Geometry Problems
Anand
∗
October 17, 2019
Problems
1. If ABC is an equilateral triangle, let D be a point on AC such that AD =
1
1
3 AC; similarly E is a point on AB such that BE = 3 AB. Prove that if
◦
BD ∩ CE = F , then prove that ∠AF C = 90 .
2. Let I be the incentre of 4ABC. Let ` be a line through I, parallel to
AB, that intersects the sides CA and CB at M, N , respectively. Prove that
AM + BN = M N .
3. Let ABC be an acute triangle inscribed in a circle. Let X be the mid point of
arc BC not containing A. Y, Z are defined similarly. Prove that the orthocenter
of XY Z is the incenter I of 4ABC
4. In 4ABC, AC = BC. Let P be a point inside ∆ABC such that ∠P AB =
∠P BC. Denote by M the midpoint of the segment AB.Show that ∠AP M +
∠BP C = 180◦
5. Let ABC be an acutetriangle. The points M and N are taken on the sides
AB and AC respectively. The circles with diameters BN and CM intersect at
points P and Q. Prove that P, Q and the orthocenter H of 4ABC are collinear.
6. Let D, E be points on sides AC and AB respectively of 4ABC such that
BD and CE are the angle bisectors of ∠B and ∠C respectively. Let X, Y be
points on BD, CE respectively such that AX ⊥ BD and AY ⊥ CE. Let A0 be
the reflection of A over XY . Show that A0 lies on BC.
7. Let ABCbe a triangle, and I the incenter, M midpoint of BC, D the
touch point of incircle and BC. Prove that perpendiculars from M, D, A to
AI, IM, BC respectively are concurrent.
8. In a quadrilateral ABCD diagonal AC is a bisector of ∠BAD and ∠ADC =
∠ACB. The points X and Y are the feet of the perpendiculars from A to BC
and CD respectively. Prove that the orthocenter of 4AXY lies on the line BD.
∗ email:
anand27112003@rediffmail.com
1
9. Let ABC be an acute-angled triangle. The A-excircle touches BC, AB and
AC at D, E and F respectively.EF ∩ BC = K. Let the A-symmedian intersect
the circumcircle of ABC at X.Prove that the reflection of A across BC lies on
the circle passing through K,X and D.
10. Let ABC be a acute triangle and let P be any point.Let the orthocenter of
ABC be H. Let U denote the intersection of BC and perpendicular from H to
AP . Similarly define V, W . Prove that U, V, W are collinear.
11. Let quadrilateral ABCD be inscribed in a circle. Suppose lines AB and CD
intersect at P and lines AD and BC intersect at Q, construct the two tangents
QE and QF to the circle where E and F are the points of tangency. Prove that
the three points P, E, F are collinear.
12. Let H be an orhocenter of an acute triangle ABC and M midpoint of side
BC. If D and E are foots of perpendicular of H on internal and external angle
bisector of angle ∠BAC, prove that M , D and E are collinear.
13. In ABC the A-bisector intersects BC and the circumcircle at D and E
respectively.Let the incircle touch BC at F and the A-mixtilliniear incircle touch
the circumcircle at G.Prove that DEF G is concyclic.
14. Acute triangle ABC is inscribed in circle ω. Let H and O denote its
orthocenter and circumcenter, respectively. Let M and N be the midpoints
of sides AB and AC, respectively. Rays M H and N H meet ω at P and Q,
respectively. Lines M N and P Q meet at R. Prove that OA ⊥ RA.
15. Consider a semicircle of center O and diameter AB. A line intersects AB
at M and the semicircle at C and D in such a way that M B < M A and
M D < M C. The circumcircles of triangles AOC and DOB intersect second
time at K. Show that M K and KO are perpendicular.
16. Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D
on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB
at X and AC at Y . Let T be the intersection of EF with BC and let M be the
midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
17. Let ABC be an acute-angled triangle with circumcircle ω and orthocenter
H. Let K be a point of ω on the other side of BC from A. Let L be the
reflection of K across AB, and let M be the reflection of K across BC. Let E
be the second point of intersection of ω with the circumcircle of triangle BLM .
Show that the lines KH, EM , and BCare concurrent.
