = 2α ⇒ ∠LKA = 90◦ − 2α
Let T be a point on BC such that LT ⊥BC. We know that the line CL is the
internal bisector of angle C, so LT = LA also we have LB = LK therefore two
triangles BT L and KAL are equal.
⇒ ∠LBT = ∠LKA = 90◦ − 2α ⇒ ∠CBK = ∠BKC = 90◦ − α
On the other hand we have:
BK = CK
⇒ ∠CBK = ∠BKC = 60◦ ⇒ α = 30◦
Therefore ∠B = 90◦ − 2α = 30◦ and ∠C = 60◦
Case iii. AC = AB. In this case, K ≡ A and L is the midpoint of AB. Let T
be a point on BC such that LT ⊥BC. We know that the line CL is the internal
bisector of angle C, so LT = LA = LB which is impossible.
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5. Let ABCD be a convex quadrilateral with these properties:
√
∠ADC = 135◦ and ∠ADB−∠ABD = 2∠DAB = 4∠CBD. If BC = 2CD
prove that AB = BC + AD.
P roposed by M ahdi Etesami F ard
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Solution.
Suppose that ∠CBD = α, so ∠DAB = 2α, therefore:
∠ADB − ∠ABD = 4α , ∠ADB + ∠ABD = 180◦ − 2α
⇒ ∠ADB = 90◦ + α , ∠ABD = 90◦ − 3α ⇒ ∠DAB + ∠CBA = 90◦
Let P be intersection point of AD and BC. So we have ∠AP
B = 90◦ . On the
√
other hand we know that ∠P DC = 45◦ , therefore P D = 22 CD = BC
2
Let the point Q be the reflection of point D in point P , Thus QD = 2P D = BC.
We know that two triangles DP B and QP B are congruent. So ∠CBD =
∠CBQ = α, therefore ∠ABQ = 90◦ − α. On the other hand ∠DAB = 2α, so
the triangle ABQ is isosceles.
⇒ AB = AQ ⇒ AB = DQ + AD = BC + AD
.
11
Solutions of 3nd Iranian Geometry Olympiad 2016 (Medium)
1. In trapezoid ABCD with AB k CD, ω1 and ω2 are two circles with diameters AD and BC, respectively. Let X and Y be two arbitrary points on ω1
and ω2 , respectively. Show that the length of segment XY is not more than
half of the perimeter of ABCD.
P roposed by M ahdi Etesami F ard
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First solution.
Let O1 and O2 be the centers of circles ω1 and ω2 , respectively. It’s clear that
O1 and O2 are the midpoints of AD and BC, respectively.
BC
AB + CD
AD
, Y O2 =
, O 1 O2 =
2
2
2
AB + BC + CD + DA
⇒ XY ≤ XO1 + O1 O2 + Y O2 =
2
XO1 =
12
Second solution.
The farthest points of two circles lie on their center line.
And it’s clear in the figure that:
XO1 =
AD
AB + CD
BC
, O1 O2 =
, Y O2 =
2
2
2
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2. Let two circles C1 and C2 intersect in points A and B. The tangent to C1
at A intersects C2 in P and the line P B intersects C1 for the second time in Q
(suppose that Q is outside C2 ). The tangent to C2 from Q intersects C1 and
C2 in C and D, respectively (The points A and D lie on different sides of the
line P Q). Show that AD is bisector of the angle CAP .
P roposed by Iman M aghsoudi
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Solution.
We know that:
∠CAB = ∠CQB , ∠DAB = ∠BDQ
⇒ ∠CAD = ∠CAB + ∠DAB = ∠CQB + ∠BDQ = ∠P BD = ∠P AD
Therefore AD is the bisector of ∠CAP .
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3. Find all positive integers N such that there exists a triangle which can be
dissected into N similar quadrilaterals.
P roposed by N ikolai Beluhov (Bulgaria) and M orteza Saghaf ian
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Solution.
For N = 1 it’s clear that this is impossible. Also for N = 2 this dissection is
impossible too, because one of the two quadrilaterals is convex and the other is
concave. For N ≥ 3 we can do this kind of dissection in equilateral triangle.
15
4. Let ω be the circumcircle of right-angled triangle ABC (∠A = 90◦ ). Tangent
to ω at point A intersects the line BC in point P . Suppose that M is the
midpoint of (the smaller) arc AB, and P M intersects ω for the second time in
Q. Tangent to ω at point Q intersects AC in K. Prove that ∠P KC = 90◦ .
P roposed by Davood V akili
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Solution.
Suppose that AB < AC. It’s enough to show that P K k AB.
PQ
MB
PB
AQ
=
, 4P M B ∼ 4P CQ ⇒
=
MA
PA
QC
PQ
PA
AC
=
4P BA ∼ 4P AC ⇒
BA
PB
We know that M A = M B, so according to above three equations we can say
that:
AQ
BA
=
(1)
QC
AC
4P M A ∼ 4P AQ ⇒
4KAQ ∼ 4KQC
⇒
4P BA ∼ 4P AC
⇒
KA
KQ
AQ
KA
AQ 2
=
=
⇒
=(
) (2)
KQ
KC
QC
KC
QC
PA
BA
PB
BA 2
PB
=
=
⇒
=(
) (3)
PA
PC
AC
PC
AC
(1), (2), (3) ⇒
KA
PB
=
KC
PC
⇒
P K k AB
16
The solution is the same in case of AB > AC.
17
5. Let the circles ω and ω 0 intersect in points A and B. Tangent to circle ω at A
intersects ω 0 in C and tangent to circle ω 0 at A intersects ω in D. Suppose that
the internal bisector of ∠CAD intersects ω and ω 0 at E and F , respectively, and
the external bisector of ∠CAD intersects ω and ω 0 in X and Y , respectively.
Prove that the perpendicular bisector of XY is tangent to the circumcircle of
triangle BEF .
P roposed by M ahdi Etesami F ard
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Solution.
Suppose that P is the intersection point of XE and Y F . We know that:
∠EXA = ∠EAC = ∠EAD = ∠F Y A = α ⇒ P X = P Y
∠ABE = ∠EXA = α , ∠ABF = 180◦ − ∠F Y A = 180◦ − α
⇒ ∠EBF = ∠XP Y = 180◦ − 2α ⇒ P EBF : cyclic
EF ⊥XY
⇒ ∠P EF = ∠AEX = ∠AF Y
⇒ PE = PF
We proved that P E = P F and the quadrilateral P EBF is cyclic. Therefore, P
is the midpoint of arc EF in the circumcircle of triangle BEF . Also we know
that the perpendicular bisector of XY is parallel to EF and passes through P .
So the perpendicular bisector of XY is tangent to the circumcircle of triangle
BEF at P .
18
Solutions of 3nd Iranian Geometry Olympiad 2016 (Advanced)
1. Let the circles ω and ω 0 intersect in A and B. Tangent to circle ω at A
intersects ω 0 in C and tangent to circle ω 0 at A intersects ω in D. Suppose that
CD intersects ω and ω 0 in E and F , respectively (assume that E is between F
and C). The perpendicular to AC from E intersects ω 0 in point P and perpendicular to AD from F intersects ω in point Q (The points A, P and Q lie on
the same side of the line CD). Prove that the points A, P and Q are collinear.
P roposed by M ahdi Etesami F ard
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Solution.
We know that:
∠AF C = ∠AED = 180◦ − ∠CAD , ∠AEF = 180◦ − ∠AQD
⇒ ∠AF D = ∠AQD
So the point Q is the reflection of the point F in the line AD. Similarly we can
say the point P is the reflection of the point E in the line AC. Therefore:
∠DAQ = ∠DAF = ∠ACD , ∠CAP = ∠CAE = ∠CDA
⇒ ∠DAQ + ∠CAD + ∠CAP = ∠ACD + ∠CAD + ∠CDA = 180◦
So the points A, P and Q are collinear.
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2. In acute-angled triangle ABC, altitude of A meets BC at D, and M is
midpoint of AC. Suppose that X is a point such that ∠AXB = ∠DXM =
90◦ (assume that X and C lie on opposite sides of the line BM ). Show that
∠XM B = 2∠M BC.
P roposed by Davood V akili
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First solution.
Let N be the midpoint of side AB. So M N k BC and ∠M BC = ∠N M B.
Therefore it’s enough to show that the line M N is the bisector of ∠XM B.
∠ADB = ∠AXB = 90◦ ⇒ AXDB : cyclic
⇒ ∠BXD = ∠BAD = 90◦ −∠ABC ⇒ ∠BXM = 180◦ −∠ABC = ∠BN M
⇒ BN XM ; cyclic , AN = N X = BN
⇒ ∠BM N = ∠XM N
20
Second solution.
Let P be the intersection point of XM and BC. Suppose that Q is the point
such that the quadrilateral ADBQ be a rectangle. We know that:
∠DXP = ∠ADP = 90◦ ⇒ ∠ADX = ∠XP D
Also we know that AXDBQ is cyclic, so:
∠ADX = ∠AQX
⇒ ∠AQX = ∠XP D
So Q, X and P are collinear because AQ k BP .
AM = M C and AQ k BP
⇒ QM = M P
Now we know that ∠QBC = 90◦ , thus:
QM = BM = M P
⇒ ∠XM B = 2∠M BC
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3. In a convex qualrilateral ABCD, let P be the intersection point of AC and
BD. Suppose that I1 and I2 are the incenters of triangles P AB and P DC
respectively. Let O be the circumcenter of P AB, and H the orthocenter of
P DC. Show that the circumcircles of triangles AI1 B and DHC are tangent
together if and only if the circumcircles of triangles AOB and DI2 C are tangent
together.
P roposed by Hooman F attahimoghaddam
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Solution.
Suppose that the circumcircles of triangles AI1 B and DHC is tangent together
at point K. Let Q be the second intersection point of circumcircles of triangles
AKD and BKC. we know that:
∠DHC = ∠DKC = 180◦ − ∠P
∠P + ∠P DK + ∠P CK = ∠DKC
⇒ ∠P DK + ∠P CK = 180◦ − 2∠P
AQKD : cyclic ⇒ ∠AQK = 180◦ − ∠P DK
BQKC : cyclic ⇒ ∠BQK = 180◦ − ∠P CK
⇒ ∠AQB = ∠P DK+∠P CK = 180◦ −2∠P = 180◦ −∠AOB ⇒ AOBQ : cyclic
Also we have ∠AKD = ∠AQD , ∠BKC = ∠BQC and ∠AQB = ∠DKC −∠P .
So ∠CQD = ∠AKB + ∠P = 180◦ − ∠AI1 B + ∠P = 90◦ + ∠P
2 = ∠CI2 D.
22
So the qudrilateral CDQI2 is cyclic. So we have to show that circumcircles of
triangles AOB and DI2 C is tangent together at the point Q. It’s enough to
show that:
∠ABQ + ∠DCQ = ∠AQD
We know that the circumcircles of triangles AI1 B and DHC are tangent together at the point K, so we have:
∠ABK + ∠DCK = ∠AKD
⇒ (∠ABQ + ∠KBQ) + (∠DCQ − ∠KCQ) = ∠AKD
We know that ∠KBQ = ∠KCQ and ∠AKD = ∠AQD, So:
∠ABQ + ∠DCQ = ∠AQD
Therefore the circumcircles of triangles AOB and DI2 C are tangent together
at point Q.
On the other side of the problem, Suppose that the circumcircles of triangles
CI2 D and AOB are tangent together at point Q. Let the point K be the second
intersection of circumcircles of triangles AQD and BQC. Similarly we can show
that the circumcircles of triangles AI1 B and DHC are tangent together at the
point K.
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Comment.
Also there is another solution using inversion with respect to a circle with
Michel’s point of the quadrilateral as its center.
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23
4. In a convex quadrilateral ABCD, the lines AB and CD meet at point E
and the lines AD and BC meet at point F . Let P be the intersection point
of diagonals AC and BD. Suppose that ω1 is a circle passing through D and
tangent to AC at P . Also suppose that ω2 is a circle passing through C and
tangent to BD at P . Let X be the intersection point of ω1 and AD, and Y
be the intersection point of ω2 and BC. Suppose that the circles ω1 and ω2
intersect each other in Q for the second time. Prove that the perpendicular
from P to the line EF passes through the circumcenter of triangle XQY .
P roposed by Iman M aghsoudi
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First solution.
Lemma 1. In the convex quadrilateral ABCD, the lines AB and CD meet
at point E and the lines AD and BC meet at point F . Let point P be the
intersection of AC and BD. Suppose that X and Y be two arbitrary points on
AD and BC, respectively. If BC ∩ P X = U and AD ∩ P Y = V , then the lines
XY , U V and EF are concurrent.
proof.
Let point Z be the intersection of XY and U V . Suppose that P F ∩ U V = L
and P F ∩ CD = K. We know that:
(Z, L, V, U ) = −1 , (E, K, D, C) = −1
If ZF intersects CD in E 0 , so we can say that (E 0 , K, D, C) = −1. Therefore
E ≡ E 0 , so the lines XY , U V and EF are concurrent.
24
Lemma 2. In the cyclic quadrilateral ABCD with circumcenter O, the lines
AB and CD meet at point E and the lines AD and BC meet at point F . If
point P be the intersection of AC and BD, then P O⊥EF .
proof.
Let ω be the circumcircle of quadrilateral ABCD. Suppose that point R is the
intersection of tangents to circle ω at A and C, and point S is the intersection
of tangents to circle ω at B and D.
According to Pascal’s theorem in Hexagonal AABCCD and ABBCDD, we can
say that points R and S lie on line EF .
We know that polar of the point R with respect to circle ω passes through P .
So polar of the point P with respect to circle ω passes through R. Similarly,
we can say that polar of the point P with respect to circle ω passes through S.
Therefore polar of the point P with respect to circle ω is EF . So P O⊥EF .
25
Suppose that P X intersects BC in point U , and P Y intersects AD in point V .
∠XQP = α ⇒ ∠XDP = ∠XP A = ∠U P C = α
∠Y QP = θ ⇒ ∠Y CP = ∠Y P B = ∠V P D = θ
⇒ ∠XV Y = ∠XQY = ∠XU Y = α + θ ⇒ QV XY U : cyclic
Let point O be the circumcenter of QV XY U . According to lemma 1, we can
say that XY , U V and EF are concurrent at point Z. Now according to lamme
2, we can say that P O⊥EF . So the perpendicular from P to EF passes through
the circumcircle of triangle XQY .
26
Second solution.
Suppose that point O is the circumcenter of triangle XQY . The inversion with
respect to a circle with center P trasnforms the problem into this figure. Suppose
that X 0 is the inversion of point X wrt P . We have to show that the line P O0
is the diameter of circumcircle of triangle E 0 P F 0 . Let O00 be the circumcenter
of triangle X 0 Q0 Y 0 . We know that the points P , O0 and O00 are collinear. So
we have to show that he line P O00 passes through the circumcenter of triangle
E0P F 0.
Suppose that O1 , O2 , O3 and O4 are the centers of circles in the above figure
and K be the intersection point of O1 O3 and O2 O4 . We know that point K
lies on perpendicular bisector of P E 0 and P F 0 , thus K is the circumcenter of
triangle P E 0 F 0 . So we have to show that P , K and O00 are collinear. On the
other hand, we know that the quadrilateral D0 B 0 Y 0 Q0 is isosceles trapezoid. So
the point O00 lies on perpendicular bisector of B 0 D0 . Similarly, the point O00 lies
on perpendicular bisector of A0 C 0 . Therefore, the point O00 is the intersection
of A0 C 0 and B 0 D0 .
27
Suppose that:
A0 C 0 ∩ O1 O2 = M , A0 C 0 ∩ O3 O4 = T
B 0 D0 ∩ O2 O3 = N , B 0 D0 ∩ O1 O4 = L
Let points U and V be on A0 C 0 such that KU ⊥A0 C 0 and O00 V ⊥A0 C 0 . Also let
points R and S be on B 0 D0 such that KR⊥B 0 D0 and O00 S⊥B 0 D0 .
We know that O1 O2 and O3 O4 are perpendicular to A0 C 0 . So O1 O2 k O3 O4
Similarly O2 O3 k O1 O4 , therefore the quadrilateral O1 O2 O3 O4 is a parallelogram. It means that the point K lies on the midpoint of the segments O1 O3
and O2 O4 . So U M = U T . Also we have A0 M = P M and C 0 T = P T
⇒ P V = A0 V − A0 P = (P M + P T ) − 2P M = P T − P M
⇒ TV = PT − PV = PM
⇒ UP = UV
Similarly, we can show that RP = RS, so point K lies on the perpendicular
bisector of P V and P S. It means that K is the citcumcenter of triangle P SV .
Therefore the points P , K and O00 are collinear.
28
5. Do there exist six points X1 , X2 , Y1 , Y2 , Z1 , Z2 in the plane such that all of
the triangles Xi Yj Zk are similar for 1 ≤ i, j, k ≤ 2.
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution. (by Ilya Bogdanov f rom Russia)
Suppose a triangle XY Z, in such a way that XY = 1, Y Z = t2 , ZX = t3 and
∠Z = ∠X + 2∠Y .
