Length Bashing in Olympiad Geometry
Sina Ghaseminejad
January 2019
1
Introduction
Geometry is an important part of Mathematics Olympiad and Competitions.
In many problems you need to compute some of the lengths or angles. Angle
chasing is simple enough for a lot of us. But sometimes finding the lengths is a
boring and hard job to do; especially if you don’t like calculations in geometry.
Here are some important theorems, lemmas and important triangle lengths that
will appear a lot in bashing geometry problems. In the last section I’ve provided
a bunch of computational problems for you. At the end I have to thank my
dear friend Matin Yousefi for helping me by gathering many of those wonderful
problems and Amir Hossein Parvardi who helped me a lot in writing this article
and improving it. I hope you like it and Happy bashing!
2
Notation and Definition
In triangle 4ABC, we have:
• A, B, C, as the vertices of the triangle.
• a, b, c, as the sides BC, AC, AB, respectively.
• r as the inraduis of the triangle.
• R as the circumradius of the triangle.
• [ABC] as the area of the triangle.
• p as the semiperimeter of the triangle; In other words:
p=
1
a+b+c
2
Length Bashing in Olympiad Geometry
3
Sina Ghaseminejad
Useful Trigonometry Formulas
In this section, there are some useful trigonometry formulas that can help you
while bashing. Sum and difference formulas are the most important ones:
sin (α + β) = sin α cos β + sin β cos α
sin (α − β) = sin α cos β − cos α sin β
cos (α + β) = cos α cos β − sin α sin β
cos (α − β) = cos α cos β + sin α sin β
tan α + tan β
tan (α + β) =
1 − tan α tan β
tan α − tan β
tan (α − β) =
.
1 + tan α tan β
Using the equations above, you will have double and half angle formulas:
sin (2α) = 2 sin α cos α
cos (2α) = cos2 α − sin2 α
cos (2α) = 2 cos2 α − 1
cos (2α) = 1 − 2 sin2 α
2 tan α
tan (2α) =
1 − tan2 α
r
α
1 − cos α
sin
=±
2
2
r
α
1 + cos α
cos
=±
2
2
α 1 − cos α
tan
=
2
sin α
α
sin α
tan
=
2
1 + cos α
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
Factorization may help you during the bashing as well:
1
(cos (α + β) + cos (α − β))
2
1
= (cos (α − β) − cos (α + β))
2
1
= (sin (α + β) + sin (α − β))
2 α+β
α−β
= 2 sin
cos
2
2
α−β
α+β
= 2 sin
cos
2
2
α−β
α+β
sin
= −2 sin
2
2
α+β
α−β
= 2 sin
sin
2
2
cos α cos β =
sin α sin β
sin α cos β
sin α + sin β
sin α − sin β
cos α − cos β
cos α + cos β
4
Some Important Theorems
Here are some theorems which are the bases of computational geometry and
will be used in this article a lot.
Theorem 4.1. (Law of Sines) In the triangle 4ABC we have:
b
c
a
=
=
= 2R.
sin A
sin B
sin C
Theorem 4.2. (Law of Cosines) In the triangle 4ABC we have:
a2 = b2 + c2 − 2bc cos A.
We have the exact same formula for b and c.
Corollary 4.3. We can calculate the cosine of an angle in a triangle directly
using the Law of Cosines:
b2 + c2 − a2
2bc
cos B and cos C can be written in the exact same way.
cos A =
As an exercise, try to prove that the equations below are true:
r
A
p(p − a)
cos
=
2
bc
r
A
(p − b)(p − c)
sin
=
2
bc
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
Theorem 4.4. (Stewart’s Theorem) In the triangle 4ABC, if a line passing
through A meets BC at X, we have:
b2 · BX + c2 · CX = a(AX 2 + BX · CX).
Theorem 4.5. (Angle Bisector Theorem) In the triangle 4ABC, we have:
CDa
sin B
b
=
=
c
BDa
sin C
If and only if ADa is an angle bisector.
Theorem 4.6. (Medians Theorem) In the triangle 4ABC, we have:
b
sin B
sin (BAMa )
=
=
c
sin C
sin (CAMa )
If and only if AMa bisects BC.
Theorem 4.7. (Ptolemy’s Inequality) If the vertices of the quadrilateral are A,
B, C, and D in order, then the theorem states that:
AB · CD + AD · BC ≥ AC · BD.