18. Let ABC be a triangle inscribed in circle Γ with center O. Let H be the
orthocenter of triangle ABC and let K be the midpoint of OH. Tangent of
Γ at B intersects the perpendicular bisector of AC at L. Tangent of Γ at C
intersects the perpendicular bisector of AB at M . Prove that AK and LM are
perpendicular.
2
19. Let DEF be orthic triangle of ABC. Let AO meet EF at Z where O
is circumcentre of ABC. Let N be midpoint of DZ. Prove that AN is the
A − symmedian of ABC.
20. Let ABC be an acute triangle.The lines l1 and l2 are perpendicular to AB
at the points A and B, respectively.The perpendicular lines from the midpoint
M of AB to the lines AC and BC intersect l1 and l2 at the points E and F ,
respectively.If D is the intersection point of the lines EF and M C, prove that
∠ADB = ∠EM F .
21. Let DEF be intouch triangle of ABC.Let M be midpoint of AC and K be
orthocenter of BIC(I is incentre of ABC). Prove that KE ⊥ M I.
22. A trapezoid ABCD (AB||CF ,AB > CD) is circumscribed.The incircle of
the triangle ABC touches the lines AB and AC at the points M and N ,respectively.Prove
that the incenter of the trapezoid ABCD lies on the line M N .
23. The point H is the orthocentre of a triangle ABC, and the segments
AD, BE, CF are its altitudes. The points I1 , I2 , I3 are the incentres of the
triangles EHF, F HD, DHE respectively. Prove that the lines AI1 , BI2 , CI3
intersect at a single point.
24. Given a hexagon ABCDEF such that AB = BC, CD = DE, EF = F A
and ∠A = ∠C = ∠E. Prove that AD, BE, CF are concurrent.
25. Let P, Q be isogonal conjugates in ABC. Let Pb , Pc be reflection of P on
AC, AB. Let Qa be reflection of Q on BC and T = Pb Pc ∩ BC. Prove that
AQa ⊥ P T .
26. Consider in ∆ABC the set of rectangles with one side on line BC and the
other two vertices on lines AB, AC. Find the locus of the center of the rectangle
as it varies.
27. 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
28. Let ABC be a triangle and let the external angle bisector of the angle
∠BAC intersect the lines perpendicular to BC and passing through B and C
at the points D and E, respectively. Prove that the lines BE, CD, AO are
concurrent, where O is the circumcenter of ABC.
29. Let ABC be a triangle and let D be a point on the segment BC, D 6= B
and D 6= C. The circle ABD meets the segment AC again at an interior point
E. The circle ACD meets the segment AB again at an interior point F . Let A0
be the reflection of A in the line BC. The lines A0 C and DE meet at P , and
the lines A0 B and DF meet at Q. Prove that the lines AD, BP and CQ are
concurrent (or all parallel).
3
30. Triangle ABC.The incircle (I) tangent BC, CA, AB at D, E, F .
AD ∩ BE = P .
X, Y, Z is reflection of P through EF, DF, DE.
Prove that: AX, BY, CZ are concurrent at OI.
31. Show that the Symmedian point lies on the line whose existence was demanded in P26.
32. A circle ω with center I is inscribed into a segment of the disk, formed by
an arc and a chord AB. Point M is the midpoint of this arc AB, and point
N is the midpoint of the complementary arc. The tangents from N touch ω in
points C and D. The opposite sidelines AC and BD of quadrilateral ABCD
meet at point X, and the diagonals of ABCD meet at point Y . Prove that
points X, Y, I and M are collinear.
33. Let ABC be a triangle and consider points D, E on sides BC, CA respectively. Let F be the intersection of the circumcircle of triangle CDE and the
line parallel from to AB passing through C. Now let G be the intersection of
lines AB and F D. Point H is selected on line AB such that ∠BEG = ∠HDA.
Given that HE = DG, prove that Q is on the angle bisector of ∠BCA, where Q
is the intersection of lines BE and AD. There are two possibilities for position
of H on AB. Please consider points H, A, B to be in the order H − A − B.
34. Let ABC be a triangle with external bisector AD, incenter I and circumcenter O. Perpendicular bisector of AI cuts OA at J.K, L are circumcenter of
triangle ABD, ACD. S, T are reflection of A through JK, JL. AE, AF are
symmedian of triangle AIC, AIB with E, F lie on IC, IB. Prove that SF, T E
and BC are concurrent.