Such a triangle exists, because for the minimum possible value of t, we have
∠Z > ∠X + 2∠Y and for t = 1 we have ∠Z < ∠X + 2∠Y . So there exists a
triangle with the above properties. Now consider the following 6 points, these
points have the properties of the problem.
29
So there exist the points X1 , X2 , Y1 , Y2 , Z1 , Z2 in the plane such that Xi Yj Zk
be the similar triangles for all of 1 ≤ i, j, k ≤ 2
4th Iranian Geometry Olympiad
Problems and Solutions - 2017
This booklet is prepared by Hirad Aalipanah, Iman Maghsoudi.
With special thanks to Morteza Saghafian, Mahdi Etesami Fard,
Davood Vakili, Hooman Fattahimoghaddam.
Copyright c Young Scholars Club 2017. All rights reserved.
Ministry of education, Islamic Republic of Iran.
www.ysc.ac.ir - www.igo-official.ir
1
Problems of 4th Iranian Geometry Olympiad 2017 (Elementary)
1. Each side of square ABCD with side length of 4 is divided into equal parts
by three points. Choose one of the three points from each side, and connect the
points consecutively to obtain a quadrilateral. Which numbers can be the area of
this quadrilateral? Just write the numbers without proof.
P roposed by Hirad Aalipanah
2. Find the angles of triangle ABC.
P roposed by M orteza Saghaf ian
3. In the regular pentagon ABCDE, the perpendicular at C to CD meets AB
at F . Prove that AE + AF = BE.
P roposed by Alireza Cheraghi
2
4. P1 , P2 ,..., P100 are 100 points on the plane, no three of them are collinear. For
each three points, call their triangle clockwise if the increasing order of them is in
clockwise order. Can the number of clockwise triangles be exactly 2017?
P roposed by M orteza Saghaf ian
5. In the isosceles triangle ABC (AB = AC), let l be a line parallel to BC through
A. Let D be an arbitrary point on l. Let E, F be the feet of perpendiculars through
A to BD, CD respectively. Suppose that P , Q are the images of E, F on l. Prove
that AP + AQ ≤ AB.
P roposed by M orteza Saghaf ian
3
Problems of 4th Iranian Geometry Olympiad 2017 (Intermediate)
1. Let ABC be an acute-angled triangle with A = 60◦ . Let E, F be the feet of
altitudes through B, C respectively. Prove that CE − BF = 23 (AC − AB).
P roposed by F atemeh Sajadi
2. Two circles ω1 , ω2 intersect at A, B. An arbitrary line through B meets ω1 ,
ω2 at C, D respectively. The points E, F are chosen on ω1 , ω2 respectively so that
CE = CB, BD = DF . Suppose that BF meets ω1 at P , and BE meets ω2 at Q.
Prove that A, P , Q are collinear.
P roposed by Iman M aghsoudi
3. On the plane, n points are given (n > 2). No three of them are collinear. Through
each two of them the line is drawn, and among the other given points, the one nearest
to this line is marked (in each case this point occurred to be unique). What is the
maximal possible number of marked points for each given n?
P roposed by Boris F renkin (Russia)
4. In the isosceles triangle ABC (AB = AC), let l be a line parallel to BC through
A. Let D be an arbitrary point on l. Let E, F be the feet of perpendiculars through
A to BD, CD respectively. Suppose that P , Q are the images of E, F on l. Prove
that AP + AQ ≤ AB.
P roposed by M orteza Saghaf ian
5. Let X, Y be two points on the side BC of triangle ABC such that 2XY = BC.
(X is between B, Y ) Let AA0 be the diameter of the circumcircle of triangle AXY .
Let P be the point where AX meets the perpendicular from B to BC, and Q be
the point where AY meets the perpendicular from C to BC. Prove that the tangent
line from A0 to the circumcircle of AXY passes through the circumcenter of triangle
AP Q.
P roposed by Iman M aghsoudi
4
Problems of 4th Iranian Geometry Olympiad 2017 (Advanced)
1. In triangle ABC, the incircle, with center I, touches the side BC at point D.
Line DI meets AC at X. The tangent line from X to the incircle (different from AC)
intersects AB at Y . If Y I and BC intersect at point Z, prove that AB = BZ.
P roposed by Hooman F attahimoghaddam
2. We have six pairwise non-intersecting circles that the radius of each is at least
one. Prove that the radius of any circle intersecting all the six circles, is at least one.
P roposed by M ohammad Ali Abam− M orteza Saghaf ian
3. Let O be the circumcenter of triangle ABC. Line CO intersects the altitude
through A at point K. Let P , M be the midpoints of AK, AC respectively. If P O
intersects BC at Y , and the circumcircle of triangle BCM meets AB at X, prove
that BXOY is cyclic.
P roposed by Ali Daeinabi − Hamid P ardazi
4. Three circles ω1 , ω2 , ω3 are tangent to line l at points A, B, C (B lies between
A, C) and ω2 is externally tangent to the other two. Let X, Y be the intersection
points of ω2 with the other common external tangent of ω1 , ω3 . The perpendicular
line through B to l meets ω2 again at Z. Prove that the circle with diameter AC
touches ZX, ZY .
P roposed by Iman M aghsoudi − Siamak Ahmadpour
5. Sphere S touches a plane. Let A, B, C, D be four points on this plane such
that no three of them are collinear. Consider the point A0 such that S is tangent to
the faces of tetrahedron A0 BCD. Points B 0 , C 0 , D0 are defined similarly. Prove that
A0 , B 0 , C 0 , D0 are coplanar and the plane A0 B 0 C 0 D0 touches S.
P roposed by Alexey Zaslavsky (Russia)
5
Solutions of 4th Iranian Geometry Olympiad 2017 (Elementary)
1. Each side of square ABCD with side length of 4 is divided into equal parts
by three points. Choose one of the three points from each side, and connect the
points consecutively to obtain a quadrilateral. Which numbers can be the area of
this quadrilateral? Just write the numbers without proof.
P roposed by Hirad Aalipanah
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
To find the area of the quadrilaterals, it’s enough to find sum of the areas of four
right triangles and subtract it from area of the square. Finally, these numberes will
be found:
6, 7, 7.5, 8, 8.5, 9, 10
6
2. Find the angles of triangle ABC.
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Let ∠ACB = x. The quadrilateral with equal sides in triangle ABC is a rhombus so it has parallel sides.
According to the angles shown in the figure, we can say that sum of the angles of
triangle ABC is equal to 180◦ .
x
(90◦ − ) + 2x + x = 180◦
2
⇒
x = 36◦
⇒ ∠A = 72◦ , ∠B = 72◦ , ∠C = 36◦
7
3. In the regular pentagon ABCDE, the perpendicular at C to CD meets AB at F .
Prove that AE + AF = BE.
P roposed by Alireza Cheraghi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Suppose that P is the intersection of AE and F C. We know that:
∠ECD = 36◦
⇒
∠ECP = 54◦ , ∠AEC = 72◦
⇒
∠EP C = 54◦
Therefore, CE = P E. On the other hand, we know that ∠ECB = ∠EBC = 72◦ .
Therefore, BE = CE = P E. Also we have ∠EAB = 108◦ so ∠AF P = ∠AP F = 54◦ .
It means that AF = AP .
⇒
AE + AF = AE + AP = P E = CE = BE 8
4. P1 , P2 ,..., P100 are 100 points on the plane, no three of them are collinear. For
each three points, call their triangle clockwise if the increasing order of them is in
clockwise order. Can the number of clockwise triangles be exactly 2017?
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
First, suppose that P1 , P2 ,..., P100 are located on a circle in a counter-clockwise order.
In this case, the number of clockwise triangles is zero. Now, start to move the points,
when a point Pi passes through a line Pj Pk , the state of triangle Pi Pj Pk (clockwise
or counter-clockwise) changes and the state of the other triangles does not change
(assume that the points move in such a way that no point passes through two lines
at the same time). So the number of clockwise triangles changes by one. Suppose
that the points P1 , P2 ,..., P100 move to obtain their
clockwise order on a circle. In this
100
case, the number of clockwise triangles is 3 which is greater than 2017. In this
process, the number of clockwise triangles changes one by one. So there is a moment
in which the number of clockwise triangles is exactly 2017. 9
5. In the isosceles triangle ABC (AB = AC), let l be a line parallel to BC through
A. Let D be an arbitrary point on l. Let E, F be the feet of perpendiculars through
A to BD, CD respectively. Suppose that P , Q are the images of E, F on l. Prove
that AP + AQ ≤ AB.
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution1.
Suppose that M , N are the midpoints of AB , AC respectively. We have:
∠AEB = ∠AF C = 90◦ ⇒
ME = NF =
AC
AB
=
2
2
Let X, Y be the feet of perpendiculars through M , N to line l respectively. We know
that two triangles AM A and AN Y are equal so AX = AY .
XP = AX + AP ≤ M E =
AB
AC
, AQ − AY = Y Q ≤ N F =
2
2
⇒ AP + AQ ≤ AB 10
Solution2.
Let P 0 , E 0 , be the reflection of P , E with respect to the perpendicular bisector
of BC. Therefore, we have ∠AE 0 C = ∠AP 0 E = 90◦ and AP 0 = AP . We have:
AP + AQ = AP 0 + AQ = QP 0 ≤ F E 0 ≤ AC = AB 11
Solutions of 4th Iranian Geometry Olympiad 2017 (Intermediate)
1. Let ABC be an acute-angled triangle with A = 60◦ . Let E, F be the feet of
altitudes through B, C respectively. Prove that CE − BF = 23 (AC − AB).
P roposed by F atemeh Sajadi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
∠A = 60◦
⇒
⇒ ∠ABF = ACF = 30◦ ⇒ AE =
AC
AB
, AF =
2
2
CE − BF = (AC − AE) − (AB − AF )
3
= (AC − AB) + (AF − AE) = (AC − AB) 2
12
2. Two circles ω1 , ω2 intersect at A, B. An arbitrary line through B meets ω1 , ω2
at C, D respectively. The points E, F are chosen on ω1 , ω2 respectively so that
CE = CB, BD = DF . Suppose that BF meets ω1 at P , and BE meets ω2 at Q.
Prove that A, P , Q are collinear.
P roposed by Iman M aghsoudi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
We know that:
∠BF D = ∠DBF = 180◦ − ∠CBP = ∠CEP
⇒
∠CEB + ∠BEP = ∠BF Q + ∠QF D
∠CEB = ∠CBE = ∠QBD = ∠QF D
⇒
∠BEP = ∠BF Q
⇒
∠BAP = ∠BEP = ∠BF Q = ∠BAQ
So A, P , Q are collinear. 13
3. On the plane, n points are given (n > 2). No three of them are collinear. Through
each two of them the line is drawn, and among the other given points, the one nearest
to this line is marked (in each case this point occurred to be unique). What is the
maximal possible number of marked points for each given n?
P roposed by Boris F renkin (Russia)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
The case n = 3 is obvious. For n > 4, take a regular n-gon and slightly deform
it to make the assumptions of the problem valid. Each vertex is the nearest to the
line connecting two adjacent vertices (this is directly proved by computing angles).
If n = 4 then two cases are possible.
1) The given points form a convex quadrilateral (say ABCD). Since for each side
the nearest vertex is unique, no two sides are parallel. Let lines AD and BC meet
beyond A and B, and lines AB and CD meet beyond B and C respectively. Then
points A, B and C are marked. Suppose D is marked as well. Then it is the nearest
vertex to AC, and the area of 4 ACD is less than that of 4 ACB. On the other
hand, draw the line through C parallel to AD. Let it meet AB at point E. Then
CE < AD, so S4 ACD > S4 ACE > S4 ACB , a contradiction.
2) One of the given points (say O) is inside the triangle ABC formed by the remaining points. Obviously O is nearest of the given points to lines AB, BC and AC.
Without loss of generality, the nearest point to line AO is B. If the nearest point
to line BO is A then C cannot be marked. Suppose the nearest point to line BO is
C. If the remaining point A is also marked then it is nearest to line CO. Let line
AO intersect BC at point A1 , and let points B1 , C1 be defined similarly. We have
S4 AA1 B < S4 AA1 C , hence A1 B < A1 C. Similarly B1 C < B1 A, C1 A < C1 B. Then
A1 B · B1 C · C1 A < A1 C · B1 A · C1 B, a contradiction with Ceva theorem. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Comment.
In fact, it is not necessary to apply Ceva theorem. Let MA , MB , MC be the bases
of the corresponding medians, and M be the centroid. Then O must belong to the
interior of each of triangles M MA B, M MB C, M MC A which is impossible.
14
4. In the isosceles triangle ABC (AB = AC), let l be a line parallel to BC through
A. Let D be an arbitrary point on l. Let E, F be the feet of perpendiculars through
A to BD, CD respectively. Suppose that P , Q are the images of E, F on l. Prove
that AP + AQ ≤ AB.
P roposed by M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution1.
Suppose that M , N are the midpoints of AB , AC respectively. We have:
∠AEB = ∠AF C = 90◦ ⇒
ME = NF =
AC
AB
=
2
2
Let X, Y be the feet of perpendiculars through M , N to line l respectively. We know
that two triangles AM A and AN Y are equal so AX = AY .
XP = AX + AP ≤ M E =
AB
AC
, AQ − AY = Y Q ≤ N F =
2
2
⇒ AP + AQ ≤ AB 15
Solution2.
Let P 0 , E 0 , be the reflection of P , E respect to the perpendicular bisector of BC.
Therefore, we have ∠AE 0 C = ∠AP 0 E = 90◦ and AP 0 = AP . We have:
AP + AQ = AP 0 + AQ = QP 0 ≤ F E 0 ≤ AC = AB 16
5. Let X, Y be two points on the side BC of triangle ABC such that 2XY = BC.
(X is between B, Y ) Let AA0 be the diameter of the circumcircle of triangle AXY .
Let P be the point where AX meets the perpendicular from B to BC, and Q be
the point where AY meets the perpendicular from C to BC. Prove that the tangent
line from A0 to the circumcircle of AXY passes through the circumcenter of triangle
AP Q.
P roposed by Iman M aghsoudi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Let O, M , N be the circumcenter of triangle AP Q, midpoint of AP , midpoint of AQ
respectively. We know that ∠OM A = ∠ON A = 90◦ so the quadrilateral AM ON is
cyclic. We have to prove ∠OA0 A = 90◦ . Therefore, we have to show that AM OA0 N
is cyclic. It’s enought to show that the quadrilateral AM A0 N is cyclic. We should
prove ∠A0 XM = ∠A0 N Y . We will show that 4A0 N Y ∼ 4A0 XM .
We know that the quadrilateral AXA0 Y is cyclic so ∠A0 XM = ∠A0 Y N . Therefore,
it’s enough to show that:
A0 X
A0 Y
=
(1)
MX
NY
17
Let H, K, M 0 , N 0 be the feet of perpendiculars through A, A0 , M , N to BC respectively. We know that M , N are the midpoints of AP , AQ so we have:
BM 0 = HM 0
CN 0 = HN 0
⇒
XM 0 = HM 0 − XH =
⇒
BH
− XH
2
(2)
CH
BC − BH
= (XY − XH) −
2
2
BC
BH
BH
Y N 0 = (XY −
)+(
− XH) =
− XH (3)
2
2
2
⇒
Y N 0 = HY − HN 0 = HY −
According to the equations (2), (3) we can say that XM 0 = Y N 0 . On the other hand,
we have AA0 is the diameter of the circumcircle of triangle AXY . Therefore:
∠A0 Y A = ∠A0 XA = 90◦
⇒
∠KA0 Y = ∠N Y N 0 , ∠KA0 X = ∠M XM 0
⇒
4A0 KX ∼ 4XM 0 M
A0 X
A0 K
=
⇒
MX
XM 0
So equation (1) is proved. ,
,
4A0 KY ∼ 4Y N 0 N
A0 K
A0 Y
=
NY 0
NY
18
⇒
A0 X
A0 Y
=
MX
NY
Solutions of 4th Iranian Geometry Olympiad 2017 (Advanced)
1. In triangle ABC, the incircle, with center I, touches the side BC at point D.
Line DI meets AC at X. The tangent line from X to the incircle (different from AC)
intersects AB at Y . If Y I and BC intersect at point Z, prove that AB = BZ.
P roposed by Hooman F attahimoghaddam
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
In triangle AXY , the point I is excenter so ∠XIY = 90◦ − A2 .
⇒
∠DIZ = 90◦ −
A
2
⇒
∠IZB =
A
= ∠BAI
2
Also we know that ∠IBZ = ∠IBA and BI = BI so two triangles ABI and ZBI are
equal. Thus, AB = BZ. 19
2. We have six pairwise non-intersecting circles that the radius of each is at least one.
Prove that the radius of any circle intersecting all the six circles, is at least one.