And the equality holds if and only if ABCD is cyclic.
Theorem 4.8. (Ceva’s Theorem) We call a line drawn from one of the vertices of triangle meeting the opposite side, a Cevian. In triangle 4ABC, three
Cevians AA0 , BB 0 , and CC 0 are concurrent if and only if we have:
AC 0 BA0 CB 0
·
·
= 1.
C 0 B A0 C B 0 A
Theorem 4.9. (Trigonometric Ceva) In triangle 4ABC, three Cevians AA0 ,
BB 0 , and CC 0 are concurrent if and only if we have:
sin CAA0 sin ABB 0 sin BCC 0
·
·
= 1.
sin A0 AB sin B 0 BC sin C 0 CA
Theorem 4.10. (Menelaus’s Theorem) Given a triangle 4ABC, and a transversal line ` that meets BC, AC, and AB at points D, E, and F respectively, then
we will have:
AF BD CE
·
·
= 1.
F B DC EA
The inverse of this theorem is also true.
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
Theorem 4.11. (Area of the Triangle) The area of a triangle can be calculated
using following formulas which can be proved easily.
Let ha be the altitude from vertex A, then the area of 4ABC can be written as
these equations:
a · ha
[ABC] =
2
bc · sin A
[ABC] =
2
Let R, r and p be the circumradius, inradius and semiperimeter of 4ABC
respectively, we will have the following equations as well:
[ABC] = pr
abc
4R
2
[ABC] = 2R sin A sin B sin C
p
[ABC] = p(p − a)(p − b)(p − c)
[ABC] =
Theorem 4.12. (Euler Line) In the triangle 4ABC, let O be the circumcenter, G be the centroid, N9 be the center of Nine-Point Circle, and H be the
orthocenter.
Then, O, G, N9 and H are colinear and they share the ratio:
OG : GN9 : N9 H = 2 : 1 : 3.
Theorem 4.13. (Nagel Line) In the triangle 4ABC, let I be the incenter, G
be the centroid, and N be the Nagel Point.
Then, I, G, and N are colinear and they share the ratio:
IG : GN = 1 : 2.
Theorem 4.14. In every triangle 4ABC, with orthocenter H and circumcenter
O, if M is the midpoint of BC, we have:
AH : OM = 2 : 1.
Theorem 4.15. In every triangle 4ABC, with Centroid G, if M is the midpoint of BC, we have:
AG : GM = 2 : 1.
5
Some Lemmas for Problem Solving
Here are some important lemmas (which are not always considered trivial and
in many situations you need to mention and prove them) that will be useful for
real Olympiad-based problem solving. Using these lemmas you can tackle many
hard problems.
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
Lemma 5.1. (Lemma of Ratio) In the triangle 4ABC, if a line passing through
A meet BC at X, we have:
BX
c · sin (BAX)
=
.
CX
b · sin (CAX)
Corollary 5.2. In the quadrilateral ABCD, we have:
sin BAD
sin B BD
=
·
.
sin CAD
sin C CD
Lemma 5.3. (The Lemma of Perpendicularly) For a line BC, Point A on the
plane and point H on BC are assumed. For a point P that lies on AH, We
have:
AB 2 − AC 2 = P B 2 − P C 2 .
If and only if AH⊥BC.
Lemma 5.4. For four angles α , α0 , β , β 0 , if we know that:
sin β
sin α
=
0
sin α
sin β 0
and α + α0 = β + β 0
Then we can conclude that: α = β and α0 = β 0 .
Lemma 5.5. For four angles α , α0 , β , β 0 , if we know that:
sin β
sin α
=
sin α0
sin β 0
and α − α0 = β − β 0
Then we can conclude that: {α, β} = {α0 , β 0 }.
Note that sin θ = sin γ, doesn’t give us θ = γ right away. It has two cases:
One of the cases is that θ and γ are equal angles, the other case is that θ and
γ are supplementary (their sum is equal to 180o ).
Lemma 5.6. (Trigonometric Ptolemy) Let A, B, C, D, be four points on the
plane. ABCD is a cyclic quadrilateral of this order, if and only if we have:
AB · sin BAC + AD · sin CAD = AC · sin BAD.