35. Quadrilateral ABCD with perpendicular diagonals is inscribed in a circle
with centre O. The tangents to this circle at A and C together with line BD
form the triangle 4. Prove that the circumcircles of BOD and 4 are tangent.
36. Consider a quadrilateral ABCD inscribed in a circle in which AB is a
diameter. Draw the tangents to the circle at A and B. Let E be the midpoint
of segment CD. Draw the perpendicular from the midpoint of segment AD to
AE and extend this perpendicular to meet the tangent at A at M . Similarly,
draw the perpendicular from the midpoint of segment BC to BE and extend
this perpendicular to meet the tangent at B at N . Prove that M N is parallel
to CD.
37. Let DEF be intouch triangle of ABC. Let S be any point on EF . And let
SD meet the incircle again at X. Let line through S parallel to BC meet sides
AB, AC at Y, Z.Prove that (XY Z) is tangent to incircle.
38. In 4ABC, let M be the mid-point of H and O. The line through M parallel
to BC meets AB, AC at D, E respectively. It is known that O is the incenter
of 4ADE. Find angles of the triangle ABC.
4
39. Let ABCD be a cyclic quadrilateral with center O. Let M, N be midpoint
of AB, CD, E = AD ∩ BC, F = AB ∩ CD, P = M N ∩ AC. Let Q be foot of
perpendicular from F upon OE. Prove that M, A, P, Q cyclic.
40. Let ABC an acute triangle and Γ its circumcircle. The bisector of BAC
intersects Γ at M 6= A. A line r parallel to BC intersects AC at X and AB at
Y . Also, M X and M Y intersect Γ again at S and T , respectively. If XY and
ST intersect at P , prove that P A is tangent to Γ.
41. A triangle ABC is inscribed in a circle (C) .Let G the centroid of 4ABC
. We draw the altitudes AD, BE, CF of the given triangle .Rays AG and GD
meet (C) at M and N .Prove that points F, E, M, N are concyclic.
42. Given a triangle 4ABC with circumcircle (O), incenter (I) and intouch
triangle 4DEF , let ωA denote the circle externally tangent to (I) at D and
tangent to (O) at a point KA ; similarly define ωB , ωC , KB , KC . Prove that the
lines AKA , BKB , CKC are concurrent at a point T .[/hide]
43. In triangle ABC let M be the midpoint of BC. Let ω be a circle inside
of ABC and is tangent to AB, AC at E, F , respectively. The tangents from
M to ω meet ω at P, Q such that P and B lie on the same side of AM . Let
X ≡ P M ∩ BF and Y ≡ QM ∩ CE. If 2P M = BC prove that XY is tangent
to ω.
44. ABC is a triangle with circumcentre O. M is the midpoint of BC. Let
X be the intersection of the angle bisector of ∠OM B and the perpendicular
bisector of AB. Let Y be the intersection of the angle bisector of ∠OM C and
the perpendicular bisector of AC. Prove that ∠XAB + ∠Y AC = 90◦ .
5
Solutions/Hints
1. First notice that 4CEB ∼
= 4BDA, then ∠EF D = 120◦ and ∠BAC = 60◦ .
So, DAF E is a cyclic quadrilateral. So, we have to prove ∠EDA = 90◦ . Now
let AP = P E = AD where P is a point on AB. So, P AD is an equilateral
triangle. So, ∠AED = 30◦ . So, ∠EDA = ∠AF C = 90◦ .
2. Join A to I. You get ∠IAB = ∠IAM = ∠M IA =⇒ AM = M I. Similarly
BN = N I. So, AM + BN = M I + IN = M N .
3. Drop perpendiculars from O to the sides which intersects the circumcircle
at X, Y, Z . Now it becomes INMO 2001 P1. Since O is the circumcentre, so
BC1 ||AO||CB1 and BC1 = AO = CB1 , therefore BCB1 C1 is a paralellogram
hence B1 C1 ||BC. Hence A1 O ⊥ B1 C1 . Similarly, we can get that O is the
orthocentre of 4A1 B1 C1 .