P roposed by M ohammad Ali Abam− M orteza Saghaf ian
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Call the centers of these six circles O1 , O2 ,..., O6 , and their radii R1 , R2 ,..., R6 . Suppose that a circle with center O and radius R intersects these six circles. Obviously
there exist i, j such that Oi OOj < 60◦ . The length of Oi Oj is at least Ri + Rj and the
lengths of OOi , OOj are less than or equal to Ri + R, Rj + R, respectively. If R < 1
then considering Ri , Rj > 1 we conclude that in triangle Oi OOj the longest side is
Oi Oj , so Oi OOj ≥ 60◦ which yields a contradiction. So we must have R ≥ 1. 20
3. Let O be the circumcenter of triangle ABC. Line CO intersects the altitude
through A at point K. Let P , M be the midpoints of AK, AC respectively. If P O
intersects BC at Y , and the circumcircle of triangle BCM meets AB at X, prove
that BXOY is cyclic.
P roposed by Ali Daeinabi − Hamid P ardazi
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
We want to show that ∠XOP = ∠B. We know that:
∠B = ∠XM A
⇒
∠XM O = 90◦ − ∠B = ∠XAK
We will prove that two triangles XOP and XM A are similar. It’s enough to show
that two triangles XP A and XOM are similar. We have ∠XM O = ∠XAK so it’s
enough to prove:
AP
AX
=
XM
OM
21
The quadrilateral BXM C is cyclic. Therefore 4AXM ∼ 4ACB.
⇒
AC
AX
=
XM
BC
(1)
The point O is the circumcenter of triangle ABC. Thus, ∠OCA = 90◦ − ∠B and
∠AKC = 180◦ − ∠A. According to law of sines in triangles AKC, OM C and ABC
we can say that:
AC
AK
OM
=
, OC =
◦
sin∠A
sin(90 − ∠B)
sin(90◦ − ∠B)
⇒
AC
AK
=
OM
OC.sin∠A
⇒
AP
AC
AC
=
=
OM
2OC.sin∠A
BC
(2)
According to equations (1) and (2), we can conclude that two triangles XP A and
XOM are similar. Therefore two triangles XOP and XM A are similar so:
∠XBP = ∠XM A = ∠B 22
4. Three circles ω1 , ω2 , ω3 are tangent to line l at points A, B, C (B lies between
A, C) and ω2 is externally tangent to the other two. Let X, Y be the intersection
points of ω2 with the other common external tangent of ω1 , ω3 . The perpendicular
line through B to l meets ω2 again at Z. Prove that the circle with diameter AC
touches ZX, ZY .
P roposed by Iman M aghsoudi − Siamak Ahmadpour
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Let S be the intersection of XY and l. Suppose that ω1 , ω2 are tangent to each
other at point E and ω2 , ω3 are tangent to each other at point F . We know that S is
the external homothetic center for ω1 , ω3 and E, F are antihomologous points so E,
F , S are collinear. Suppose that XY meets ω1 , ω2 at P , Q respectively so we have:
SE · SF = SP · SQ = SA · SC
Therefore, the quadrilaterals P EF Q and AEF C are cyclic. Let T be the midpoint
of arc XY in circle ω2 (the arc which has not Z).We know that the line tangent to
ω2 at T is parallel to XY so T , E, P are collinear. Similarly T , Q, F are collinear.
TE · TP = TF · TQ
⇒
Pω1 (T ) = Pω3 (T ) (1)
On the other hand, the line which is tangent to ω2 at Z is parallel to l so Z, E, A
are collinear. Similarly Z, F , C are collinear.
ZE · ZA = ZF · ZC
⇒
23
Pω1 (Z) = Pω3 (Z) (2)
According to the equations (1), (2) we can say that ZT is the radical axis of ω1 , ω3 .
Therefore, if M is the midpoint of AC then Z, T , M are collinear because M lies on
radical axis of ω1 , ω3 too. Let D be the intersection of ZM , P Q and H be the feet
of perpendiculars through Z to P Q. We know that:
∠HZD = ∠M ZB =
|∠ZXY − ∠ZY X|
=α
2
Let x = ZY , y = ZX, t = XY . We can say that:
AM =
ZB
ZB
ZB
ZB 2xy
(∠XZY )
= ZH =
· ZD =
·
· cos
cos α
ZH
ZH x + y
2
ZD
To prove the problem, it’s enough to show that the distances of M from two lines
. Thus, we have to show that:
ZX, ZY are equal to AC
2
∠XZY
AC
=
2
2
(∠XZY )
ZB 2xy
2AM · sin
=
·
· sin (∠XZY ) = AC
2
ZH x + y
AM · sin
⇐⇒
On the other hand, we have:
1
1
S4XY Z = t · ZH = xy sin (∠XZY )
2
2
So it’s enough to show that:
AC =
2t · ZB
x+y
24
⇒
t=
xy sin (∠XZY )
ZH
We know that:
∠BF Z = ∠BEZ = 90◦
⇒
ZB 2 = ZE · ZA = ZF · ZC
Therefore, lenghts of tangents from Z to ω1 , ω3 are equal to ZB. Now, According
to Casey’s theorem (generalization of Ptolemy’s theorem) in two cyclic quadrilaterals
ZY ω1 X and ZXω3 Y we can say that:
ZY ω1 X : x · P X + y · (t + P X) = t · ZB
⇒
(x + y)P X + yt = t · ZB
ZXω3 Y : y · Y Q + x · (t + Y Q) = t · ZB
⇒
(x + y)Y Q + xt = t · ZB
⇒
PX + Y Q + t =
2t · ZB
x+y
25
⇒
AC =
2t · ZB
x+y
5. Sphere S touches a plane. Let A, B, C, D be four points on this plane such that
no three of them are collinear. Consider the point A0 such that S is tangent to the
faces of tetrahedron A0 BCD. Points B 0 , C 0 , D0 are defined similarly. Prove that A0 ,
B 0 , C 0 , D0 are coplanar and the plane A0 B 0 C 0 D0 touches S.
P roposed by Alexey Zaslavsky (Russia)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Solution.
Let S touch the plane at point P . D is the point of concurrency of the planes
passing through AB, BC, CA and touching S (similarly for A0 , B 0 , C 0 ). Suppose
that the plane passing through X, Y and touching S, touches it at Pxy . The points
P , Pab , Pac , Pad are on a circle Wa , because when we connect them to P , the resulted
line touches S. Similarly we can find other triples which are concyclic with P , name
the circles with Wb , Wc , Wd . Now use inversion with center P and arbitrary radius,
we get 4 lines with three points on each line. Considering the concurrency of four
circles at Michels Point we conclude that the circles Pab Pbc Pca , Pab Pbd Pda , Pac Pcd Pda ,
Pbc Pcd Pdb (call them Wd0 , Wc0 , Wb0 , Wa0 ) are concurrent on S at a point P 0 . Wd0 is
the locus of feet of tangencies from D0 to S, so D0 P 0 is tangent to S and D0 is on
the tangent plane at P 0 to S. Similarly we conclude that A0 , B 0 , C 0 are also on the
tangent plane at P 0 to S. So A0 , B 0 , C 0 , D0 are coplanar and this plane is tangent to
S. 26
Fifth Iranian Geometry
Olympiad
IGO 2018
The problems along with their solutions
Contents
Elementary Level
2
Problems
2
Solutions
4
Intermediate Level
11
Problems
11
Solutions
13
Advanced Level
20
Problems
20
Solutions
21
Elementary Level
1
Problems
1. As shown below, there is a 40 × 30 paper with a filled 10 × 5 rectangle inside of it. We want
to cut out the filled rectangle from the paper using four straight cuts. Each straight cut is a
straight line that divides the paper into two pieces, and we keep the piece containing the filled
rectangle. The goal is to minimize the total length of the straight cuts. How to achieve this
goal, and what is that minimized length? Show the correct cuts and write the final answer.
There is no need to prove the answer.
2. Convex hexagon A1 A2 A3 A4 A5 A6 lies in the interior of convex hexagon B1 B2 B3 B4 B5 B6 such
that A1 A2 ∥ B1 B2 , A2 A3 ∥ B2 B3 ,..., A6 A1 ∥ B6 B1 . Prove that the areas of simple hexagons
A1 B2 A3 B4 A5 B6 and B1 A2 B3 A4 B5 A6 are equal. (A simple hexagon is a hexagon which does
not intersect itself.)
3. In the given figure, ABCD is a parallelogram.
√ We
◦
know that ∠D = 60 , AD = 2 and AB = 3 + 1.
Point M is the midpoint of AD. Segment CK is
the angle bisector of C. Find the angle CKB.
2
4. There are two circles with centers O1 , O2 lie inside of circle ω and are tangent to it. Chord AB
of ω is tangent to these two circles such that they lie on opposite sides of this chord. Prove
that ∠O1 AO2 + ∠O1 BO2 > 90◦ .
5. There are some segments on the plane such that no two of them intersect each other (even at
the ending points). We say segment AB breaks segment CD if the extension of AB cuts CD
at some point between C and D.
(a) Is it possible that each segment when extended from
both ends, breaks exactly one other segment from each
way?
(b) A segment is called surrounded if from both sides of
it, there is exactly one segment that breaks it.
(e.g. segment AB in the figure.) Is it possible to have
all segments to be surrounded?
3
Solutions
1. As shown below, there is a 40 × 30 paper with a filled 10 × 5 rectangle inside of it. We want
to cut out the filled rectangle from the paper using four straight cuts. Each straight cut is a
straight line that divides the paper into two pieces, and we keep the piece containing the filled
rectangle. The goal is to minimize the total length of the straight cuts. How to achieve this
goal, and what is that minimized length? Show the correct cuts and write the final answer.
There is no need to prove the answer.
Proposed by Morteza Saghafian
-----------------------------------------------------------------Solution. The answer is 65. Here is an example of the solution:
.
■
4
2. Convex hexagon A1 A2 A3 A4 A5 A6 lies in the interior of convex hexagon B1 B2 B3 B4 B5 B6 such
that A1 A2 ∥ B1 B2 , A2 A3 ∥ B2 B3 ,..., A6 A1 ∥ B6 B1 . Prove that the areas of simple hexagons
A1 B2 A3 B4 A5 B6 and B1 A2 B3 A4 B5 A6 are equal. (A simple hexagon is a hexagon which does
not intersect itself.)
Proposed by Mahdi Etesamifard - Hirad Aalipanah
-----------------------------------------------------------------Solution. As you can see, we have divided the area between two polygons into 6 trapezoids.
In each trapezoid is it easy to see that the triangles which have the same area (like B1 A1 A2 and
B2 A1 A2 ) each belongs to one of the simple hexagons. Therefore, if we add up their areas and
add the common area (the area of A1 A2 A3 A4 A5 A6 ) to them, we can conclude that the areas of
the two simple hexagons are equal.
■
5
3. In the given figure, ABCD is a parallelogram.
√ We
know that ∠D = 60◦ , AD = 2 and AB = 3 + 1.
Point M is the midpoint of AD. Segment CK is
the angle bisector of C. Find the angle CKB.
Proposed by Mahdi Etesamifard
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -√
-------------Solution 1. Let X be a point on AB such that AX = 1 and XB = 3.√ We know that
∠M AX = 120◦ . Therefore by Pythagoras theorem we know that M X = 3. So we have
∠M BX = 15◦ and ∠CBK = 45◦ . Hence, ∠CKB = 180◦ − 60◦ − 45◦ = 75◦ .
■
Solution 2. Let N be the midpoint of side BC. M N intersects CK at L. It’s clear that
the triangle CN L is equilateral. Therefore, we have LN = CN =√N B. So, BCL is a rightangled triangle.
√ Because of Pythagoras’s theorem we have BL = 3. On the other hand, we
have M L = 3 and ∠BLN = 30◦ . Because of that, we have ∠LBM = 15◦ and so we have
∠CBK = 30◦ + 15◦ = 45◦ . Hence, ∠CKB = 180◦ − 60◦ − 45◦ = 75◦ .
■
6
4. There are two circles with centers O1 , O2 lie inside of circle ω and are tangent to it. Chord AB
of ω is tangent to these two circles such that they lie on opposite sides of this chord. Prove
that ∠O1 AO2 + ∠O1 BO2 > 90◦ .
Proposed by Iman Maghsoudi
-----------------------------------------------------------------Solution. Let AC, BC be tangents from A, B to the circle with center O1 and AD, BD be
tangents from A, B to the circle with center O2 . It’s enough to show that ∠CAD + ∠CBD >
180◦ . Or to show that ∠ACB + ∠ADB < 180◦ .
We know that C, D lie on the outside of circle ω. Therefore, we can always say that ∠ACB <
∠AXB and ∠ADB < ∠AY B because of the exterior angles. But we know that ∠AXB +
∠AY B = 180◦ . Hense, we can conclude that ∠ACB + ∠ADB < 180◦ and the statement is
proven.
■
7
5. There are some segments on the plane such that no two of them intersect each other (even at
the ending points). We say segment AB breaks segment CD if the extension of AB cuts CD
at some point between C and D.
(a) Is it possible that each segment when extended from
both ends, breaks exactly one other segment from each
way?
(b) A segment is called surrounded if from both sides of
it, there is exactly one segment that breaks it.
(e.g. segment AB in the figure.) Is it possible to have
all segments to be surrounded?
Proposed by Morteza Saghafian
------------------------------------------------------------------
8
Solution.
(a) No. Consider the convex hull of the endpoints of these segments. Let A be a vertex of the
convex hull, where AB is one of the segments.
We know that there exist segments CD, EF as in the figure. So A lies inside the convex hull
of C, D, E, F and therefore it cannot be a vertex of the main convex hull. Contradiction!
□
(b) Yes. The figure below shows that it is possible for all segments to be surrounded.
□
■
9
Intermediate
Level
10
Problems
1. There are three rectangles in the following figure. The lengths of some segments are shown.
Find the length of the segment XY .
2. In convex quadrilateral ABCD, the diagonals AC and BD meet at the point P . We know
that ∠DAC = 90◦ and � 2∠ADB = ∠ACB. If we have ∠DBC + 2∠ADC = 180◦ prove that
2AP = BP .
3. Let ω1 , ω2 be two circles with centers O1 and O2 , respectively. These two circles intersect each
other at points A and B. Line O1 B intersects ω2 for the second time at point C, and line O2 A
intersects ω1 for the second time at point D . Let X be the second intersection of AC and ω1 .
Also Y is the second intersection point of BD and ω2 . Prove that CX = DY .
4. We have a polyhedron all faces of which are triangle. Let P be an arbitrary point on one of
the edges of this polyhedron such that P is not the midpoint or endpoint of this edge. Assume
that P0 = P . In each step, connect Pi to the centroid of one of the faces containing it. This
line meets the perimeter of this face again at point Pi+1 . Continue this process with Pi+1 and
the other face containing Pi+1 . Prove that by continuing this process, we cannot pass through
all the faces. (The centroid of a triangle is the point of intersection of its medians.)
11
5. Suppose that ABCD is a parallelogram such that ∠DAC = 90◦ . Let H be the foot of perpendicular from A to DC, also let P be a point along the line AC such that the line P D is tangent
to the circumcircle of the triangle ABD. Prove that ∠P BA = ∠DBH.
12
Solutions
1. There are three rectangles in the following figure. The lengths of some segments are shown.
Find the length of the segment XY .
Proposed by Hirad Aalipanah
-----------------------------------------------------------------Let us continue the rectangular sides to get the ABC triangle. Because AB = BC we can say
that ∠BCA = ∠BAC = 45◦ . Therefore,
of the segments
using the
√
√
√ we can√determine some
2
Pythagoras’s theorem such as AD = 2 2, DX = 2, CE = 2 2 and EY = 2 . So, we have
√
√
√
√
XY = AC − AD − DX − CE − EY = 10 2 − 2 2 − 2 − 2 2 −
13
√
√
2
9 2
=
2
2
14
2. In convex quadrilateral ABCD, the diagonals AC and BD meet at the point P . We know
that ∠DAC = 90◦ and � 2∠ADB = ∠ACB. If we have ∠DBC + 2∠ADC = 180◦ prove that
2AP = BP .
Proposed by Iman Maghsoudi
-----------------------------------------------------------------Solution.
Let M be the intersection point of the angle bisector of ∠P CB with segment P B. Since
∠P CM = ∠P DA = θ and ∠AP D = ∠M P C, we get that △P M C ∼ △P AD, which means
∠P M C = 90◦ .
Now in triangle CP B, the angle bisector of vertex C is the same as the altitude from C, this
means CP B is an isosceles triangle and so P M = M B, P C = CB.
In triangle DBC, we have
\ + 2θ + P
\
\
DBC
CD + P
DC = 180◦ .
This along with the assumption that ∠DBC + 2∠ADC = 180◦ , implies ∠P CD = ∠P DC.
■
Therefore P C = P D and so △P M C ∼
= △P AD, hence AP = P M = P2B .
15
3. Let ω1 , ω2 be two circles with centers O1 and O2 , respectively. These two circles intersect each
other at points A and B. Line O1 B intersects ω2 for the second time at point C, and line O2 A
intersects ω1 for the second time at point D . Let X be the second intersection of AC and ω1 .
Also Y is the second intersection point of BD and ω2 . Prove that CX = DY .
Proposed by Alireza Dadgarnia
-----------------------------------------------------------------Solution. First, we use a well-known lemma.
Lemma. Let P QRS be a convex quadrilateral with RQ = RS, ∠RP Q = ∠RP S and P Q ̸= P S.