Lemma 5.7. (Trigonometric Colinearity) Let A, B, C, D, be four points on
the plane. B, C, D are colinear if and only if we have:
sin BAC
sin CAD
sin BAD
+
=
.
AD
AB
AC
Lemma 5.8. (Isogonal Lines Lemma) In the triangle 4ABC, P is a point on
>
segment a and Q is a point on arc BC, if we know that ∠P AB = ∠CAQ, we
will have:
AP · AQ = bc.
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
Proof. Since we know that ∠P AB = ∠CAQ, it’s obvious that ∠ABP = ∠AQC =
∠B, so we get:
b
AQ
4ABP ∼ 4AQC ⇒
=
AP
c
=⇒ AP · AQ = bc.
Corollary 5.9. Let I be the incenter of 4ABC and Ia be the center of the
excircle relative to the vertex A, then this lemma can be written for I and Ia as
well.
In the triangle 4ABC, we have:
AI · AIa = bc.
Proof. We know that ∠BAIa = ∠IAC = ∠ A2 ; with some angle chasing we can
see that ∠ABIa = ∠AIC = ∠B + ∠ A+C
2 . So we have:
4ABIa ∼ 4AIC ⇒
c
AI
=
AIa
b
=⇒ AI · AIa = bc.
6
Geometric Calculations
6.1
Median
In the triangle 4ABC, we have:
• Ma , Mb , Mc , as the midpoints of sides a, b, c respectively.
• ma , mb , mc , as the medians from midpoints of a, b, c, respectively.
6.1.1
Length of ma
Using the Law of Cosines, we have:
m2a
2
=c +
a 2
2
a2
− 2c
cos B = c +
− ac
2
4
a
m2a = c2 +
2
a2
a2
c2
b2
−
−
+
4
2
2
2
c2
b2
a2
+
−
2
2
4
r
2
2
2b + 2c − a2
=⇒ ma =
.
4
m2a =
Page 7
a2 + c2 − b2
2ac
Length Bashing in Olympiad Geometry
6.2
Sina Ghaseminejad
Altitude
In the triangle 4ABC, we have:
• H as the Orthocenter of triangle.
• ha , hb , hc , as the altitudes from A, B, C, respectively.
• Ha , Hb , Hc are where the altitudes ha , hb , hc , touch a, b, c, respectively.
6.2.1
Length of ha
i) Using the Law of Sines, we have:
sin B
sin 90
=
ha
c
=⇒ ha = c · sin B.
ii) Using the formulas of the area, we will get:
bc · sin A
a · ha
=
2
2
bc · sin A
⇒ ha =
.
a
a
By the Law of Sines we know that sin A = 2R , using that, we get:
=⇒ ha =
6.2.2
bc
.
2R
Length of BHa
In the triangle 4ABHa , we have:
sin (90 − B)
sin B
=
.
BHa
c
Since we know that sin (90 − B) = cos B, we will get:
=⇒ BHa = c · cosB.
We can also change it this way:
BHa = c
a2 + c2 − b2
2ac
=⇒ BHa =
a2 + c2 − b2
2a
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Length Bashing in Olympiad Geometry
6.2.3
Sina Ghaseminejad
Length of AH
We know that HHb CHa is a cyclic quadrilateral; So:
AH · AHa = AHb · AC.
Knowing that AHa = b · sin C and AHb = c · cos A, we get:
AH · b · sin C = c · cos A · b
=⇒ AH = a · cot A.
Using
a
sin A
= 2R, it can be written like:
=⇒ AH = 2R · cos A.
6.2.4
Length of HHa
Since we know that HHb CHa is a cyclic quadrilateral; So we get ∠BHHa = ∠C.
Using the law of sines in triangle 4BHHa and BH = b · cot B, we have:
=⇒ HHa = b · cot B · cos C.
Which can be transformed as:
=⇒ HHa = 2R · cos B · cos C.
6.2.5
Length of Hb Hc
With some angle chasing we will know that ∠AHb Hc = ∠ABC and ∠AHc Hb =
∠ACB, this will lead us to the similarity 4AHb Hc ∼ 4ABC; so:
AHb
Hb Hc
=
= cos A
a
c
=⇒ Hb Hc = a · cos A.
Page 9
Length Bashing in Olympiad Geometry
6.3
Sina Ghaseminejad
Internal and Exterior Angle Bisectors
In the triangle 4ABC, we have:
• I as the incenter of the triangle.