4. Let ∠P AB = ∠P BC = β
And ∠CAB = ∠CBA = α
Since P M ⊥ AB and ∠AP M = ∠BP M ⇒ α − β = β → α = 2β
So P is the incenter of 4ABC
∠CP A = ∠CP B = x
∠AP M = ∠BP M = y
2(x + y) = 360◦ → x + y = ∠AP C + ∠BP M = ∠AP M + ∠BP C = 180◦
5. Let D , E , F be the feet of the altitudes from vertices A , B , C respectively.
Then by intersecting chords :
BH × HE = P H × HQ
CH × HF = P H × HQ
Note : B, P, E, N, Q and C, P, F, M, Q are concyclic .(by cyclic quadrilaterals).
It’s well known that AH ×HD = BH ×HE = CH ×HF (by cyclic quadrilaterals) ⇒ BH × HE = P H × HQ = CH × HF = P H × HQ ⇒ P H × HQ = P H × HQ ⇒
HQ = HQ. Which implies that Q–H–P
0
6. By angle chasing ∠IXY = ∠B
2 ,so we have XY k BC. Let X = AX ∩
∠B
0
0
BC =⇒ ∠AX B = 2 = ∠XAB =⇒ ABX is isosceles triangle =⇒ X is
midpoint of AX 0 =⇒ A0 lies on BC.
7. As we see many cyclic quadrilaterals, we imediatelly think of radical axis...
Letting E be the tangency point of A-excircle, and S the foot of altitude from
D to AE, we easily find that IM bissects DS. So, ∠ISM = ∠IDM = 90, so if
L is the foot of altitude from M to AI, IM DSL is cyclic, and also if H is foot
of perpendicular from A to BC, quadrilaterals AHDS and AHLM are cyclic
from easy angle chase. So their radical axes are concurrent, so AH, M L, and
DS are concurrent as required.
6
8. Let P be the foot of the altitude from X onto AY. Let XP meet BD at Q. We
have to prove that Q is the orthocenter. Obviously XQ||CD. So BQ/DQ =
BX/XC. Also Triangles ADY and ACX are similar, so AX/AY = CX/DY .
Triangles AY C and AXB are similar, so AX/AY = BX/CY = CX/DY , so
BX/XC = DY /CY means Y Q||BC so Q is the orthocenter.
9. Let the midpoint of BC be M. Since (B, C; D, K) = 1 =⇒ M B 2 =
M D.M K. Also let the reflection of X on BC be X 0 . Notice that this is the
AHM point. So, A, M, X 0 collinear. Now the reflection of X 0 on M is the reflection of X on perpendicular bisector of BC. This lies on (ABC). By power
of point , M A.M X 0 = M B.M C = M D.M K. So, A, X 0 , K, Dcyclic, reflect this
circle on BC to get the desired result
10. Let the foot of the perpendicular from H to AP be U 0 .Define V 0 and W 0
similarly. Let AD, BE, CF be the altitudes of the triangle. Then quadrilaterals
AU DU 0 ,BV EV 0 ,CW EW 0 are cyclic. By PoP,
HU · HU 0 = HA · HD = HB · HE = HV · HV 0
So, we get three cyclic quadrilaterals U V U 0 V 0 ,V W V 0 W 0 ,U W U 0 W 0 . Note that
HU 0 V 0 W 0 is also cyclic. We use directed angles modulo 1800
]HU V = ]U 0 U V = ]U 0 V 0 V = ]U 0 V 0 H = ]U 0 W 0 H = ]U 0 W 0 W = ]U 0 U W = ]HU W
11. Let R be AC ∩ BD. Then by Brocard’s theorem , P R is polar of Q w.r.t to
ABC. Also by definition of polar,EF is the polar of Q. So , P, E, F collinear.
12. WLOG let ∠B > ∠C Let K be the foot of the altitude fromA,N be the
midpoint of AB. Observe that ADHE is a rectangle. Therefore, DE passes
through the midpoint of AH(point P )
∠DP K = ∠DP H = 2∠DAH = 2(A/2 − (90 − B) = B − C
∠M P K = ∠M N K = ∠KN A − ∠M N A = 2B − (1800 − A) = B − C
Therefore points D,P ,M and E are collinear.