Then P QRS is cyclic.
Proof. Assume the contrary, and let P ′ ̸= P be the intersection point of the circle passing
through R, S, Q with line P R.
Since P ′ QRS is cyclic and RQ = RS, we get ∠SP ′ R = ∠QP ′ R. Now let’s considerate
on triangles SP ′ P and QP ′ P . In these two triangles we have ∠SP ′ P = ∠QP ′ P and also
∠P ′ P Q = ∠P ′ P S. This means these two triangles are congruent, hence P Q = P S, which is a
contradiction. So the lemma is proved.
Back to the problem.
Triangles ADY and BXC are similar, because
\ = BXC
\ = 180◦ − BXA,
\
ADY
and
\
\ = 180◦ − AY
[
DY
A = BCX
B.
Note that O2 lies on the angle bisector of ∠AO1 B, O2 A = O2 C and also O1 A ̸= O1 C. So we
can use the lemma and conclude that O1 AO2 C is cyclic. Similarly, we get that O2 BO1 D is
cyclic.
\
[
\
\
\
AY
D = 180◦ − AY
B=O
1 CA = O1 O2 A = O1 BD.
Which means AC ∥ BD and so AY = BC. But since △ADY ∼ △BXC, we get that these
two triangles are congruent and so CX = DY .
■
16
4. We have a polyhedron all faces of which are triangle. Let P be an arbitrary point on one of
the edges of this polyhedron such that P is not the midpoint or endpoint of this edge. Assume
that P0 = P . In each step, connect Pi to the centroid of one of the faces containing it. This
line meets the perimeter of this face again at point Pi+1 . Continue this process with Pi+1 and
the other face containing Pi+1 . Prove that by continuing this process, we cannot pass through
all the faces. (The centroid of a triangle is the point of intersection of its medians.)
Proposed by Mahdi Etesamifard - Morteza Saghafian
-----------------------------------------------------------------Solution. Suppose that AB is the edge that P lies on. Let M be the midpoint of AB
and without loss of generality, assume that P lies between B and M . We will prove that it is
impossible to pass through a face which doesn’t contain A. (Such face exists in any polyhedron)
Let B = B0 , B1 , B2 , . . . be the vertices adjacent to A in this order. Let Mi be the midpoint of
ABi . By using induction, we prove that for each i, Pi lies on edge ABi , between Bi and Mi .
For i = 0 the claim is true. Now assume the claim for i and consider the triangle ABi Bi+1 with
centroid Gi .
Since Pi lies between Mi and Bi , we get that Pi Gi lies between Mi Gi and Bi Gi , which are the
medians of this triangle. So Pi+1 lies on ABi+1 , between Mi+1 and Bi+1 . So the claim is proved.
We proved that Pi ’s lie on ABi ’s, so the sequence of points Pi goes around A and therefore does
not pass through a face which doesn’t contain A.
■
17
5. Suppose that ABCD is a parallelogram such that ∠DAC = 90◦ . Let H be the foot of perpendicular from A to DC, also let P be a point along the line AC such that the line P D is tangent
to the circumcircle of the triangle ABD. Prove that ∠P BA = ∠DBH.
Proposed by Iman Maghsoudi
------------------------------------------------------------------
Suppose that AB, AD meet the circumcircle of triangle P DB for the second time at points
X, Y respectively. Let ∠CDB = α and ∠ADB = θ. Therefore, we have ∠ABD = α, and so
∠ADP = α.
Also ∠P DB = ∠P XB = α + θ, and ∠P AX = ∠ACD = ∠DAH. Which implies
△
△
AP
AX
=
,
AH
AD
△
△
AX
AY
=
,
XAD ∼ Y AB =⇒
AB
AD
AP
AY
=⇒
=
.
AH
AB
AP X ∼ ADH =⇒
△
△
Now since ∠HAB = ∠P AY = 90◦ , It can be written that AP Y ∼ AHB.
\ = P[
\
\
=⇒ HBA
YA=P
BD =⇒ P[
BA = DBH.
■
18
Advanced Level
19
Problems
1. Two circles ω1 , ω2 intersect each other at points A, B. Let P Q be a common tangent line of
these two circles with P ∈ ω1 and Q ∈ ω2 . An arbitrary point X lies on ω1 . Line AX intersects
ω2 for the second time at Y . Point Y ′ ̸= Y lies on ω2 such that QY = QY ′ . Line Y ′ B intersects
ω1 for the second time at X ′ . Prove that P X = P X ′
2. In acute triangle ABC, ∠A = 45◦ . Points O, H are the circumcenter and the orthocenter of
ABC, respectively. D is the foot of altitude from B. Point X is the midpoint of arc AH of the
circumcircle of triangle ADH that contains D. Prove that DX = DO.
3. Find all possible values of integer n > 3 such that there is a convex n-gon in which, each
diagonal is the perpendicular bisector of at least one other diagonal.
4. Quadrilateral ABCD is circumscribed around a circle. Diagonals AC, BD are not perpendicular
to each other. The angle bisectors of angles between these diagonals, intersect the segments
AB, BC, CD and DA at points K, L, M and N . Given that KLM N is cyclic, prove that so is
ABCD.
5. ABCD is a cyclic quadrilateral. A circle passing through A, B is tangent to segment CD at
point E. Another circle passing through C, D is tangent to AB at point F . Point G is the
intersection point of AE, DF , and point H is the intersection point of BE, CF . Prove that the
incenters of triangles AGF, BHF, CHE, DGE lie on a circle.
20
Solutions
1. Two circles ω1 , ω2 intersect each other at points A, B. Let P Q be a common tangent line of
these two circles with P ∈ ω1 and Q ∈ ω2 . An arbitrary point X lies on ω1 . Line AX intersects
ω2 for the second time at Y . Point Y ′ ̸= Y lies on ω2 such that QY = QY ′ . Line Y ′ B intersects
ω1 for the second time at X ′ . Prove that P X = P X ′
Proposed by Morteza Saghafian
-----------------------------------------------------------------Solution.
QY = QY ′ implies ∠QY Y ′ = ∠QY ′ Y . Considering circle ω2 , we have ∠QY Y ′ = ∠Y ′ QP . This
means Y Y ′ ∥ P Q.
We also have ∠Y ′ Y A = ∠Y ′ BA and ∠ABX ′ = ∠AXX ′ . This means XX ′ ∥ Y Y ′ ∥ P Q.
Therefore ∠P XX ′ = ∠X ′ P Q = ∠P X ′ X, so P X = P X ′
■
21
2. In acute triangle ABC, ∠A = 45◦ . Points O, H are the circumcenter and the orthocenter of
ABC, respectively. D is the foot of altitude from B. Point X is the midpoint of arc AH of the
circumcircle of triangle ADH that contains D. Prove that DX = DO.
Proposed by Fatemeh Sajadi
-----------------------------------------------------------------Solution.
Since ∠AXH = 90◦ and XA = XH, we conclude that ∠AHX = 45◦ = ∠ADX.
Also ∠BOC = 2∠A = 90◦ , therefore points O, D lie on a circle with diameter BC. This implies
\ = OBC
\ = 45◦ =⇒ ODX
\ = 90◦ .
ODA
But note that
b = 45◦ = 1 AXH.
\ = 90◦ − A
\
ACH
2
This alongside with XA = XH means X is the circumcenter of triangle ACH and so XA = XC.
Thus OX is the perpendicular bisector of AC and so OX ⊥ AC. Now in triangle ODX, the
angle bisector of vertex D is the same as the altitude from D, hence it is an isosceles triangle
with DX = DO.
■
22
3. Find all possible values of integer n > 3 such that there is a convex n-gon in which, each
diagonal is the perpendicular bisector of at least one other diagonal.
Proposed by Mahdi Etesamifard
-----------------------------------------------------------------Solution. Let m be the total number of the perpendicular bisectors of all diagonals in the given
n-gon. The statement of the problem implies that m is not less than the number of diagonals.
But it is clear that the total number of perpendicular bisectors of the diagonals does not exceed
the number of diagonals! Hence, we conclude that each diagonal is the perpendicular bisector
of exactly one other diagonal. Since the perpendicular bisector of a diagonal is a unique line, we
get that for each diagonal d, there is exactly one diagonal d′ such that d′ is the perpendicular
bisector of d.
Consider three adjacent vertices B, A, C of the n-gon, where A lies between B and C. BC is a
diagonal of the n-gon, and only diagonals that contain A have an intersection point with BC.
Specially, the diagonal which is the perpendicular bisector of BC passes through A. Hence
AB = AC. Using this similar idea, it is deduced that all sides of this n-gon have the same
length.
Similar to the previous part, consider four adjacent points of the n-gon, A, B, C, D with the
given order. If n > 4, then AD is a diagonal of the n-gon, and the only diagonals that contain
B or C, have an intersection point with AD. Therefore either B or C lie on the perpendicular
bisector of AD. Without loss of generality, assume that BA = BD. According to the previous
argument, BA = BC = CD. Thus triangle BCD is an equilateral and so ∠BCD = 60◦ . (In
the other case we would have ∠ABC = 60◦ .)
23
This implies that between any two adjacent vertices, there is one that has a 60 degree angle.
Hence there is at least n2 angles of 60◦ in this n-gon.
It is known that the total number of 60 degree angles in an n-gon with n > 3 is at most 2. So
we must have n2 ≤ 2 which means n ≤ 4, a contradiction.
Clearly, any rhombus satisfies the desired property. So the answer is n = 4 .
■
24
4. Quadrilateral ABCD is circumscribed around a circle. Diagonals AC, BD are not perpendicular
to each other. The angle bisectors of angles between these diagonals, intersect the segments
AB, BC, CD and DA at points K, L, M and N . Given that KLM N is cyclic, prove that so is
ABCD.
Proposed by Nikolai Beluhov (Bulgaria)
-----------------------------------------------------------------Solution. Let P be the intersection point of AC, BD. First we claim that KL and M N are not
parallel. Assume the contrary, that KL ∥ M N . Since KLM N is cyclic, we have KN = M L,
and P K = P L, P M = P N . We also have
KP
PL
=
.
PM
PN
Let AP = x, BP = y, CP = z and DP = t. Also let ∠AP B = 2α and ∠BP C = 2θ. We have
xy
zt
KP
KP =
cos α, P M =
cosα =⇒
=
x+y
z+t
PM
1
z
1
x
+ 1t
.
+ y1
Similarly,
PL
=
PN
Since
KP
PM
=
1
x
1
y
+ 1t
.
+ z1
PL
,
PN
with a little of calculation, we shall have
(
)(
)
1
1
1
1
1
1 1
1
1 1 1 1
+
+
=
+
+
=⇒
−
+ + +
= 0.
yz z 2 zt
tx x2 xy
x y
x y z
t
Which means x = z. But note that P K = P L implies
yz
xy
cos α =
cos θ.
x+y
y+z
But θ = 90◦ − α, so we must have α = θ = 45◦ , and so AC ⊥ BD, which is a contradiction.
Therefore the claim is proved. With the similar idea, we can show that KN and LM are not
parallel.
25
By Menelaus’ theorem, KL and M N meet at a point Q on AC such that
BR
= PBPD .
and N K meet at a point R on BD such that RD
AQ
QC
=
AP
PC
and LM
Let the incircle ω of ABCD touch its sides at K ′ , L′ , M ′ , and N ′ . By Brianchon’s theorem,
AL′ , CK ′ , and BD are concurrent. By Ceva’s and Menelaus’ theorems, K ′ , L′ , and Q are
collinear. Analogously, M ′ , N ′ , and Q are collinear and L′ M ′ and N ′ K ′ meet at R.
By Brianchon’s theorem, K ′ M ′ and L′ N ′ meet at P . It follows the diagonals and opposite
sides of both KLM N and K ′ L′ M ′ N ′ intersect at the vertices of △P QR. Therefore, both the
circumcircle of KLM N and ω coincide with the polar circle of △P QR.
Since K is a common point of AB and ω, K ≡ K ′ . Analogously, L ≡ L′ , M ≡ M ′ , and N ≡ N ′ .
Hence the angle bisector KM of AC and BD makes equal angles with AB and CD and ABCD
is cyclic, as needed.
■
26
5. ABCD is a cyclic quadrilateral. A circle passing through A, B is tangent to segment CD at
point E. Another circle passing through C, D is tangent to AB at point F . Point G is the
intersection point of AE, DF , and point H is the intersection point of BE, CF . Prove that the
incenters of triangles AGF, BHF, CHE, DGE lie on a circle.
Proposed by Le Viet An (Vietnam)
-----------------------------------------------------------------Solution.
Let I, J, K, L be the incenters of the triangles AGF, BHF, CHE, DGE respectively. Let ω be
the circumcircle of ABCD. In case of AB ∥ CD, we would conclude that ABCD is an isosceles
trapezoid and it is easy to see that IJKL is also an isosceles trapezoid.
So assume that AB ̸∥ CD and let M be the intersection point of rays BA and CD. Since
ABCD is cyclic, it is obtained that
M A · M B = M D · M C = PM (ω)
Since M E is tangent to ⊙ABE, we get
\
\
M
EA = M
BE.
We also have M E 2 = M A · M B = PM (⊙ABE) and M F 2 = M D · M C = PM (⊙CDF ), which
\
\
implies M E = M F , and so M
EF = M
F E. Therefore,
[ =M
\
\
\
\
\.
AEF
EF − M
EA = M
FE − M
BE = BEF
The latest equation means that EF is the interior angle bisector of ∠AEB. Similarly, F E is
the interior angle bisector of ∠CF D.
27
Note that H, J, K are collinear and ∠F JH = 90◦ +
∠F BH
.
2
Thus
\
\
M
BE
M
EA
F[
JK = 90◦ +
= 90◦ +
2
2
◦
[
[
180 − AEC
AEC
= 90◦ +
= 180◦ −
2
2
(
)
[ + BEC
\
AEB
◦
◦
\
\
= 180 −
= 180 − F EB + BEK
2
\
= 180◦ − F
EK
This results in that EF JK is cyclic. With similar arguments, EF IL is also cyclic.
Since EF is the interior angle bisector of ∠GEH and ∠GF H, it is easy to see that triangles
= HE
= k.
GEF and HEF are congruent. Therefore EG = EH and F G = F H, and so GE
GF
HF
Consider three lines, the exterior angle bisector of vertex G in △GEF , the exterior angle bisector of vertex H in △HEF and the line EF . According to the latest equation, there is two cases:
• These three lines are pairwise parallel. This means EF JK and EF IL are isosceles trapezoids. Hence the segments EF , JK and IL have the same perpendicular bisector and so
IJKL is an isosceles trapezoid.
□
• These three points are concurrent at a point P where
PE
PF
= k. Now we simply have
P J · P K = PP (⊙EF JK) = P E · P F = PP (⊙EF IL) = P I · P L.
□
Which means IJKL is cyclic.
■
28
6th Iranian Geometry
Olympiad
IGO 2019
Contest problems with solutions
6th Iranian Geometry Olympiad
Contest problems with solutions.
This booklet is prepared by Alireza Dadgarnia and Benyamin Ghaseminia.
Copyright c Iranian Geometry Olympiad Secretariat 2018-2019. All rights reserved.
Contents
Elementary Level
3
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Intermediate Level
15
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Advanced Level
29
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Elementary Level
1
Problems
1) There is a table in the shape of a 8 × 5 rectangle with four holes on its
corners. After shooting a ball from points A, B and C on the shown paths,
will the ball fall into any of the holes after 6 reflections? (The ball reflects
with the same angle after contacting the table edges.)
B
4
◦
45
4
45
◦
2
45◦
4
A
45◦
3
45◦
C
45◦ 1
(→ p.5)
2) As shown in the figure, there are two rectangles ABCD and P QRD with
the same area, and with parallel corresponding edges. Let points N, M
and T be the midpoints of segments QR, P C and AB, respectively. Prove
that points N, M and T lie on the same line.
3
4
Elementary Level
P
Q
N
A
D
M
R
T
B
C
(→ p.8)
3) There are n > 2 lines on the plane in general position; Meaning any two
of them meet, but no three are concurrent. All their intersection points
are marked, and then all the lines are removed, but the marked points
are remained. It is not known which marked point belongs to which two
lines. Is it possible to know which line belongs where, and restore them
all?
(→ p.9)
4) Quadrilateral ABCD is given such that
∠DAC = ∠CAB = 60◦ ,
and
AB = BD − AC.
Lines AB and CD intersect each other at point E. Prove that
∠ADB = 2∠BEC.
(→ p.10)
5) For a convex polygon (i.e. all angles less than 180◦ ) call a diagonal bisector if its bisects both area and perimeter of the polygon. What is the
maximum number of bisector diagonals for a convex pentagon?
(→ p.11)
Solutions
1) There is a table in the shape of a 8 × 5 rectangle with four holes on its
corners. After shooting a ball from points A, B and C on the shown paths,
will the ball fall into any of the holes after 6 reflections? (The ball reflects
with the same angle after contacting the table edges.)
B
4
◦
4
45◦
45
2
45◦
4
A
45◦
3
45◦
C
45◦ 1
Proposed by Hirad Alipanah
------------------------------------------------Solution. It’s easy to track the trajectory of the ball.