• da , db , dc , as the angle bisectors of ∠A, ∠B, ∠C.
• Da , Db , Dc are where the bisectors da , db , dc , meet a, b, c, respectively.
• Ea , Eb , Ec are where the incircle touch a, b, c, respectively.
• d0a , d0b , d0c as the exterior angle bisectors of ∠A, ∠B, ∠C, respectively.
• Da0 , Db0 , Dc0 are where d0a , d0b , d0c meet the extension of BC, AC, AB,
respectively.
6.3.1
Length of da
Using the Stewart’s Theorem, we will have:
a2 bc
ab
ac
2
2
2
a da +
+c
+b
(b + c)2
b+c
b+c
2
da = bc 1 −
6.3.2
⇒ d2a = bc − BDa · CDa
2
a2
2b2 c2
(b + c2 − a2 ) + 2bc
=
(b + c)2
(b + c)2
2bc
2 2
4b c
A
d2a =
· cos2
2
(b + c)
2
2bc
A
=⇒ da =
· cos
.
b+c
2
d0a and Da0
We can calculate the length of d0a , just like da with the Stewart’s Theorem.
The results are quite similar:
⇒ (d0a )2 = BDa0 · CDa0 − bc
A
2bc
0
· sin
=⇒ da =
|b − c|
2
Page 10
Length Bashing in Olympiad Geometry
6.3.3
Sina Ghaseminejad
Length of BDa
Using the Angle Bisector Theorem, we have:
BDa
c
=
CDa
b
⇒
6.3.4
BDa
c
=
BDa + CDa
b+c
BDa
c
=
a
b+c
a·c
=⇒ BDa =
.
b+c
Length of AI
i) Using the Angle Bisector Theorem in the triangle 4ABDa , we will get:
c
AI
=
IDa
BD
c
b+c
AI
= ac =
IDa
a
b+c
AI
b+c
=
AI + IDa
a+b+c
b+c
AI
=
⇒
ADa
2p
b+c
2bc
A
b+c
AI = ADa ·
=
· cos
·
2p
b+c
2
2p
bc
A
=⇒ AI =
· cos
.
p
2
ii) If we connect I to Ec , it’s obvious that IEc ⊥AB and IEc = r; in the triangle
4AIEc , we will get:
r
.
⇒ AI =
sin A2
iii) Having Ec and the incircle again, it’s easy to prove that AEc = p − a; in
the triangle 4AIEc , we will get:
=⇒ AI =
p−a
.
cos A2
Page 11
Length Bashing in Olympiad Geometry
6.4
Sina Ghaseminejad
Mixtilinear Incircles
A Mixtilinear Incircle is a circle tangent to two of a triangle’s segments, and the
circumcircle of that triangle. There are three Mixtilinear Incircles in a triangle,
we show Mixtilinear Incircle opposite to vertex A (the circle that is tangent to
AB, AC and circumcircle of 4ABC) with Γa .
Γb and Γc are defined is the same way.
Let Γa touch AB and AC at E and F respectively; it can be proved that I lies
on EF (which is actually an special case of Sawayama Thebault’s Theorem).
Using that, since AI⊥EF we will have:
AE = AF =
AI
.
cos A2
Using the length of AI directly, we get:
=⇒ AE = AF =
bc
.
p
Lengths of BE and CF are easy to find as well:
BE = c −
=⇒ BE =
c(p − b)
p
CF = b −
=⇒ CF =
6.5
bc
p
bc
p
b(p − c)
p
T and T 0
We defined T and T 0 as the points where the internal and exterior bisectors
of ∠A meet the circumcircle of the triangle 4ABC; or we can say that T is
>
>
the midpoint of arc BC not containing A, and T 0 is the midpoint of arc BC
containing A. We use the same notations as the previous parts in this section:
• da , db , dc , as the angle bisectors of ∠A, ∠B, ∠C.
• Da , Db , Dc are where the bisectors da , db , dc , meet a, b, c, respectively.
• Ea , Eb , Ec are where the incircle touch a, b, c, respectively.
• d0a , d0b , d0c as the exterior angle bisectors of ∠A, ∠B, ∠C, respectively.
• Da0 , Db0 , Dc0 are where d0a , d0b , d0c meet the extension of BC, AC, AB,
respectively.