13. Let us assume that E and D intersect BC at N and let F be the foot of
the perpendicular from C to AB. To show that M = N it is enough to show
F N = F C = N B or ETST ∠N F C = F CN = 90 − ∠B or ETST ∠F CA =
∠C + ∠B + 90 , which is true.
14. Let the reflection of H across AB and AC be K and L respectively. It is a
well known fact that K and L would lie on ω. So, we have
KH · HP = LH · HQ =⇒
LH
KH
· HP =
· HQ =⇒ M H · HP = N H · HQ
2
2
Hence, M, N, P, Q are concyclic. Also notice that (M N A) is internally tangent
to ω. Now we consider the Radical Center of (M N A), (M N P Q), ω which is
indeed R described in the problem. So, R lies on the radical axis of ω and
(M N A) and since, both are internally tangent at A implies that RA is tangent
to ω which implies RA ⊥ OA.
7
15. Since K is the miquel point of ACBD (refer to EGMO proposition 10.14).
By Brocard’s theorem , we get the desired condition
16. By angle chasing you can show BCXY cyclic. So, by Power of point,we
get DB.DC = DX.DY . Also since (B, C; D, T ) = −1 we get DM.DT =
DB.DC = DX.DY We get the desired result by converse of Power of point.
17. Let HA and HC be the reflections of H across BC and BA, which lie on ω.
Let E 0 be the second intersection of line HA M with ω. By construction, lines
E 0 M and HK concur on BC. Firstly, we will claim that L, HC , and E 0 are
collinear. Notice that
]LHC B = −]KHB = ]M HA B
by reflections, and that
]M HA B = ]E 0 HA B = ]E 0 HC B
as desired. Now,
]LE 0 M = ]HC E 0 HA = ]HC BHA = 2]ABC
and
]LBM = ]LBK + ]KBM = 2]ABK + 2]KBC = 2]ABC
so B, L, E 0 , M are concyclic. Hence E ≡ E 0 . Hence, proved.
18. Let B 0 be the point on circumcircle such that BB 0 k AC. Since LO ⊥ AC.
We get that LB 0 is tangent to (ABC). Similarly do M C 0 . Let G be centroid
of ABC and let DEF be orthic triangle of ABC. Then it is well known(can be
proved by homothety) that G, E, B 0 and G, F, C 0 collinear. Perform a homothety
0
centered at G with ratio −1
2 .The image of L (L ) is the intersection of tangents
to nine-point circle at E, Mb (midpoint of AC).Now A lies on polar of L0 , M 0
w.r.t nine-point circle. We get the desired result.
19.
Lemma 1. AH and AO are isogonal.
Lemma 2. The A-symmedian of ABC is the A-median of AEF .
Lemma 3. Z is the perp. from A to EF
Lemma 4. If AP and AQ are isogonal wrt triangle ABC and AX and AY are
isogonal wrt AP Q,AX and AY are isogonal wrt ABC.
√
We invert at A with radius AB · AC After Inversion, B 0 and C 0 are the foot
of the E 0 and F 0 altitudes of triangle AE 0 F 0 . D0 is the orthocentre of AE 0 F 0
Z 0 is the point diametrically opposite A in (AE 0 F 0 ) . AN 0 is a symmedian of
AZD (since AN was a median) But the midpoint of D0 Z 0 is the midpoint of
E 0 F 0 .(reflecting orthocentre) Therefore, AN 0 and the A-median of AE 0 F 0 are
isogonal wrt ADZ. AN 0 is the symmedian of AE 0 F 0 . (by lemmas 1 and 4)
Inverting back AN is the median of AEF . By lemma 2 we get the desired
conclusion.
8
20. Let Ha Hb Hc be the orthic triangle of ABC.It can be easily seen that
Ha AHb B is cyclic with center M . Also since EM ⊥ BHb we get that AC
is the polar of E w.r.t AHa BHb . Similarly for F . We get that the polar of C
is EF by La Hire. Also by brocard the orthocenter(H) of ABC lies on EF .
Hence D is the C − Humpty point. Since ∠EM F = 180 − ∠C (Considering the
quadrilateral Ha Hb M C)= ∠AHB = ∠ADB( as ADBH is cyclic)
21. Let P is tangency point of B-excircle to AC. Then ∠KIE = ∠C = ∠BCP
and
KI
BC
=
.