Point A:
5
6
Elementary Level
2
2
1
2
4
45◦
A
A
45
◦
3
5
1
3
3
2
6
1
1
1
1
(a) The ball goes through a hole with one (b) The ball doesn’t go through a hole afreflection.
ter six reflections.
Point B:
B
2
4
3
B
4
45◦
2
45◦
3
2
2
4
4
1
2
5
4
1
1
2
1
1
5
1
4
(a) The ball goes through a hole with five (b) The ball goes through a hole with five
reflections.
reflections.
4
2
2
4
1
2
2
5
2
3
45◦
2
3
1
1
6
4
C
1
4
3
1
C
45 1
1
◦
(a) The ball goes through a hole with six (b) The ball goes through a hole with four
refelctions.
reflections.
There are two point of views to the problem:
Solutions
7
(a) Looking for the trajectories where the ball goes through a hole with
at most 6 reflections:
In this case, all cases except A(b) are desired.
(b) Looking for the trajectories where the ball goes through a hole with
exactly 6 reflections:
In this case, C(a) is the only answer to the problem.
8
Elementary Level
2) As shown in the figure, there are two rectangles ABCD and P QRD with
the same area, and with parallel corresponding edges. Let points N, M
and T be the midpoints of segments QR, P C and AB, respectively. Prove
that points N, M and T lie on the same line.
P
Q
N
D
A
R
M
T
B
C
Proposed by Morteza Saghafian
------------------------------------------------Solution. Let L be the intersection point of P Q and CR, and let K be
the intersection point of BC and AP .
Q
P
A
D
N
M
R
T
K
B
L
C
Since QR k P C, it is deduced that L, N and M are collinear; Similarly,
K, T and M are collinear. Therefore it suffices to prove that P LCK is
a parallelogram to deduce that K, M and L are collinear, and get the
desired result of the problem. Since P L k CK, it suffices to show that
P K k CL, or P A k CR. Since the areas of the two rectangles are equal,
it is implied that
P D · DR = AD · CD =⇒
AD
PD
=
.
CD
DR
Which implies AP k CR, and the proof is complete.
Solutions
9
3) There are n > 2 lines on the plane in general position; Meaning any two
of them meet, but no three are concurrent. All their intersection points
are marked, and then all the lines are removed, but the marked points are
remained. It is not known which marked point belongs to which two lines.
Is it possible to know which line belongs where, and restore them all?
Proposed by Boris Frenkin - Russia
------------------------------------------------Answer. Yes, it is.
Solution. Draw the lines which each of them contains n−1 marked points,
at least. All the original lines are among these lines. Conversely, let some
line ` contains some n − 1 marked points. They are points of meet of some
pairs of the original lines (`1 , `2 ) , (`3 , `4 ) , . . . , (`2n−3 , `2n−2 ). Since n > 2,
we have 2n − 2 > n, so `i coincides with `j for some 1 ≤ i < j ≤ 2n − 2.
Then these lines belong to distinct pairs in the above list, and the two
corresponding marked points belong to `i = `j . But then also ` = `i , and
we are done.
10
Elementary Level
4) Quadrilateral ABCD is given such that
∠DAC = ∠CAB = 60◦ ,
and
AB = BD − AC.
Lines AB and CD intersect each other at point E. Prove that
∠ADB = 2∠BEC.
Proposed by Iman Maghsoudi
------------------------------------------------Solution. Consider point F on ray BA such that AF = AC.
F
A
B
D
E
C
Knowing that AB = BD − AC, it is implied that BF = BD. Therefore

AF = AC

AD = AD
=⇒ 4F AD ∼
(1)
= 4CAD.

∠F AD = ∠CAD = 60◦
Note that
(1)
∠BEC = ∠F AD − ∠ADC = 60◦ − ∠ADF.
(2)
On the other hand
∠ADB = ∠F DB − ∠ADF = ∠AF D − ∠ADF
= (120◦ − ∠ADF ) − ∠ADF
= 120◦ − 2∠ADF
(2)
= 2∠BEC.
So the claim of the problem is proved.
Solutions
11
5) For a convex polygon (i.e. all angles less than 180◦ ) call a diagonal bisector if its bisects both area and perimeter of the polygon. What is the
maximum number of bisector diagonals for a convex pentagon?
Proposed by Morteza Saghafian
------------------------------------------------Answer. The maximum number of bisector diagonals is 2.
Solution. Note that for each vertex, there is at most one bisector diagonal
that passes through it; Therefore there are at most 2 bisector diagonals in
the pentagon. The following figure shows an example where the pentagon
has two bisector diagonals.
C
B
A
D
E
12
Elementary Level
Intermediate Level
13
Problems
1) Two circles ω1 and ω2 with centers O1 and O2 respectively intersect each
other at points A and B, and point O1 lies on ω2 . Let P be an arbitrary
point lying on ω1 . Lines BP, AP and O1 O2 cut ω2 for the second time
at points X, Y and C, respectively. Prove that quadrilateral XP Y C is a
parallelogram.
(→ p.17)
2) Find all quadrilaterals ABCD such that all four triangles DAB, CDA,
BCD and ABC are similar to one-another.
(→ p.19)
3) Three circles ω1 , ω2 and ω3 pass through one common point, say P . The
tangent line to ω1 at P intersects ω2 and ω3 for the second time at points
P1,2 and P1,3 , respectively. Points P2,1 , P2,3 , P3,1 and P3,2 are similarly defined. Prove that the perpendicular bisector of segments P1,2 P1,3 , P2,1 P2,3
and P3,1 P3,2 are concurrent.
(→ p.20)
4) Let ABCD be a parallelogram and let K be a point on line AD such
that BK = AB. Suppose that P is an arbitrary point on AB, and the
perpendicular bisector of P C intersects the circumcircle of triangle AP D
at points X, Y . Prove that the circumcircle of triangle ABK passes
through the orthocenter of triangle AXY .
(→ p.23)
◦
5) Let ABC be a triangle with ∠A = 60 . Points E and F are the foot of
angle bisectors of vertices B and C respectively. Points P and Q are considered such that quadrilaterals BF P E and CEQF are parallelograms.
Prove that ∠P AQ > 150◦ . (Consider the angle P AQ that does not contain side AB of the triangle.)
(→ p.25)
15
16
Intermediate Level
Solutions
1) Two circles ω1 and ω2 with centers O1 and O2 respectively intersect each
other at points A and B, and point O1 lies on ω2 . Let P be an arbitrary
point lying on ω1 . Lines BP, AP and O1 O2 cut ω2 for the second time
at points X, Y and C, respectively. Prove that quadrilateral XP Y C is a
parallelogram.
Proposed by Iman Maghsoudi
------------------------------------------------Solution.
ω2
A
ω1
X
O1
O2
C
P
B
Y
One can obtain
> >
AO1 B
XCY
∠AP B =
+
.
(1)
2
2
Note that ∠AO2 O1 = ∠BO2 O1 , therefore ∠AO1 O2 = ∠BO1 O2 . Now
17
18
Intermediate Level
since O1 is the circumcenter of triangle AP B, it is deduced that
∠AP B = 180◦ −
>
>
∠AO1 B
AXC
ABC
= 180◦ −
=
.
2
2
2
(2)
Hence, according to equations 1 and 2 it is obtained that
> >
>
> >
AO1 B
XCY
ABC
+
=
=⇒ XCY = BY C,
2
2
2
thus BX k CY , similarly AY k XC. This imlplies that XP Y C is a
parallelogram and the claim of the problem is proved.
Solutions
19
2) Find all quadrilaterals ABCD such that all four triangles DAB, CDA,
BCD and ABC are similar to one-another.
Proposed by Morteza Saghafian
------------------------------------------------Answer. All rectangles.
Solution. First assume that ABCD is a concave quadrilateral. Without
loss of generality one can assume ∠D > 180◦ , in other words D lies inside
of triangle ABC. Again without loss of generality one can assume that
∠ABC is the maximum angle in triangle ABC. Therefore
∠ADC = ∠ABC + ∠BAD + ∠BCD > ∠ABC.
Thus ∠ADC is greater than all the angles of triangle ABC, so triangles
ABC and ADC cannot be similar. So it is concluded that ABCD must
be convex.
Now let ABCD be a convex quadrilateral. Without loss of generality one
can assume that the ∠B is the maximum angle in the quadrilateral. It
can be written that
∠ABC > ∠DBC,
∠ABC ≥ ∠ADC ≥ ∠BCD.
Since triangles ABC and BCD are similar, it is implied that ∠ABC =
∠BCD and similarly, all the angles of ABCD are equal; Meaning ABCD
must be a rectangle. It is easy to see that indeed, all rectangles satisfy
the conditions of the problem.
20
Intermediate Level
3) Three circles ω1 , ω2 and ω3 pass through one common point, say P . The
tangent line to ω1 at P intersects ω2 and ω3 for the second time at points
P1,2 and P1,3 , respectively. Points P2,1 , P2,3 , P3,1 and P3,2 are similarly defined. Prove that the perpendicular bisector of segments P1,2 P1,3 , P2,1 P2,3
and P3,1 P3,2 are concurrent.
Proposed by Mahdi Etesamifard
------------------------------------------------Solution.
P1,2
P2,1
P3,1
P
P3,2
P1,3
P2,3
First assume that no two of the lines `1 ≡ P2,1 P3,1 , `2 ≡ P1,2 P3,2 and
`3 ≡ P1,3 P2,3 are parallel; Consider triangle XY Z made by intersecting
these lines, where
X ≡ `2 ∩ `3 ,
Y ≡ `1 ∩ `3 ,
Z ≡ `2 ∩ `1 .
Note that
∠P3,2 P1,2 P = ∠P3,2 P P2,3 = ∠P P1,3 P2,3 ,
meaning XP1,2 = XP1,3 . Similarly, it is implied that Y P2,1 = Y P2,3 and
ZP3,1 = ZP3,2 . Therefore, the angle bisectors of angles Y XZ, XY Z and
Solutions
21
Y ZX are the same as the perpendicular bisectors of segments P1,2 P1,3 ,
P2,1 P2,3 and P3,1 P3,2 ; Thus, these three perpendicular bisectors are concurrent at the incenter of triangle XY Z, resulting in the claim of the
problem.
Now assume that at least two of the lines `1 = P2,1 P3,1 , `2 = P1,2 P3,2 and
`3 = P1,3 P2,3 are parallel; Without loss of generality assume that `1 and
`2 are parallel. Similar to the previous case,
∠P1,2 P3,2 P = ∠P1,2 P P2,1 = ∠P2,1 P3,1 P.
But since `1 k `2 , it is also true that
∠P1,2 P3,2 P + ∠P2,1 P3,1 P = 180◦ ,
Hence ∠P1,2 P3,2 P = ∠P2,1 P3,1 P = 90◦ . This equation immediately implies `3 ∦ `2 , because otherwise it would be deduced that `3 ⊥ P1,3 P1,2
and `2 ⊥ P1,3 P1,2 , resulting in P1,3 P1,2 k P3,1 P3,2 ; Which is clearly not
possible. Now consider trapezoid XY P2,1 P1,2 . The problem is now equivalent to show that the angle bisector of ∠X, angle bisector of ∠Y and the
perpendicular bisector of P3,1 P3,2 concur. Note that `1 and `2 are parallel
to the perpendicular bisector of P3,1 P3,2 , and in fact, the perpendicular
bisector of P3,1 P3,2 connects the midpoints of XY and P3,1 P3,2 . Now the
claim of the problem is as simple as follows.
P1,2
P2,1
K
0
P3,1
P3,1
0
P3,2
P3,2
Y
K0
X
Claim. In trapezoid XY P2,1 P1,2 , the angle bisector of ∠X, the angle
bisector of ∠Y , and the mid-line of the trapezoid are concurrent.
Proof. Let K be the intersection of the angle bisector of ∠X and the angle
0
0
bisector of ∠Y . Let P3,2
, P3,1
and K 0 be the foot of perpendicular lines
22
Intermediate Level
from K to lines P1,2 X, P2,1 Y and XY , respectively. Since K lies on the
0
angle bisector of ∠X, it is deduced that KP3,2
= KK 0 , and similarly since
0
0
0
K lies on the angle bisector of ∠Y , KP3,1 = KK 0 ; Thus KP3,1
= KP3,2
,
meaning K lies on the mid-line of trapezoid XY P2,1 P1,2 .
This result leads to the conclusion of the problem.
Solutions
23
4) Let ABCD be a parallelogram and let K be a point on line AD such
that BK = AB. Suppose that P is an arbitrary point on AB, and the
perpendicular bisector of P C intersects the circumcircle of triangle AP D
at points X, Y . Prove that the circumcircle of triangle ABK passes
through the orthocenter of triangle AXY .
Proposed by Iman Maghsoudi
------------------------------------------------Solution. Let AN be the altitude of triangle AXY . Suppose that the
circumcircle of triangle ABK intersects AN at the point R. It is enough
to show that R is the orthocenter of triangle AXY .
K
P
A
B
R
X
N
D
T
C
Q
Y
S
Suppose that P C and AN intersects the circumcircle of triangle AP D for
the second time at point T and S, respectively and AN intersects CD at
Q. We know that
∠BRS = ∠AKB = ∠KAB = ∠P T D =⇒ ∠BRS = ∠P T D
(1)
Notice that XY is perpendicular to AN and P C, so AN k P C. Also
AP k CQ so AP CQ is parallelogram. Now we have

∠BAR = ∠T CD

(1) =⇒ ∠ARB = ∠CT D
=⇒ 4ARB ∼
= 4CT D =⇒ CT = AR.

AB = CD
24
Intermediate Level
So P T QR is also a parallelogram. AP T S is an isosceles trapezoid so
CQ = AP = T S.
Therefore T QSC is also an isosceles trapezoid. Finally
CS = T Q = P R =⇒ CSRP is an isosceles trapezoid.
Therefore, S and R are symmetric with respect to XY . So R should be
the orthocenter of triangle AXY .
Solutions
25
5) Let ABC be a triangle with ∠A = 60◦ . Points E and F are the foot of
angle bisectors of vertices B and C respectively. Points P and Q are considered such that quadrilaterals BF P E and CEQF are parallelograms.
Prove that ∠P AQ > 150◦ . (Consider the angle P AQ that does not contain side AB of the triangle.)
Proposed by Alireza Dadgarnia
------------------------------------------------Solution. Let I and be the intersection point of lines BE and CF , and
let R be the intersection point of lines QE and P F . It is easy to see that
∠BIC = 120◦ . Thus AEIF is a cyclic quadrilateral and so
CE · CA = CI · CF
(1)
Also ∠P RQ = ∠BIC = 120◦ , therefore it suffices to show that at least
on of the angles ∠AP R or ∠AQR is greater than or equal to 30◦ .
K
Q
A
P
R
E
F
I
B
C
Assume the contrary, meaning both of these angles are less than 30◦ .
Hence there exists a point K on the extension of ray CA such that
∠KQE = 30◦ . Since ∠IAC = 30◦ ∠ACI = ∠KEQ, it is deduced that
4AIC ∼ 4QKE. This implies
CI
KE
AE
CF · CI (1)
=
>
=⇒ AE <
= CE.
CA
QE
CF
CA
26
Intermediate Level
Similarly, it is obtained that AF < BF . On the other hand at least
on of the angles ∠ABC or ∠ACB are not less than 60◦ . Without loss
of generality one can assume that ∠ABC ≥ 60◦ thus AC ≥ BC and
according to angle bisector theorem it is obtained that AF ≥ BF , which
is a contradiction. Hence the claim of the problem.
Advanced Level
27
Problems
1) Circles ω1 and ω2 intersect each other at points A and B. Point C lies on
the tangent line from A to ω1 such that ∠ABC = 90◦ . Arbitrary line `
passes through C and cuts ω2 at points P and Q. Lines AP and AQ cut
ω1 for the second time at points X and Z respectively. Let Y be the foot
of altitude from A to `. Prove that points X, Y and Z are collinear.
(→ p.31)
2) Is it true that in any convex n-gon with n > 3, there exists a vertex and a
diagonal passing through this vertex such that the angles of this diagonal
with both sides adjacent to this vertex are acute?
(→ p.33)
3) Circles ω1 and ω2 have centres O1 and O2 , respectively. These two circles
intersect at points X and Y . AB is common tangent line of these two
circles such that A lies on ω1 and B lies on ω2 . Let tangents to ω1 and ω2
at X intersect O1 O2 at points K and L, respectively. Suppose that line
BL intersects ω2 for the second time at M and line AK intersects ω1 for
the second time at N . Prove that lines AM, BN and O1 O2 concur.
(→ p.34)
4) Given an acute non-isosceles triangle ABC with circumcircle Γ. M is the
>
midpoint of segment BC and N is the midpoint of BC of Γ (the one that
doesn’t contain A). X and Y are points on Γ such that BX k CY k AM .
Assume there exists point Z on segment BC such that circumcircle of
triangle XY Z is tangent to BC. Let ω be the circumcircle of triangle
ZM N . Line AM meets ω for the second time at P . Let K be a point
on ω such that KN k AM , ωb be a circle that passes through B, X and
tangents to BC and ωc be a circle that passes through C, Y and tangents
to BC. Prove that circle with center K and radius KP is tangent to 3
circles ωb , ωc and Γ.