Page 12
Length Bashing in Olympiad Geometry
6.5.1
Sina Ghaseminejad
Length of AT
An special case of the Isogonal Lines Lemma that we proved, is when A, P , Q
are collinear; And that is when P ≡ Da and Q ≡ T .
Using the lemma, we get:
ADa · AT = bc.
2bc
We know that ADa = b+c · cos A2 , by putting that in our equation we have:
2bc
A
AT ·
· cos
= bc
b+c
2
=⇒ AT =
6.5.2
b+c
.
2 cos ( A2 )
Length of AT 0
Here, we need to use Da0 again. it’s obvious that T 0 , A and Da0 are collinear;
with angle chasing we know that ∠ABDa0 = ∠AT 0 C and ∠Da0 AB = ∠CAT 0 ,
so:
c
d0a
4ABDa0 ∼ 4AT 0 C ⇒
=
AT 0
b
2bc
0
Just like the length of AT , by putting da = |b−c| · sin A2 in the equation, we
get:
|b − c|
.
=⇒ AT 0 =
2 sin A2
6.5.3
Length of BT, CT and IT
Reading the title, you might have guessed the point of this part: BT , CT and
IT are all equal. with some angle chasing, it’s simple to show that ∠T BI =
∠T IB = ∠ A+B
and ∠T CI = ∠T IC = ∠ A+C
2
2 ; so BT = CT = IT (and it’s not
hard to prove they are all equal to T Ia as well).
Adding T Ma to the structure, where Ma is the midpoint of side BC, we will
have:
A
a
T B · cos
=
2
2
a
.
=⇒ T B = T C = T I =
2 cos A2
Page 13
Length Bashing in Olympiad Geometry
6.5.4
Sina Ghaseminejad
T and T’ images on AC
In this part, without loss of generality, we assume that AC > AB.
S and S 0 will be images of T and T 0 on the segment AC.
By the definition, T 0 S 0 ⊥AC and T S⊥AC, using the Law of Sines, we will have
the following results:
b+c
=⇒ AS = CS 0 =
2
|b
−
c|
=⇒ AS 0 = CS =
2
6.6
Excircles and Excenters
Just like Mixtilinear Incircles, there are three Excircles in a triangle. We define
the Excircle opposite to vertex A with ωa .
ωb and ωc are defined in the same way. In this section, in triangle 4ABC, we
have:
• Ia , Ib , Ic , as the centers of ωa , ωb , ωc , respectively.
• Xa , Xb , Xc , are where ωa , ωb , ωc , touch a, b, c, respectively.
• ra , rb , rc , as inradii of excircles ωa , ωb , ωc , respectively.
• Ea , Eb , Ec are where the incircle touch a, b, c, respectively.
• I as the incenter of the triangle.
6.6.1
Length of AIa
i) Using the Isogonal Lines Lemma again, and having AI =
AIa ·
bc
· cos
p
bc
p
· cos
A
2
, we get:
A
= bc.
2
With some calculations, we have:
=⇒ AIa =
bc
· cos
p−a
A
.
2
ii) Let ωa touch AB at Zc .
It’s obvious that and Ia Zc ⊥AZc and Ia Zc = ra ; using that in the triangle
4AZc Ia , we get:
ra
.
=⇒ AIa =
sin A2
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Length Bashing in Olympiad Geometry
Sina Ghaseminejad
iii) Having Zc again; it’s easy to prove that AZc = p; using that in the triangle
4AZc Ia , we get:
p
.
=⇒ AIa =
cos A2
6.6.2
Length of AIb
If we connect Ib to Xb , it’s obvious that Ib Xb ⊥AC and Ib Xb = rb ; using that,
we will have:
rb
AIb =
sin 90 − A2
rb
.
=⇒ AIb =
cos A2
6.6.3
Length of ra
i) Calculating the length of ra is quite like proving r =
of 4ABC to calculate it.
Consider ra ; we know that:
[ABC]
;
p
we use the area
[ABC] = [AIa B] + [AIa C] − [CIa B].
p · ra
p · ra
a · ra
⇒ [ABC] =
+
−
2
2
2
With a little calculations, we get:
[ABC] = (p − a) · ra
=⇒ ra =
[ABC]
.
p−a
In the exact same way, we can prove that rb =
[ABC]
p−b
and rc =
[ABC]
p−c .
ii) Let ωa touch AB at Zc . It’s obvious that and Ia Zc ⊥AZc ; If we connect I
to Ec , we know that IEc ⊥AB as well. It is well known that A, I and Ia are
colinear, so we can prove that 4AIEc ∼ 4AIa Zc .