CP
IE
So 4CBP ∼ 4EKI and ∠IKE = ∠CBP ⇒ KE ⊥ BP .
We have that BP k IM , now KE ⊥ IM .
22. Let I be the incenter of triangle ABC and R be the common point of the
lines AI and M N . it is enough to show that CR is the angle bisector of ∠ACD.
now notice that ∠CIN = 90 - 1/2∠ABC and ∠CN I = 90 + 1/2∠ABC . So
CNIR is a cyclic quadrilateral. Now ∠N IR = ∠N CR = 1/2(∠ACN - ∠ABC)
so ∠ACR = 1/2(∠ACN + ∠ABC) so CR is the angle bisector of ∠ACD.
23. Say reflecting about angle bisector of ∠A sends E, F, H to E ∗ , F ∗ , H ∗ respectively. Consider a homothety from A sending E ∗ F ∗ to BC. Since AE ∗ F ∗ H ∗
is obviously cyclic, H ∗ is sent to a point on (ABC), say, P . Let the incenter
of ∆P BC be X. Define Y, Z analogously. It clearly suffices to prove that
AX, BY, CZ concur.
Note that triangles XBC, Y CA, ZAB are directly similar and our result
follows from Jacobi.
24. Let O be the circumcentre of ACE. It can be easily seen that BO, DO, F O
are angle bisectors.It is easy to show by angle chasing that OA bisects ∠BAF .
Let the perpendicular from O to AB be H then OH = OAsin( ∠BAF
). It can
2
be shown that ABCDEF has a incircle centered at O. After this, applying
Brianchon gives that AD,BE,CF are concurrent.
25. Qa T = QT, Qa P = QPc =⇒ Qa T 2 −Qa P 2 = QT 2 −QPc2 = AT 2 −APc2 =
AT 2 − AP 2
26. Let the rectangle be DEF G where D, E lie on AB, AC. Let X be the
midpoint of DE and Y be the midpoint of F G. Then X moves on the Amedian of ∆ABC while Y moves on BC. Also, XY ⊥ BC. Clearly, if M is the
midpoint of BC and W is the foot of A- altitude of ∆ABC, then the midpoint
of XY clearly moves along the M -median of ∆AW M .
27. Since the quadrilateral has an incircle, let the tangancy points of this circle
with sides AB.BC, CD, DA be K1 , L1 , M1 , N1 . Also, let N L ∩ KM = Z Let
X and Y be the intersection points of AC and N M , and BD and LM . Thus,
AX
by Menelaus Theorem (in triangle DBC with traversal CX) we get XC
=
AP
(using
angle
bisector
theorem)
which
is
symmetric
about
line
XC.
Hence,
PC
9
KL also passes through X. Similarly, N K, DY, M Y are concurrent. Thus,
AL1 , CK1 , BD are also concurrent and same for the other side. Now using
ceva’s and Menelaus, we get l1 M1 ∩N1 K1 = Y and similarly, L1 K1 ∩M1 N1 = X
and K1 M1 ∩ N1 L1 = Z =⇒ (KLM N ) and (K1 L1 M1 N1 ) coincide with polar
circle of 4XY Z =⇒ KLN M ≡ K1 L1 N1 M1 =⇒ ABCD is cyclic.
28. Let A0 be antipode of A in ABC. By Jacobi on triangle ABC we get that
AA0 , BE, CD concur. Since A, O, A0 are collinear we are done
29. RMM Problem
30. Let the reflection of P about F E be PA . Thus, ∠AF PA = ∠AF E −
∠PA F E = ∠AEF −∠P F E = ∠AEF −∠EF C = ∠ACP . Similarly, ∠AEPA =
∠ABP . Now by sine rule in triangle PA F A and PA EA, we get
sin(∠PA F A)
sin(∠F AG)
=
sin(∠PA AE)
sin(∠PA EA)
. Thus, applying trignometric form of ceva’s theorem, we get,
Y sin(∠F APA )
sin(∠PA AE)
=
Y sin(∠PA F A)
sin(∠PA EA)
=
Y sin(∠ACP )
sin(∠ABP )
=1
31. By P26 we know that the locus is a line joining the midpoint of the Aaltitude and midpoint of BC. From here, the problem is well-known. Indeed
using the fact that the antiparallels to sides CB and CA that meet on the
symmedian from A have equal lengths the conclusion follows easily.