29
30
Advanced Level
(→ p.35)
5) Let points A, B and C lie on the parabola ∆ such that the point H,
orthocenter of triangle ABC, coincides with the focus of parabola ∆.
Prove that by changing the position of points A, B and C on ∆ so that the
orthocenter remain at H, inradius of triangle ABC remains unchanged.
(→ p.38)
Solutions
1) Circles ω1 and ω2 intersect each other at points A and B. Point C lies on
the tangent line from A to ω1 such that ∠ABC = 90◦ . Arbitrary line `
passes through C and cuts ω2 at points P and Q. Lines AP and AQ cut
ω1 for the second time at points X and Z respectively. Let Y be the foot
of altitude from A to `. Prove that points X, Y and Z are collinear.
Proposed by Iman Maghsoudi
------------------------------------------------Solution.
ω1
A
ω2
Z
P
Y
B
Q
C
X
Since ∠AY C = ∠ABC = 90◦ , it is concluded that AY BC is a cyclic
quadrilateral. Hence
∠BY C = ∠BAC = ∠BXA = ∠BXP.
31
32
Advanced Level
So P Y BX, and similarly QBY Z are cyclic quadrilaterals. Therefore it is
implied that
∠BY X = ∠BP X = ∠AQB = ∠ZQB = 180◦ − ∠ZY B.
Meaning points X, Y and Z are collinear.
Solutions
33
2) Is it true that in any convex n-gon with n > 3, there exists a vertex and a
diagonal passing through this vertex such that the angles of this diagonal
with both sides adjacent to this vertex are acute?
Proposed by Boris Frenkin - Russia
------------------------------------------------Answer. Yes, it is true.
Solution. Suppose the answer is no. Given a convex n-gon (n > 3), consider its longest diagonal AD (if the longest diagonal is not unique, choose
an arbitrary on among them). Let B and C be the vertices neighboring
to A. Without loss of generality assume that ∠BAD ≥ 90◦ . This means
BD > AD, so BD is not a diagonal and hence is a side of the n-gon.
Furthermore, ∠ADB < 90◦ . Let C 0 be the vertex neighboring to D and
distinct from B. Then ∠ADC 0 ≥ 90◦ . Similarly, AC 0 > AD, so AC 0 is
a side, C 0 ≡ C and n = 4. Angles BAC and BDC are obtuse, so BC
is longer than AC and BD, hence BC > AD and AD is not the longest
diagonal, a contradiction. Hence the claim.
34
Advanced Level
3) Circles ω1 and ω2 have centres O1 and O2 , respectively. These two circles
intersect at points X and Y . AB is common tangent line of these two
circles such that A lies on ω1 and B lies on ω2 . Let tangents to ω1 and ω2
at X intersect O1 O2 at points K and L, respectively. Suppose that line
BL intersects ω2 for the second time at M and line AK intersects ω1 for
the second time at N . Prove that lines AM, BN and O1 O2 concur.
Proposed by Dominik Burek - Poland
------------------------------------------------Solution.
Let P be the midpoint of AB; Since P has the same power with respect
to both circles, it lies on the radical axis of them, which is line XY .
A0
X
L
B0
O1
O2
M
K
N
Y
A
P
B
According to the symmetry, KY is tangent to ω1 , therefore XY is the
polar of K with respect to ω1 . Since P lies on XY , the polar of P passes
through K, and similarly, it also passes through A; Meaning AK is the
polar of P with respect to ω1 and P N is tangent to ω1 . Similarly, P M is
tangent to ω2 ; Thus points A, B, M and N lie on a circle with center P
and ∠AM B = ∠AN P = 90◦ . Let A0 be the antipode of A in circle ω1 ,
and let B 0 be the antipode of B. Line BN passes through A0 and line
AM passes through B 0 . Note that AA0 B 0 B is a trapezoid and O1 and O2
are the midpoints of its bases; Hence A0 B, B 0 A and O1 O2 are concurrent,
resulting in the claim of the problem.
Solutions
35
4) Given an acute non-isosceles triangle ABC with circumcircle Γ. M is the
>
midpoint of segment BC and N is the midpoint of BC of Γ (the one that
doesn’t contain A). X and Y are points on Γ such that BX k CY k AM .
Assume there exists point Z on segment BC such that circumcircle of
triangle XY Z is tangent to BC. Let ω be the circumcircle of triangle
ZM N . Line AM meets ω for the second time at P . Let K be a point
on ω such that KN k AM , ωb be a circle that passes through B, X and
tangents to BC and ωc be a circle that passes through C, Y and tangents
to BC. Prove that circle with center K and radius KP is tangent to 3
circles ωb , ωc and Γ.
Proposed by Tran Quan - Vietnam
------------------------------------------------Solution. Let I, J and O be the centers of circles ωb , ωc and Γ, respectively. It’s easy to see that I, J and O are collinear. Let S be the
intersection of IJ and BC. Since X and Y are symmetrical to B and C
with respect to IJ, it is implied that S, X and Y are collinear. Let T
>
be a point on BC of Γ (the one that doesn’t contain A) such that ST is
tangent to Γ. T Z meets Γ again at Q.
Q
Y
A
Γ
J
R
O
X
D
I
S
Z
K
Ω
M
B
C
P
T
ω
N
L
36
Advanced Level
Since
ST 2 = SX · SY = SZ 2 ,
ZT is the interior angle bisector of ∠BT C, it is concluded that Q is the
>
midpoint of BAC of Γ. This leads to ZT ⊥ N T , resulted in T lies on ω.
Let R be the intersection of AM and T Q, and Ω be the circumcircle of
triangle RT P . Since
∠RP T = ∠M P T = ∠SZT = ∠ST R,
ST is tangent to Ω. Therefore Ω is tangent to Γ at T . We will prove that
K is the center of Ω. Now


SOM T is cyclic =⇒ ∠OSM = ∠OT M
OS ⊥ AM and M S ⊥ OM =⇒ ∠OM A = ∠OSM = ∠OT M


∠OT Q = ∠OQT
Which implies
∠M T R = ∠OT Q + ∠OT M = ∠OQT + ∠OM A = ∠M RT,
and thus M R = M T . Let L be the intersection of AM and T N . Since
4RT L is a right triangle at T , it is concluded that M is the midpoint of
RL.
KN k M L =⇒ ∠M T N = ∠M LT = ∠KN T
=⇒ KM k T N
=⇒ KM ⊥ RT
Thus, M K is the perpendicular bisector of RT . Note that
∠ZP R = ∠ZT M = ∠ZRP =⇒ ZR = ZP.
Since ZN is the diameter of ω it is implied that
ZK ⊥ KN =⇒ ZK ⊥ RP.
Therefore ZK is the perpendicular bisector of RP . Hence K is the center
of Ω.
We will prove Ω is tangent to ωb and ωc . Let D be the intersection of
T N and BC, and denote (S, SZ) to be circle with center S and radius
SZ. Since T Z is the interior angle bisector of ∠BT C, we have T D is the
Solutions
37
exterior angle bisector of the same angle. This leads to (DZ, BC) = −1.
Since M is the midpoint of BC, we have
M B 2 = M Z · M D,
which implies M lies on the radical axis of ωb and (S, SZ). Combining
with M A ⊥ SI, we have M A is the radical axis of ωb and (S, SZ). Thus,
the powers of point R with respect to ωb and (S, SZ) are equal. Invert
about the circle centered at R with radius
√
r = RZ · RT ,
which inverts Ω 7→ ZM ≡ BC and ωb 7→ ωb . Since circle ωb is tangent to
BC, we have Ω is tangent to ωb .
Analogously, we have Ω is tangent to ωc . This completes the proof. 38
Advanced Level
5) Let points A, B and C lie on the parabola ∆ such that the point H,
orthocenter of triangle ABC, coincides with the focus of parabola ∆.
Prove that by changing the position of points A, B and C on ∆ so that the
orthocenter remain at H, inradius of triangle ABC remains unchanged.
Proposed by Mahdi Etesamifard
------------------------------------------------Solution. Since H coincides with the focus of parabola ∆, the circles
wA = (A, AH), wB = (B, BH) and wC = (C, CH) are tangent to line `,
the directrix of ∆.
C
∆
ωC
Directrix
A
ωA
Axis of Symmetry
H
`
B
ωB
Now consider triangle ABC.
A
B0
C0
H
B
A0
C
Solutions
39
It is well-known that
HA · HA0 = HB · HB 0 = HC · HC 0 = t.
Also
HA = 2R cos A
HA0 = 2R cos B cos C
=⇒ t = 4R2 cos A cos B cos C.
(1)
Inversion with center H and inversion radius −2t, inverts the three circles
wA , wB and wC to lines BC, AC and AB respectively. In this inversion,
line ` inverts to incircle of triangle ABC. Therefore IH ⊥ `, thus point I
lies on axis of symmetry of ∆. Also point H lies on the incircle of triangle
ABC. Hence HI = r.
C
∆
Directrix
A H
I
F
Axis of Symmetry
K
B
As a result, if orthocenter of ABC lies on its incircle; Also
HI 2 = 2r2 − 4R2 cos A cos B cos C
=⇒ r2 = 2r2 − 4R2 cos A cos B cos C
=⇒ r2 = 4R2 cos A cos B cos C
40
Advanced Level
According to (1), it is implied that r2 = t = HA · HA0 . In inversion,
points K and F are invert points, thus
−−→ −−→
HK · HF = −2t = −2r2 =⇒ HK = r.
Which gives the result that inradius of triangle ABC is constant.
7th Iranian Geometry
Olympiad
IGO 2020
Contest problems with solutions
7th Iranian Geometry Olympiad
Contest problems with solutions.
This booklet is prepared by Sina Qasemi Nezhad, Alireza Dadgarnia, Hesam
Rajabzadeh, Siavash Rahimi, Mahdi Etesamifard and Morteza Saghafian.
With special thanks to Matin Yousefi and Alireza Danaei.
Copyright ©Iranian Geometry Olympiad Secretariat 2019-2020. All rights
reserved.
Contents
Elementary Level
3
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Intermediate Level
15
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Advanced Level
29
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Elementary Level
1
Problems
Problem 1. By a fold of a polygon-shaped paper, we mean drawing a
segment on the paper and folding the paper along that. Suppose that a paper
with the following figure is given. We cut the paper along the boundary of
the shaded region to get a polygon-shaped paper.
Start with this shaded polygon and make a rectangle-shaped paper from it
with at most 5 number of folds. Describe your solution by introducing the
folding lines and drawing the shape after each fold on your solution sheet.
(Note that the folding lines do not have to coincide with the grid lines of the
shape.)
(→ p.5)
Problem 2. A parallelogram ABCD is given (AB 6= BC). Points E and
G are chosen on the line CD such that AC is the angle bisector of both
angles ∠EAD and ∠BAG. The line BC intersects AE and AG at F and H,
respectively. Prove that the line F G passes through the midpoint of HE.
(→ p.8)
Problem 3. According to the figure, three equilateral triangles with side
3
4
Elementary Level
lengths a, b, c have one common vertex and do not have any other common
point. The lengths x, y and z are defined as in the figure. Prove that
3(x + y + z) > 2(a + b + c).
x
b
c
y
z
a
(→ p.9)
Problem 4. Let P be an arbitrary point in the interior of triangle ABC.
Lines BP and CP intersect AC and AB at E and F , respectively. Let K
and L be the midpoints of the segments BF and CE, respectively. Let the
lines through L and K parallel to CF and BE intersect BC at S and T ,
respectively; moreover, denote by M and N the reflection of S and T over
the points L and K, respectively. Prove that as P moves in the interior of
triangle ABC, line M N passes through a fixed point.
(→ p.10)
Problem 5. We say two vertices of a simple polygon are visible from each
other if either they are adjacent, or the segment joining them is completely
inside the polygon (except two endpoints that lie on the boundary). Find all
positive integers n such that there exists a simple polygon with n vertices in
which every vertex is visible from exactly 4 other vertices.
(A simple polygon is a polygon without hole that does not intersect itself.)
(→ p.11)
Solutions
Problem 1. By a fold of a polygon-shaped paper, we mean drawing a
segment on the paper and folding the paper along that. Suppose that a paper
with the following figure is given. We cut the paper along the boundary of
the shaded region to get a polygon-shaped paper.
Start with this shaded polygon and make a rectangle-shaped paper from it
with at most 5 number of folds. Describe your solution by introducing the
folding lines and drawing the shape after each fold on your solution sheet.
(Note that the folding lines do not have to coincide with the grid lines of the
shape.)
Proposed by Mahdi Etesamifard
--------------------------------------------------Solution. There are different ways of folding to get a rectangle. For instance,
a solution can be given with only 4 number of folds, as following
First fold:
5
6
Second fold:
Third fold:
Fourth fold:
Elementary Level
Solutions
7
Comment. that by moving the folding lines slightly in 3rd and 4th folds (to
down and up respectively), all the folding segments will be internal.
8
Elementary Level
Problem 2. A parallelogram ABCD is given (AB 6= BC). Points E and
G are chosen on the line CD such that AC is the angle bisector of both
angles ∠EAD and ∠BAG. The line BC intersects AE and AG at F and H,
respectively. Prove that the line F G passes through the midpoint of HE.
Proposed by Mahdi Etesamifard
--------------------------------------------------Solution. Since AD and BC are parallel, we deduce that ∠F CA = ∠DAC =
∠F AC. So F A = F C. Similarly, GA = GC. So triangles 4GAF and
4GCF have a common side and two equal sides and are congruent. Resulting ∠GAF = ∠GCF which leads to ∠HAF = ∠ECF and ∠AF H = ∠CF E.
Therefore triangles 4AF H and 4CF E are congruent as well and we get
F E = F H. Similarly, GE = GH. So both points F and G lie on perpendicular bisector of segment HE. Resulting that F G is the perpendicular bisector
of segment HE.
Solutions
9
Problem 3. According to the figure, three equilateral triangles with side
lengths a, b, c have one common vertex and do not have any other common
point. The lengths x, y and z are defined as in the figure. Prove that
3(x + y + z) > 2(a + b + c).
x
b
c
y
z
a
Proposed by Mahdi Etesamifard
--------------------------------------------------Solution. Consider the three white triangles in the picture, rotating each
triangle 60◦ degrees, clock-wise, will make a side of it coincide with another
side of another triangle. So we can rotate one of them and glue it to the next,
then rotating the glued figure a broken path will be formed between two
points with distances like 2a which has length x + y + z. Thus x + y + z > 2a
summing up all three possible inequalities proves the desired.
10
Elementary Level
Problem 4. Let P be an arbitrary point in the interior of triangle ABC.
Lines BP and CP intersect AC and AB at E and F , respectively. Let K
and L be the midpoints of the segments BF and CE, respectively. Let the
lines through L and K parallel to CF and BE intersect BC at S and T ,
respectively; moreover, denote by M and N the reflection of S and T over
the points L and K, respectively. Prove that as P moves in the interior of
triangle ABC, line M N passes through a fixed point.
Proposed by Ali Zamani
--------------------------------------------------Solution. Since in quadrilateral EM CS, diagonals bisect each other, this
quadrilateral is a parallelogram. So, EM k BC. Let X be the intersection of
EM and CF . Note that M L k CX and L is the midpoint of CE, resulting
that M is the midpoint of EX as well. Since EX k BC, using parallel lines,
one can find that M P passes through the midpoint of BC. Similarly, N P
passes through the midpoint of BC. Hence proved.
Solutions
11
Problem 5. We say two vertices of a simple polygon are visible from each
other if either they are adjacent, or the segment joining them is completely
inside the polygon (except two endpoints that lie on the boundary). Find all
positive integers n such that there exists a simple polygon with n vertices in
which every vertex is visible from exactly 4 other vertices.
(A simple polygon is a polygon without hole that does not intersect itself.)
Proposed by Morteza Saghafian
--------------------------------------------------Solution.
First we prove there is no such polygon for n > 6. Let A1 , A2 , . . . , An be the
vertices.
Lemma 1. Let Ai be visible from Ai−1 , Aj , Ak , Ai+1 in clockwise order (note
that the first and the last one are the edge-neighbors). Then Ai−1 , Aj can see
each other, Aj , Ak can see each other and Ak , Ai+1 can see each other.
Proof. One can consider the triangulation of the three parts of polygon separated by Ai Aj and Ai Ak .
Lemma 2. Using the same naming as Lemma 1, Aj Ak is a side.
Proof. Assume that Aj Ak is an internal diagonal. By Lemma 1, Aj can see
Aj−1 . But Aj Ai and Aj Ak are internal diagonals. So Aj Ai−1 is a side. So
there is only one vertex between Ai , Aj on the perimeter of polygon. Similarly,
there is only one vertex between Aj , Ak and only one vertex between Ak , Ai
on the perimeter of polygon. This contradicts n > 6. So Aj Ak is a side and
k = j − 1.