Since in triangle 4AIEc we have AEc = p−a , IEc = r and in triangle 4AIa Zc
we have AZc = p , Ia Zc = ra , by using the similarity:
IEc
AEc
=
AZc
IZc
p−a
r
=
p
ra
By putting r =
[ABC]
,
p
we get the same result:
=⇒ ra =
[ABC]
p−a
Page 15
Length Bashing in Olympiad Geometry
7
Sina Ghaseminejad
Training Problems
Here are some problems gathered from Mathematics Olympiads all over the
world that can be solved by length bashing for some exercise. You may solve
some of these in much easier ways; But please don’t! There’s much more benefit
in solving them by calculations than other ways.
1. (Blanchet’s Theorem) In an acute triangle 4ABC, F is the foot of the
perpendicular from C to AB and P is a point on CF. Let the lines AP and BP
meet BC and AC at D and E, respectively. Show that the angles ∠EF C and
∠DF C are equal.
2. Quadrilateral XABY is inscribed in the semicircle ω with diameter XY .
Segments AY and BX meet at P . Point Z is the foot of the perpendicular from
P to line XY . Point C lies on ω such that line XC is perpendicular to line AZ.
Let Q be the intersection of segments AY and XC. Prove that:
CY
AY
BY
+
=
.
XP
XQ
AX
3. Let B1 , C1 be the midpoints of sides AC, AB of a triangle 4ABC. The
rays CC1 , BB1 meet the tangents to the circumcircle at B and C at K and L
respectively. Prove that the angles ∠BAK and ∠CAL are equal.
4. Equal circles ω1 and ω2 are passing through each others centers. The triangle
4ABC is inscribed into the circle ω1 such that lines AC and BC are tangent
to the circle ω2 . Prove that: cos A + cos B = 1.
5. In a right angled-triangle 4ABC, ∠ACB = 90o . Its incircle Γ meets BC,
AC, AB at D,E,F respectively. AD cuts Γ at P . If ∠BP C = 90o , Prove that:
AE + AP = P D.
6. In a triangle 4ABC, let Ma , Mb , Mc be the midpoints of the sides BC, CA,
> > >
AB, respectively, and Ta , Tb , Tc be the midpoints of the arcs BC, CA, AB of the
circumcircle of 4ABC, not containing the vertices A, B, C, respectively. For
i ∈ {a, b, c}, let wi be the circle with Mi Ti as diameter. Let pi be the common
external common tangent to the circles wj and wk (for all {i, j, k} = {a, b, c})
such that wi lies on the opposite side of pi than wj and wk do. Prove that the
lines pa , pb , pc form a triangle similar to 4ABC and find the ratio of similitude.
7. Let 4ABC be a triangle with circumradius R, perimeter P and area K.
Determine the maximum value of: KP
R3
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8. Bisector of side BC intersects circumcircle of triangle 4ABC in points P and
Q. Points A and P lie on the same side of line BC. Point R is an orthogonal
projection of point P on line AC. Point S is middle of line segment AQ. Show
that points A, B, R, S lie on one circle.
9. Let D, E, F be points on the sides BC, CA, AB respectively of a triangle
4ABC such that BD = CE = AF and ∠BDF = ∠CED = ∠AF E. Show that
4ABC is equilateral.
10. Let 4ABC be an acute-angled triangle with altitudes AD, BE, and CF .
Let H be the orthocentre, that is, the point where the altitudes meet. Prove
that:
AB · AC + BC · CA + CA · CB
≤ 2.
AH · AD + BH · BE + CH · CF
11. Let 4ABC be an acute angled triangle. Let D, E, F be points on BC, CA, AB
such that AD is the median, BE is the internal bisector and CF is the altitude. Suppose that ∠F DE = ∠C, ∠DEF = ∠A and ∠EF D = ∠B. Show that
4ABC is equilateral.
12. The incircle of a triangle 4ABC touches the side AB and AC at respec>
tively at X and Y . Let K be the midpoint of the arc AB on the circumcircle
of 4ABC. Assume that XY bisects the segment AK. What are the possible
measures of angle ∠BAC?