32. Let AB∩CD = Z. Now since N A2 = N K.N Q (EGMO chapter 4) = N D2 .
We get that N is center of ABCD. Now trivially AB, DC is the polar of M, I
w.r.t ABCD. By La hire, IM is the polar of Z. By brocard XY is the polar of
Z. So , X, Y, I, M collinear
33. Claim 1: AEDG is concyclic.
Proof: ∠DGA = 1800 − ∠CF D = ∠CED = 1800 − ∠AED.
Claim 2: BDEH is concyclic.
Proof: ∠BED = ∠DEG − ∠BEG = ∠DAG − ∠HDA = ∠BHD.
Claim 3: H,E,F are collinear.
Proof: ∠F ED = ∠F CD = ∠DBH = 1800 − ∠DEH
Claim 4: F ED ∼ CBA.
Proof: ∠EF D = ∠BCA ∠F ED = ∠CBA (see above).
Main proof
Since points H,E,F and G,D,F lie on two seperate lines and CF ||AB, we get
CEF ∼ AEH
CDF ∼ BDG
Therefore,
CE
FE
=
EA
EH
10
BD
DG
=
DC
FD
Multilpying these two equations and using DG = EH,
CE BD
FE
BC
·
=
=
EA DC
FD
CA
CE BD CA
·
·
=1
EA DC BC
Using Ceva’s theorem, we get the required conclusion.
=⇒
34. Unsolved
35. Let BD meet the tangent at A and C on X and Y . Also assume that the
tangents at A and C meet at point Z. Simple angle chasing show that triangle
XY Z is isosceles with ZX = ZY . Also, OAC ∼ ZXY . Now consider R as the
reflection of A wrt to O. Thus, again chasing the angle we get RCA ∼ OAZ.
Thus, we get 2AZ · ZX = XY · OZ. Now, the problem is trivial by Casey’s
Theorem by considering the circles (X), (Y ), (Z) with radius zero and circle
(BOD).
36. Let midpoint of AD, BC, AB be X, Y, O respectively. Then M A2 − M E 2 =
2
2
2
2
And, N B 2 − N E 2 = BC −BD
. Since AD2 + BD2 =
XA2 − XE 2 = AD −AC
4
4
2
2
2
2
2
2
AC +BC we get that M A −M X = N B −N Y =⇒ OA2 +M A2 −M E 2 =
OB 2 + N B 2 − N E 2 =⇒ OE ⊥ M N.
37. Let Y Z cut the incircle in Y 0 , Z 0 and let line ` be the polar of S =⇒ as
polar of S passes through A. Let Y Z cut ` in W and let EF cut ` in T . Thus
we have, (T, S; F, E) = −1 and (W, S; Y 0 , Z 0 ) = −1. Also, as (T, S; F, E) =
−1 =⇒ (AT, AS; AF, AE) = −1 =⇒ (W, S; Y, Z) = −1. So, (W, S; Y, Z) =
(W, S; Y 0 , Z 0 ) = 1. Now, ∠Y 0 XD = ∠Y 0 Z 0 D∠Z 0 Y 0 D = ∠Z 0 XD =⇒ ∠Y 0 XS =
∠Z 0 XS. Also, (W, S; Y 0 , Z 0 ) = −1 =⇒ ∠W XS = 90◦ . Thus, we get
∠Y XS = ∠ZXS. Now suppose that XO meet (XY Z) at G =⇒ ∠Y ZG =
∠Y XG = ∠ZXG = ∠ZY G =⇒ G is the mid point of the arc =⇒ tangent
at G to (XY Z) is parallel to BC =⇒ incircle and (XY Z) are tangent to each
other by taking a homothety from X.
38. Easy
39. It is easy to notice that F M QON is a cyclic pentagon. Now we wish to
prove that Q is the Miquel point of N P AF .
Claim:- BD ∩ AC is the orthocenter of 4OEF .
By Brocard’s Theorem we get that O is the orthocenter of 4EF K where
K = AC ∩ BD. So, K or AC ∩ BD is the orthocenter of 4OEF .
Now let EK ∩ F O = J. So, EK.KJ = AK.KC = KQ.KF =⇒ CQAF is
a cyclic quadrilateral.