Now let i be such that Ai−1 , Ai+1 are visible from each other. We know that
such i exists, for instance you cam take an ear triangle in the triangulation of
the polygon. By Lemma 2, Ai−1 can see Ai+2 , Ai+1 can see Ai−2 and Ai−2
can see Ai+2 . So we found the four vertices visible from Ai−1 , Ai+1 . If Ai
can see a vertex, then it is visible by either Ai−1 or Ai+1 (by Lemma 1). So
Ai should see Ai−2 , Ai+2 and this means Ai−2 Ai+2 is a side (by Lemma 2).
Any convex pentagon is an example.
The only remaining case is n = 6 which means in Lemma 2 there are vertices
Ai , Aj , Ak such that Ai Aj , Aj Ak , Ak Ai are internal diagonals. Let them be
A2 , A4 , A6 in the hexagon. So A3 is not visible from A6 , meaning that one
of the angles A2 , A4 is larger that 180◦ . But then A3 cannot see either A1 or
A5 . Which contradicts the fact that A3 is visible from 4 other vertices. So
n = 6 is also not possible and the only possible n is 5.
12
Elementary Level
Intermediate Level
13
Problems
Problem 1. A trapezoid ABCD is given where AB and CD are parallel.
Let M be the midpoint of the segment AB. Point N is located on the segment
1
1
CD such that ∠ADN = ∠M N C and ∠BCN = ∠M N D. Prove that N
2
2
is the midpoint of the segment CD.
(→ p.17)
Problem 2. Let ABC be an isosceles triangle (AB = AC) with its circumcenter O. Point N is the midpoint of the segment BC and point M is the
reflection of the point N with respect to the side AC. Suppose that T is a
1
point so that AN BT is a rectangle. Prove that ∠OM T = ∠BAC.
2
(→ p.18)
Problem 3. In acute-angled triangle ABC (AC > AB), point H is the
orthocenter and point M is the midpoint of the segment BC . The median
AM intersects the circumcircle of triangle ABC at X. The line CH intersects
the perpendicular bisector of BC at E and the circumcircle of the triangle
ABC again at F . Point J lies on circle ω, passing through X, E, and F , such
that BCHJ is a trapezoid (CB k HJ). Prove that JB and EM meet on ω.
(→ p.19)
Problem 4. Triangle ABC is given. An arbitrary circle with center J,
passing through B and C, intersects the sides AC and AB at E and F ,
respectively. Let X be a point such that triangle F XB is similar to triangle
EJC (with the same order) and the points X and C lie on the same side of
the line AB. Similarly, let Y be a point such that triangle EY C is similar to
triangle F JB (with the same order) and the points Y and B lie on the same
side of the line AC. Prove that the line XY passes through the orthocenter
of the triangle ABC.
15
16
Intermediate Level
(→ p.21)
Problem 5. Find all numbers n ≥ 4 such that there exists a convex polyhedron with exactly n faces, whose all faces are right-angled triangles.
(Note that the angle between any pair of adjacent faces in a convex polyhedron is less than 180◦ .)
(→ p.23)
Solutions
Problem 1. A trapezoid ABCD is given where AB and CD are parallel.
Let M be the midpoint of the segment AB. Point N is located on the segment
1
1
CD such that ∠ADN = ∠M N C and ∠BCN = ∠M N D. Prove that N
2
2
is the midpoint of the segment CD.
Proposed by Alireza Dadgarnia
--------------------------------------------------Solution. We have
1
∠BCN + ∠ADN = (∠M N D + ∠BDN ) = 90◦ .
2
Hence, AD and BC intersect in point P such that ∠DP C = 90◦ . Since M is
the midpoint of AB,
∠P M A = 2∠P BA = 2∠P CD = ∠M N D.
Note that AB and CD are parallel, therefore, P M and M N are parallel to
and M, N and P lie on a straight line, hence N is the midpoint of segment
CD.
P
A
D
B
M
N
C
17
18
Intermediate Level
Problem 2. Let ABC be an isosceles triangle (AB = AC) with its circumcenter O. Point N is the midpoint of the segment BC and point M is the
reflection of the point N with respect to the side AC. Suppose that T is a
1
point so that AN BT is a rectangle. Prove that ∠OM T = ∠BAC.
2
Proposed by Ali Zamani
--------------------------------------------------Since 4ABC is an isosceles triangle, we have ∠AN C = 90◦ . Therefore,
∠OCM = ∠OCA + ∠M CA = ∠OAC + ∠N CA = 90◦ = ∠T AO.
Also we have CM = CN = BN = AT and OC = OA; So triangles 4OCM
and 4OAT are congruent. Which leads to OT = OM and
∠AOT = ∠M OC =⇒ ∠T OM = ∠AOC.
Thus, 4AOC ∼ 4M OT and ∠OM T = ∠OAC = 12 ∠A.
T
A
M
O
B
N
C
Solutions
19
Problem 3. In acute-angled triangle ABC (AC > AB), point H is the
orthocenter and point M is the midpoint of the segment BC . The median
AM intersects the circumcircle of triangle ABC at X. The line CH intersects
the perpendicular bisector of BC at E and the circumcircle of the triangle
ABC again at F . Point J lies on circle ω, passing through X, E, and F , such
that BCHJ is a trapezoid (CB k HJ). Prove that JB and EM meet on ω.
Proposed by Alireza Dadgarnia
--------------------------------------------------Solution. Let D be the foot of altitude passing through A and P, K be the
intersection of lines EM, AC and JH, AM , respectively.
A
P
F
J
H
K
E
B
D
C
M
X
Q
From parallel lines, we have
ME
DH
MK
=
=
=⇒ EK k AC.
(1)
EP
HA
KA
Note that ∠XKE = ∠XAC = ∠XF E. So K lies on ω. Let Q be the second
intersection point of line EM and circle ω. We have
(1)
∠KJQ = ∠KEP = ∠EP C = ∠QP C.
Note that it suffices to prove that ∠KJQ = ∠CBQ or prove that CP BQ is
a cyclic quadrilateral. Which is equivalent to M P · M Q = M B · M C. Also,
20
Intermediate Level
P
noting the parallel lines we can write M A = M K·M
. Using this equation
ME
and power of the point M with respect to the circumcircle of triangle 4ABC,
we have
MB · MC = MA · MX =
MK · MX
· M P = M Q · M P.
ME
Where the last equation comes from power of the point M with respect to
circle ω. Hence proved.
Comment. The same proof can be used to obtain the following generalised
result:
In triangle ABC point P is an arbitrary point and point D lies on the line
BC. The line AD intersects the circumcircle of triangle ABC at X. The line
CP intersects the line parallel to AP through D at E and the circumcircle
of triangle ABC again at F . Suppose that P lies inside of circle ω, passing
through X, E, and F . Point J lies on ω such that BCP J is a trapezoid
(CB k P J). Then JB and ED meet on ω.
Solutions
21
Problem 4. Triangle ABC is given. An arbitrary circle with center J,
passing through B and C, intersects the sides AC and AB at E and F ,
respectively. Let X be a point such that triangle F XB is similar to triangle
EJC (with the same order) and the points X and C lie on the same side of
the line AB. Similarly, let Y be a point such that triangle EY C is similar to
triangle F JB (with the same order) and the points Y and B lie on the same
side of the line AC. Prove that the line XY passes through the orthocenter
of the triangle ABC.
Proposed by Nguyen Van Linh - Vietnam
--------------------------------------------------Solution. Let H be the orthocenter of triangle 4ABC, P be the intersection
of BE and CF . P H cuts the perpendicular bisector of BC at Z.
A
Z
E
B0
F
C0
P
H
X
B
J
C
Y
We have
∠HBP = ∠ABH −∠ABP = 90◦ −∠BAC−∠ABP = 90◦ −∠BEC = ∠JBC.
Then BH and BJ are isogonal lines with respect to angle ∠P BC. Similarly,
CH and CJ are isogonal lines with respect to angle ∠P CB. From this, we
deduce that H and J are isogonal conjugate with respect to triangle 4BP C.
Then ∠HP B = ∠JP C. But ZB = ZC, JF = JE and 4P F E ∼ 4P BC.
Therefore 4P F E ∪ {J} ∼ 4P BC ∪ {Z}. Which follows that 4JEF ∼
22
Intermediate Level
4ZCB.
Let B 0 , C 0 be the intersections of BH and AC, CH and AB, repectively. We
have
H
H
P(BE)
= HB · HB 0 = HC · HC 0 = P(CF
),
P
P
P(BE)
= P B · P E = P C · P F = P(CF
).
We get Z lies on HP , which is the radical axis of circles with diameters BE
and CF . Analogously, X, Y also lie on HP . Therefore XY passes through
the orthocenter of triangle 4ABC.
Solutions
23
Problem 5. Find all numbers n ≥ 4 such that there exists a convex polyhedron with exactly n faces, whose all faces are right-angled triangles.
(Note that the angle between any pair of adjacent faces in a convex polyhedron is less than 180◦ .)
Proposed by Hesam Rajabzadeh
--------------------------------------------------Solution. If such a polyhedron exists for some n, the total number of sides
of faces is from one hand equal to 3n , and on the other is twice the number
of edges. So 3n is divisible by 2 and n must be even. We will give an example
of such a polyhedron for any even number n ≥ 4.
To this purpose, we need the following lemma.
Lemma 1. Let O be the origin in the 3-dimensional space and suppose X, Y
are two distinct points (different from O) in the xy-plane so that ∠OXY =
90◦ . Then for any point O0 on the z-axis, the triangle O0 XY is right-angled
(with ∠O0 XY = 90◦ ).
Proof. The proof is based on the Pythagorean Theorem. If O0 = O, there
is nothing to prove. If O0 6= O, the line OO0 (the z-axis) is perpendicular
to the xy-plane and so is perpendicular to every line in this plane passing
through O. In particular two triangles, O0 OX and O0 OY are right-angled.
According to the Pythagorean Theorem in these two triangles together with
triangle OXY , we have
O0 Y 2 = O0 O2 + OY 2 = O0 O2 + OX 2 + XY 2 = O0 X 2 + XY 2 .
which implies ∠O0 XY = 90◦ .
24
Intermediate Level
Now we return to the main problem. If n = 4, the tetrahedron with vertices
O0 , O, X, Y as in the lemma works (above figure). So we can assume n ≥ 6.
Take m = n−2
2 ≥ 2. First, we construct a convex (m + 2)-gon OA0 A1 · · · Am
in the xy-plane (take O to be the origin) so that
• OA0 = OAm .
• All the triangles of the form OAi Ai+1 (for 0 ≤ i ≤ m − 1) are rightangled.
Consider m different rays with initial point O (denote them by l1 , . . . , lm
respectively in clockwise order) so that for a sufficiently small value of α,
∠l1 Ol2 = ∠l2 Ol3 = · · · = ∠lm−1 Olm = α.
(1)
Take an arbitrary point on the ray l1 and call it A1 . Start from A1 and inductively by drawing perpendiculars from Ai to li+1 define the points A2 , A3 , . . . , Am
so that
∠OA2 A1 = ∠OA3 A2 = · · · = OAm Am−1 = 90◦ .
(2)
By (1) and (2) all the triangles OA1 A2 , OA2 A3 , . . . , OAm−1 OAm are similar.
OA3
OAm
2
Therefore OA
= · · · = OA
= OA
OA1 . We denote this common value by
m−1
2
r < 1. Note that r can be arbitrarily close to 1 by taking α small. Now we
have
OAm
OA3 OA2
OAm =
.··· .
.
.OA1 = rm OA1 .
OAm−1
OA2 OA1
Note that since α is small all the points A2 , A3 , . . . , Am are on the same side
of the line OA1 . Take the point A0 on the other side of this line so that
∠OA0 A1 = 90◦ and OA0 = rm .OA1 (A0 is one of the intersection points
Solutions
25
of the circle with diameter OA1 and the circle with center O and radius
rm .OA1 ). If r is sufficiently close to 1 (equivalently α sufficiently close to
zero), rm will be close to one and we can ensure that ∠A0 OA1 is small and
so the polygon satisfies all desired properties.
After construction of the polygon, consider two points O0 , O00 on the z-axis (on
different sides of the xy-plane) with OO0 = OO00 = OA0 = OAm . Then the
polyhedron with vertices O0 , O00 , A0 , A1 , . . . , Am (convex hull of these points)
have exactly n = 2m + 2 faces, and all are right-angled triangles. Indeed, it
has 2m faces of the form O0 Ai Ai+1 and O00 Ai Ai+1 which are all right-angled
according to the lemma and two faces O0 A0 O00 and O0 Am O00 that are isosceles
right triangles.
26
Intermediate Level
Advanced Level
27
Problems
Problem 1. Let M , N , and P be the midpoints of sides BC, AC, and AB
of triangle ABC, respectively. E and F are two points on the segment BC
1
1
so that ∠N EC = ∠AM B and ∠P F B = ∠AM C. Prove that AE = AF .
2
2
(→ p.31)
Problem 2. Let ABC be an acute-angled triangle with its incenter I. Suppose that N is the midpoint of the arc BAC of the circumcircle of triangle
ABC, and P is a point such that ABP C is a parallelogram. Let Q be the
reflection of A over N , and R the projection of A on QI. Show that the line
AI is tangent to the circumcircle of triangle P QR.
(→ p.33)
Problem 3. Assume three circles mutually outside each other with the
property that every line separating two of them have intersection with the
interior of the third one. √Prove that the sum of pairwise distances between
their centers is at most 2 2 times the sum of their radii.
(A line separates two circles, whenever the circles do not have intersection
with the line and are on different sides of it.)
√
Note. Weaker results with 2 2 replaced
√ by some other c may be awarded
points depending on the value of c > 2 2.
(→ p.35)
Problem 4. Convex circumscribed quadrilateral ABCD with incenter I
is given such that its incircle is tangent to AD, DC, CB, and BA at K,
L, M , and N . Lines AD and BC meet at E and lines AB and CD meet
at F . Let KM intersects AB and CD at X and Y , respectively. Let LN
intersects AD and BC at Z and T , respectively. Prove that the circumcircle
of triangle XF Y and the circle with diameter EI are tangent if and only if
29
30
Advanced Level
the circumcircle of triangle T EZ and the circle with diameter F I are tangent.
(→ p.37)
Problem 5. Consider an acute-angled triangle ABC (AC > AB) with its
orthocenter H and circumcircle Γ. Points M and P are the midpoints of the
segments BC and AH, respectively. The line AM meets Γ again at X and
point N lies on the line BC so that N X is tangent to Γ. Points J and K
lie on the circle with diameter M P such that ∠AJP = ∠HN M (B and J
lie on the same side of AH) and circle ω1 , passing through K, H, and J,
and circle ω2 , passing through K, M , and N , are externally tangent to each
other. Prove that the common external tangents of ω1 and ω2 meet on the
line N H.
(→ p.43)
Solutions
Problem 1. Let M , N , and P be the midpoints of sides BC, AC, and AB
of triangle ABC, respectively. E and F are two points on the segment BC
1
1
so that ∠N EC = ∠AM B and ∠P F B = ∠AM C. Prove that AE = AF .
2
2
Proposed by Alireza Dadgarnia
--------------------------------------------------Solution. Let H be the foot of the altitude passing through Q, A be the
midpoint of N P and K be the intersection point of N E and P F .
A
Q
P
N
K
B
E
H
F
M
C
If we prove that points K, H and Q are collinear, using parallel lines ,we get
that H is the midpoint of EF which is equivalent to the problem. Clearly, AM
passes through Q and H is the reflection of A with respect to N P . Therefore,
∠P QH = ∠AQP = ∠AM B. So it suffices to show that ∠P QK = ∠AM B.
Note that
∠N EC + ∠P F B =
1
(∠AM B + ∠AM C) = 90◦ =⇒ ∠EKF = 90◦ .
2
31
32
Advanced Level
So KQ is a median on the hypotenuse in triangle 4P KN and we’ll get
∠P QK = 2∠P N K = 2∠N EC = ∠AM B
which completes the proof.
Solutions
33
Problem 2. Let ABC be an acute-angled triangle with its incenter I. Suppose that N is the midpoint of the arc BAC of the circumcircle of triangle
ABC, and P is a point such that ABP C is a parallelogram. Let Q be the
reflection of A over N , and R the projection of A on QI. Show that the line
AI is tangent to the circumcircle of triangle P QR.
Proposed by Patrik Bak - Slovakia
--------------------------------------------------Solution. Let M, S be the midpoint of segments BC, AI, respectively. By
a homothety with center A and ratio 21 , P goes to M , Q to N and R to
T ; Where T is the projection of A on SN . So it suffices to show that the
circumcircle of triangle 4M N T is tangent to AI.
N
A
T
S
I
B
M
C
D
We claim that this circle is tangent to AI at point I. We know that ∠N AS =
90◦ , So by the similarity of two triangles 4ASN , 4T SA, we’ll get
ST · SN = SA2 = SI 2 .
Therefore, SI is tangent to the circumcircle of triangle 4IT N . Now if we
show that SI is tangent to the circumcircle of triangle 4N IM as well, our
proof is completed; Because the circle passing through I and N and tangent
to SI is unique. Let D be the second intersection point of AI and circumcircle
of triangle 4ABC. Note that ∠DBM = ∠DCB = ∠DN B. Therefore,
DM · DN = DB 2 = DI 2 .