13. Let ABCD be a cyclic quadrilateral. Let P , Q, R be the feet of the perpendiculars from D to the lines BC, CA, AB, respectively. Show that P Q = QR
if and only if the bisectors of ∠ABC and ∠ADC are concurrent with AC.
14. Bisector of angle ∠BAC of triangle 4ABC intersects circumcircle of this
triangle in point D 6= A. Points K and L are orthogonal projections on line AD
of points B and C, respectively. Prove that AD ≥ BK + CL.
15. Let N be a point on the longest side AC of a triangle 4ABC. The perpendicular bisectors of AN and N C intersect AB and BC respectively in K and
M . Prove that the circumcenter O of 4ABC lies on the circumcircle of triangle
4KBM
16. Let 4ABC be an acute angled triangle. D is the midpoint of BC. E,
F are on sides AC, AB such that ∠EDF = ∠A and DE = DF . Prove that:
DE ≥
AB + AC
.
4
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17. Let ABCD be a cyclic quadrilateral inscribed in a circle. Let E, F , G,
> > > >
H be the midpoints of arcs AB, BC, CD, AD of the circle respectively. Suppose that AC ·BD = EG·F H. Show that AC, BD, EG, F H are all concurrent.
18. In a scalene triangle 4ABC the altitudes AA1 and CC1 intersect at H, O
is the circumcenter, and B0 the midpoint of side AC. The line BO intersects
side AC at P , while the lines BH and A1 C1 meet at Q. Prove that the lines
HB0 and P Q are parallel.
19. Triangle 4ABC has integer-length sides. Line ` is tangent to the circumcircle of 4ABC at A. Let S be the intersection of ` and BC. if AS is an
integer as well, Prove that: gcd(AB, AC) > 1.
20. Let AD be an angle bisector of triangle 4ABC. Perpendiculars from
D to the other sides of 4ABC, meet AB at X and AC at Y , Let P be the
intersection of segments BY and CX. Prove that: AK⊥BC.
21. Let I and G be the incenter and centroid of triangle 4ABC respectively.
Prove that IG is perpendicular to BC if and only if we have: AB + BC = 3BC.
22. (Droz-Farny Theorem) Let H be the orthocenter of the triangle 4ABC.
Let L1 and L2 be any two mutually perpendicular lines through H. Let A1 , B1
, and C1 be the points where L1 intersects the side lines BC, CA, and AB, respectively. Similarly, let Let A2 , B2 , and C2 be the points where L2 intersects
those side lines. Prove that the midpoints of the three segments A1 A2 , B1 B2 ,
and C1 C2 are collinear.
23. Let ABCD be a trapezoid with AB k CD. Lines AC and BD meet at
P . H1 and H2 are the orthocenters of triangles 4AP D and 4BP C respectively. Let M be the midpoint of H1 H2 . Prove that: M P ⊥CD.
24. Let P , Q be two points on the side BC of triangle 4ABC such that
2P Q = BC and P is closer to B than Q. Perpendiculars at P and Q to BC,
meet AB and AC at X and Y respectively. of O is the circumcenter of 4ABC,
Prove that AXOY is cyclic.
25. In a quadrilateral ABCD, Lines AC and BD meet at P . Let H1 and
H2 be the orthocenters of 4ABP and 4CDP respectively. Let G1 and G2 be
the centroids of 4ADP and 4BCP respectively. Prove that: H1 H2 ⊥G1 G2 .
26. Let ω be the incircle of triangle 4ABC with center I. Circle ω touches AB
and AC at E and F respectively. Line BI and CI intersect EF at X and Y
respectively. If M is the midpoint of BC, Prove that 4M XY is equilateral if
and only if ∠A = 60o .
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27. Let Ω be the circumcircle of triangle 4ABC and G be its centroid. Lines
AG, BG and CG meet Ω at A0 , B 0 and C 0 respectively. Prove that:
X
X
A0 G ≥
AG.
28. 4ABC is a triangle with incenter I, Midpoints of AB and AC are M
and N . Line CI meets M N at K. We call the line perpendicular to M N at K,
`1 and the line parallel to BI passing through N , `2 . Let P be the intersection
of `1 and `2 . Prove that: P I⊥AC.