Now as F M QN and CQAF are cyclic quadrilaterals and then concur at Q.
Hence, Q must be the Miquel point of N P AF . Hence, M, A, P, Q are concyclic.
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40. First note thatM is midpoint od arc BC.Let M X, M Y meet BC at P, Q
respectively. Then as M QB M BT we get M Q.M T = M B 2 .So we get SP QY
cyclic.By converse of Reim’s theorem we get T SXY cyclic. Applying radical
axis theorem on (ABC), (AXY ), (ST XY ) we get the desired result
41. Let H be midpoint of BC,Let DG interesct (ABC) at N 0 .By homothety
centered at G with ratio -2 , nine point circle goes to (ABC). So, AN 0 k DH.
By converse of Reim we get DHN M cyclic. Let I denote second intersection of (AEF ), (ABC). Its well known that I, M ,orthocenter ofABC collinear.
So (AIDH) cyclic.Applying radical axis on (AIDH), (DHN M ), (ABC) we get
that DH, N M, AI concur at a point T . Applying radical axis on (AIEF ), (EF BC), (ABC)
we get EF, BC, AI concur. This concurrency point is also T . By POP, T N.T M =
T D.T H = T A.T I = T E.T F . Thus (F EN M ) cyclic.
42. Let M1 be the midpoint of the arc BC containing A. Define M2 and M3
similarly.
Observe that ωA is tangent to BC since it is tangent to (I) at D. By the
circles inscribed in segments configuration, KA = M1 D ∩ (ABC). Also note by
the incentre excentre lemma, KA D is the angle bisector of BK1 C.
=⇒
BK1
BD
sin∠BAKA
=
=
sinCAKA
CK1
CD
We use similar reasoning for points B and C to get:
sin∠BAKA sin∠CBKB sin∠ACKC
BD CE AF
·
·
=1
·
·
=
sin∠CAKA sin∠ABKB sin∠BCKC
CD EA F B
By the trig form of Ceva’s, we get the required conclusion.
43. Observe that M P = M B = M C = M Q =⇒ BP CQ is cyclic with
circumcentre M .Also ω is orthogonal wrt (BP CQ).Let BP ∩ CQ = A0 . By the
three tangents lemma ω = (AH 0 ) where H 0 is the orthocentre of A0 BC.Also note
that BQ, P C are altitudes of A0 BC. Now, the problem reads as follows (wrt
triangle A0 BC): Let ABC be a triangle with orthocentre H and orthic triangle
DEF .M is the midpoint of BC. Let BI and CG be the tangents to (AH).(Both
tangents are outside. triangle ABC.)BG ∩ M F = J and CI ∩ M E = K.Prove
that JK is tangent to (AH). In fact JK is tangent to (AH) at H.
Let HA ∩ F E = L Applying Pascal on (HHAF F E),we see that B, L and
M F ∩ HH are collinear.
Lemma 5. B, L, G are collinear.
Proof:By Brocard’s Theorem we get that BL is the Polar of C WRT (AH).
Also as CG is tangent to (AH). So, P ∈ Polar of C. So, B−L−G are collinear.
So, B, G, M F ∩ HH are collinear. =⇒ M F ∩ HH = J and JH is tangent to
(AH).Similarly, KH is also tangent to (AH).
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44. Let Z = BP ∩ CQ, S = P C ∩ BQ. The problem condition implies that
∠BP C = ∠BQC = 90◦ . Let ω 0 the circle with diameter ZS. Since Z lies on the
S-polar wrt. (BP QC), ω 0 is orthogonal to (BP QC). It is easy to prove that
S lies on ω (straightforward angle chasing) and ω is orthogonal to (BP QC),
too, hence ω = ω 0 and Z ∈ ω. Let B 0 , C 0 be the inverse points of B, C wrt.
ω. We have B 0 , C 0 ∈ (BP QC). Since ∠F C 0 C = ∠BC 0 C = 90◦ , BF must be
the polar of C wrt. ω. Similarly, CE is the B-polar. Observe that P lies on
the C-polar and P -polar, then P C is the X-polar and analogously, BQ is the
Y -polar. Hence, S is the pole of XY wrt. ω. Since S ∈ ω, we must have XY
tangent to ω.
45.
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