34
Advanced Level
Thus, DI is tangent to the circumcircle of triangle 4N IM and we’re done.
Solutions
35
Problem 3. Assume three circles mutually outside each other with the
property that every line separating two of them have intersection with the
interior of the third one. √Prove that the sum of pairwise distances between
their centers is at most 2 2 times the sum of their radii.
(A line separates two circles, whenever the circles do not have intersection
with the line and are on different sides of it.)
√
Note. Weaker results with 2 2 replaced
√ by some other c may be awarded
points depending on the value of c > 2 2.
Proposed by Morteza Saghafian
--------------------------------------------------Solution. According to the figure, we denote the radii of the circles by
r1 , r2 , r3 and the distance Oi Oj by dij . Moreover, let l, l0 be two interior
common tangents of circles ω1 and ω2 . We denote the tangency points of l
1 +r2
and l0 as in the figure. Obviously d12 = rsin
α (α is defined in the figure).
Withour loss of generality we assume that r1 ≤ r2 .
l0
l
B0
A
r1
r2
α
A0
B
By assumption we can deduce that both lines l and l0 must intersect the third
circle (ω3 ). If the intersection point of l and ω3 lies outside between A and
B, we can find a line separating ω1 and ω2 so which does not intersect ω3 and
this is a contradiction with the assumptions. We have similar arguments for
l0 . So we can assume that the intersection of ω3 with l and l0 is below B and
A0 respectively. Therefore, r3 is at least the radius of the circle tangent to l
at B and also is tangent to l0 (why?). The radius of this circle is r2 cot2 α.
Hence
1 − sin2 α
r1 + r2
d212
2
r3 ≥ r2 cot α = r2
≥
−1 .
2
(r1 + r2 )2
sin2 α
36
Advanced Level
Consequently,
d212 ≤ (r1 + r2 )2 + 2r3 (r1 + r2 ),
(∗)
We have similar equations for d13 and d23 . Summing these three together
with Cauchy-Shwarz Inequality gives the assertion. Indeed,
X
dij
2
≤3
X
d2ij ≤ 6
X
ri2 + 18
X
X 2
ri rj ≤ 8
ri
Here the first and third inequality are coming from Cauchy-Shwarz Inequality
and the second inequality is the consequence of summing (∗) and two other
similar inequlities.
Remark. Using upper bound (r1 + r2 + r3 )2 for the right-hand side of (∗)
gives√d12 ≤ r1 + r2 + r3 . Summing these, gives a weaker result with 3 replaced
by 2 2.
Solutions
37
Problem 4. Convex circumscribed quadrilateral ABCD with incenter I
is given such that its incircle is tangent to AD, DC, CB, and BA at K,
L, M , and N . Lines AD and BC meet at E and lines AB and CD meet
at F . Let KM intersects AB and CD at X and Y , respectively. Let LN
intersects AD and BC at Z and T , respectively. Prove that the circumcircle
of triangle XF Y and the circle with diameter EI are tangent if and only if
the circumcircle of triangle T EZ and the circle with diameter F I are tangent.
Proposed by Mahdi Etesamifard
--------------------------------------------------Solution. First, let us prove these lemmas:
Lemma 1. Lines AC, BD, KM and LN are concurrent.
Proof. Using Brianchon’s Theorem in quadrilateral ABCD, one can simply
conclude the fact that AC, BD, KM and LN are concurrent.
Lemma 2. Let P be the point of concurrency of lines in Lemma 1. Therefore,
P is also the intersection point of quadrilateral ABCD’s diagonals and we
have IP ⊥EF .
Proof. We know that polar of point P is in fact line EF . Therefore, we’ll get
IP ⊥EF .
Lemma 3. A circle with diameter EI and the circumcircle of triangle 4XY J
are tangent.
E
C
J
M
D
K
I
Y
X
F
A
B
38
Advanced Level
Proof. For the proof of tangency of circumcircle of triangle 4XY J to the
circle with diameter EI (circle ω2 ), it suffices that the equation of Casey’s
Theorem hold for points X, Y, J and circle ω2 .
±XY · PωJ2 ± XJ · PωY2 ± Y J · PωX2 = 0.
Since PωJ2 = 0, Therefore,
√
√
XJ Y K · Y M = Y J XK · XM
(1)
Since X, Y lie on the radical axis of two circles ω and ω2 , We have:
Y K · Y M = Y L2
,
(1)
XK · XM = XN 2 =⇒ XJ · Y L = Y J · XN
(2)
So, we have to prove equation (2). Using Menelaus’s Theorem for triangle
4XF Y and line LN P , We have:
XN F L Y P
·
·
F N Y L XP
F N =F L
=⇒
XP
XN
=
.
YL
YP
From equation (2), we get:
XJ
XN
XP
=
=
.
YJ
YL
YP
Therefore we need to prove that JP is the exterior angle bisector of angle
∠XJY . Since JQ⊥JP , we need to prove that (XY, QP ) = −1.
NL
(XY, P Q) = F (XY, P Q) = (N L, P U ) = −1.
And since point U lies on EF (polar of P ), the last equation holds and we’re
done.
Lemma 4. AK is tangent to the circumcircle of triangle 4ABC if and only
if
2
BK
AB
=
.
KC
AC
Proof. Using The Law of Sines and Ratio Lemma, one can simply get the
desired results.
Lemma 5. If angle bisectors of angles ∠E and ∠F are perpendicular, then
ABCD is a cyclic quadrilateral.
Solutions
39
Proof. It’s trivial.
E
J
C
L
D
M
K
I
S
Y
X
F
A
N
B
Now, Let’s get back to the problem. First, we assume that two circles ω1 and
ω2 are tangent to each other. Let S be the foot of the perpendicular line to
F I passing through E. Using Casey’s Theorem for points X, F, Y and circle
ω2 , we have:
q
q
q
PωX2 ± XY PωF2 = 0
√
√
√
=⇒ ± XF Y K · Y M ± Y F XK · XM ± XY F S · F I = 0.
± XF
PωY2 ± Y F
(3)
Points X and Y lie on the radical axis of circles ω and ω2 . Therefore we have:
Y K · Y M = Y L2
,
XK · XM = XN 2 .
So equation (1) can be written as:
± XF · Y L ± Y F.XN ± XY
√
F S · F I = 0.
According to the figure, We have: ∠F1 = ∠F2 =
α+β
2 .
α+β
Y L = F L ± F Y = F I · cos (F1 ) ± F Y = F I · cos
± F Y,
2
α+β
XN = F N ∓ XF = F I · cos (F2 ) ∓ XF = F I · cos
∓ XF.
2
(4)
40
Advanced Level
Now, by putting them in equation (4), We’ll get:
α+β
± FY
± XF · F I · cos
2
√
α+β
± Y F · F I · cos
∓ XF ± XY F S · F I = 0
2
√
α+β
=⇒ ± F I XF + Y F cos
= ±XY F S · F I
2
√
XF + Y F
α+β
=⇒F I
cos
= FS · FI
XY
2
r
α+β
sin α + sin β
FS
=⇒ cos
·
=
2
sin α + β
FI
α−β
FS
=⇒ cos2
.
(5)
=
2
FI
E
C
R
D
K
Y
X
M
L
V
Q
S
β
α
F
A
I
N
B
Also, we have:
∠F RS = 90◦ −
α+β
2
α−β
⇒ ∠QV R = 90◦ −
2
α
−
β
⇒ ∠EIF = 90◦ −
.
2
So, by equation (5), we have:
sin2 (EIF ) =
FS
.
FI
(6)
Solutions
41
We consider three cases for point S on line F I:
Case 1) ∠EIF = 90◦ . Which gives us that S and I coincide.
FS
= 1.
FI
Now, by Lemma 5, ABCD is a cyclic quadrilateral. On the other
hand, ABCD is circumscribed and every equation resulted from Casey’s
Theorem for the circumcircle of triangle 4XF Y and the circle with
diameter EI, can be written for the circumcircle of triangle 4T EZ and
the circle with diameter F I as well. So by Casey’s Theorem, these two
circles are tangent to each other.
sin2 (EIF ) =
Case 2) ∠EIF < 90◦ .
ES
FS
sin (EIF ) =
=
.
EI
FI
Now by Lemma 4, we get that EF is tangent to the circumcircle of
triangle 4ESI and
2
∠F ES = ∠F IF =⇒ ∠IEF = 90◦ .
Z
J ≡E
Y
T
C
D
I
F
X
A
B
Now since ∠IEF = 90◦ , the foot of perpendicular line to EF passing
through I, (Point J) coincides with point E. By Lemma 3, the circumcircle of triangle 4T JZ (which is also the circumcircle of triangle
42
Advanced Level
4T EZ), will be tangent to the circle with diameter F I. In this case,
tangency point of the circumcircle of triangle 4T EZ and the circle with
diameter EI, will be point I and tangency point of the circumcircle of
triangle 4T EZ and the circle with diameter F I, will be point E.
Case 3) ∠EIF > 90◦ . Since
sin2 (EIF ) =
FS
> 1,
FI
this case will never happen.
Solutions
43
Problem 5. Consider an acute-angled triangle ABC (AC > AB) with its
orthocenter H and circumcircle Γ. Points M and P are the midpoints of the
segments BC and AH, respectively. The line AM meets Γ again at X and
point N lies on the line BC so that N X is tangent to Γ. Points J and K
lie on the circle with diameter M P such that ∠AJP = ∠HN M (B and J
lie on the same side of AH) and circle ω1 , passing through K, H, and J,
and circle ω2 , passing through K, M , and N , are externally tangent to each
other. Prove that the common external tangents of ω1 and ω2 meet on the
line N H.
Proposed by Alireza Dadgarnia
--------------------------------------------------Solution 1. Let D be the intersection of AH and BC. Denote Ω by the
circle with diameter P M . It’s obvious that D lies on Ω. Also since ABC is
acute, H lies on the segment P D and so inside of Ω. N lies on the extension
of DM and so outside of Ω. We claim that there are at most two possible
cases for K. The following lemma proves our claim.
Lemma. Given a circle ω and four points A, B, C, and D, such that A and
B lie on the circle, C inside and D outside of the circle. There are exactly
two points like K on ω so that the circumcircles of triangles ACK and BDK
are tangent to each other.
Proof. Invert the whole diagram at center A with arbitrary radius, the images
of points and circle are denoted by primes. Since A lies on ω, ω 0 is a line,
passes through B 0 and K 0 . Notice that C 0 and D0 lie on the different sides of
ω 0 . Since the circumcircles of triangles ACK and BDK are tangent to each
other, we have C 0 K 0 is tangent to the circumcircle of triangle B 0 D0 K 0 . It
means ∠C 0 K 0 B 0 = ∠B 0 D0 K 0 . Let X and Y be two arbitrary points, lie on
ω 0 and the different sides of B 0 .
C0
B0
K0
−→
Y
D0
ω0
X
44
Advanced Level
0
First assume that K 0 ≡ B
so ∠C 0 B 0 Y = ∠C 0 K 0 B 0 > 0 = ∠K 0 D0 B 0 and when
−−0→
0
K moves along the ray B X, ∠C 0 K 0 B 0 decreases and ∠K 0 D0 B 0 increases. It
−−→
yields there is exactly one point K 0 on the ray B 0 X so that ∠C 0 K 0 B 0 =
∠B 0 D0 K 0 . In the same way we get there is only one possible case for K 0 on
−−→
the ray B 0 Y and the result follows.
Γ
A
P
A0
L
Ω
H
J
B
D
C
M
N
X
E
Denote ω1 and ω2 by the circumcircles of triangles AJP and HN D. Let H
be the indirect homothety that sends ω1 to ω2 . Notice that J and N lie on
the different sides of AH. Now since the arc AP of ω1 is equal to the arc
HD of ω2 and AP k HD, H sends A to D and P to H therefore (A, H) and
(P, D) are anti-homologous pairs. Let L be the anti-homologous point of J
under H. It’s well-known that the pairs of anti-homologous points lie on a
circle so ALHJ and LP JD are cyclic quadrilaterals.
Let E be the reflection of A over the point M . We claim that HDEN is
cyclic. A0 lies on Γ so that AA0 k BC. We know that (A0 X, BC) = −1 hence
N A0 is tangent to Γ. Also by symmetry N E is tangent to the circumcircle
of triangle CEB. Now since HE is the diameter of this circle, we have
∠N EH = 90◦ = ∠N DH and our claim is proved. The line AM meets the
Solutions
45
circumcircle of triangle P DM again at L0 . We have
AL0 · AM = AP · AD =⇒ AL0 · AE = AH · AD
it follows that L0 HDEN is cyclic so L0 ≡ L. We have
∠P JH = ∠AJH − ∠AJP = ∠HLM − ∠HN D
= ∠HLM − ∠HLD = ∠DLM = ∠DJM
therefore ∠HJD = 90◦ . From this we can conclude that the cirucmcircles
of triangles DHJ and DM N are tangent to each other and the common
external tangents of them are concurrent at H since the tangent line to the
circumcircle of triangle DHJ through H is parallel to DM N . So the problem
is proved for K ≡ D, now suppose that K 6= D. Since ∠AHL = ∠LN M
the circumcircles of triangles LHJ and LM N are tangent to each other. So
L ≡ K. Denote O1 and O2 by the circumcenters of triangles LHJ and LM N .
It’s obvious that O1 , L, and O2 are collinear so ∠O1 LH + ∠O2 LN = 90◦ . It
yields
∠HO1 L = 180◦ − 2∠O1 LH = 2∠O2 LN = 180◦ − ∠LO2 N =⇒ O1 H k O2 N
therefore the direct homothety that sends (O1 ) to (O2 ), sends H to N and
the conclusion follows.
Solution 2. Let D be the intersection of AH and BC. Denote Ω by the
circle with diameter P M . It’s obvious that D lies on Ω.
A
P
J0
L
Γ
Ω
F
O
H
J
N0
B
D
C
M
X0
X
N
46
Advanced Level
Let F be the intersection of N H and M J. Since J and B lie on the same
side of P D, J lies on the arc P D (the one that does not contain M ) so J and
H lie on the same side of BC. Also
∠HN M = ∠AJP < ∠JP D = ∠JM D
therefore F and J lie on the same side of N M and we have 4F M N ∼
4AP J since ∠JP D = ∠JM D. It follows that A, F , H, and J are concylic.
Let J 0 and N 0 be the reflections of J and N over the points P and M ,
respectively. Since P is the midpoint of AH, AJ 0 HJ is a parallelogram. The
A−symmedian meets Γ again at X 0 . Since XX 0 k BC, by symmetry N 0 X 0
is tangent to Γ, too. Also we know that (AX 0 , BC) = −1 so N 0 A is tangent
to Γ. Now 4F M N ∼ 4AP J yields 4F M N 0 ∼ 4AP J 0 . It follows that
∠N 0 F M = ∠J 0 AP = ∠AHJ = 180◦ − ∠AF J
hence A, F , and N 0 are collinear. Again from 4F M N 0 ∼ 4AP J 0 we get
∠P JH = ∠AJ 0 P = ∠F N 0 M = 90◦ − ∠P M N 0 = ∠DP M = ∠DJM
In the third equality we used that M P ⊥ AN 0 (It’s a well-known property,
If we let O be the center of Γ then AP M O is a parallelogram). It yields
∠HJD = ∠P JM = 90◦ . Like the first solution we know that there are at
most two possible cases for K and we can conclude that D is one of them.
Now we suppose that K 6= D. Let AM meets Ω again at L. We have
∠LAH = 90◦ − ∠LM D = ∠LJD − 90◦ = ∠LJH
therefore ALHJ is cyclic. Since M P ⊥ AN 0 and AP ⊥ M N 0 , P is the
orthocenter of triangle AN 0 M and N 0 P ⊥ AM . It follows that N 0 , P and L
lie on a same line. Now since ∠ALP = ∠N 0 LM = 90◦ and ∠AP L = ∠N 0 M L,
we have 4AP L ∼ 4N 0 M L. It yields 4LM N ∼ 4LP H. Hence
∠M LN = ∠P LH =⇒ ∠HLN = ∠P LM = 90◦
so LN DH is cyclic and ∠AHL = ∠LN M . It follows that the circumcircles
of triangles LHJ and LM N are tangent to each other. So L ≡ K. Denote
O1 and O2 by the circumcenters of triangles LHJ and LM N . It’s obvious
that O1 , L, and O2 are collinear so ∠O1 LH + ∠O2 LN = 90◦ . It yields
∠HO1 L = 180◦ − 2∠O1 LH = 2∠O2 LN = 180◦ − ∠LO2 N =⇒ O1 H k O2 N
therefore the direct homothety that sends (O1 ) to (O2 ), sends H to N and
the conclusion follows.
Solutions
47
Comment. We can also prove LHDN is cyclic by angle-chasing. We have
∠DLM = ∠DP M = 90◦ − ∠P M D = ∠P JD − 90◦ = ∠P JH
also ∠HLM = ∠AJH so ∠HLD = ∠AJP = ∠HN D and it follows that
LHDN is cyclic.