29. If I is the incenter of triangle 4ABC and a circle passing through A and
I meets AB and AC at E and F , And M is the midpoint of EF , Show that:
∠BM C > 90o
30. The incircle of triangle 4ABC touches the sides AB, BC and CA at
D, E and F respectively. Perpendicular line from E to DF , meets DF at X.
Prove that two angles ∠BXE and ∠CXE are equal.
31. BB 0 and CC 0 are two altitudes of triangle 4ABC, Angle bisector of ∠A
meets B 0 C 0 and BC at E and F . Let `1 be the perpendicular line to B 0 C 0 at
E and `2 be the perpendicular line to BC at F . The intersection of `1 and `2
is X. Prove that AX is a median of 4ABC.
32. (Leibniz Theorem) Let G be the centroid of triangle 4ABC. Prove that,
for any point P on the plane, we have:
X
P A2 =
1X
AB 2 + 3P G.
3
>
33. Circumcircle of triangle 4ABC is Ω. Let K be the midpoint of arc BAC.
Prove that: KB + KC ≥ AB + AC.
34. Points P and Q are inside the triangle 4ABC such that ∠P BA = ∠QBC,
∠P CB = ∠QCA and ∠P AB = ∠QAC. Prove that:
AP · sin BP C = AQ · sin BQC.
35. AA0 , BB 0 and CC 0 a altitudes of triangle 4ABC. We extent lines AA0 ,
CC 0 from A0 , C 0 and extend BB 0 from B. Let X, Y and Z be points on lines
AA0 , BB 0 and CC 0 such that AA0 = A0 X, CC 0 = C 0 Z and BY = 3BB 0 . Prove
that angles ∠ABC and ∠XY Z are equal.
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Length Bashing in Olympiad Geometry
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36. Let M be the midpoint of side BC in triangle 4ABC. Points X, Y
are outside the triangle and ∠AXB = ∠AY C = 90o and ∠XBA = ∠Y CA.
a) Prove that: XM = Y M .
b) Prove that: ∠XM B = 2∠XBA.
37. Let r be the radius of the circle inscribed in triangle 4ABC and Ωa be
the excircle relative to vertex A and Ia be its center. Ωa touches lines BC, AB
and AC at D, E and F respectively. EF meets Ia B and Ia C at X and Y . The
intersection of XC and Y B is K. Prove that the distance of K and BC is equal
to r.
38. A circle is inscribed in Trapezoid ABCD with AB k CD. Lines AC and
BD meet at P . Prove that: ∠CP D > 90o .
39. Line ` is passing through triangle 4ABC. AX, BY and CZ are perpendiculars to `. Prove that the centroid of 4ABC lies on ` if and only if we
have: AX = BY + CZ.
40. A Line passing through the centroid of triangle 4ABC, meets AB at
X and AC at Y . Prove that:
AX AY
·
≥ 4.
BX CY
41. In triangle 4ABC, Points X, Y are chosen on AB, AC such that BX = CY .
If M is the midpoint of BC and N is the midpoint of XY , Prove that M N is
parallel to the angle bisector of ∠A.
42. Let Ω be the circumcircle of triangle 4ABC. A line passing through the
>
>
centroid of 4ABC meets arcs AB and AC of Ω at X and Y . Prove that:
AX · AY = BX · BY + CX · CY.
43. Let AH, AD and AM be an altitude, an angle bisector and a median
through A in triangle 4ABC respectively. If we know that DH < DM , Prove
that: BC < 2AD
44. ABCD is a quadrilateral and points E, F are on BC, AD such that
BC
AD
BE = AF = n for some real number n. Prove that:
AD · BC + n(n − 1) · AB 2 + n · AB · DC
.
2n2
(Here [ABCD] represents the area of quadrilateral ABCD).
[ABEF ] ≤
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Length Bashing in Olympiad Geometry
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45. In the triangle 4ABC, points X, X 0 are on BC, Y , Y 0 are on CA and
Z, Z 0 are on AB such that BX = X 0 C, CY = Y 0 A and AZ = Z 0 B. Prove
that: [XY Z] = [X 0 Y 0 Z 0 ].
(Here [ABC] represents the area of triangle 4ABC).
46. The incircle of triangle 4ABC with center I touches AB, BC and CA
at D, E and F . ` is a line passing through A parallel to DF . Lines ED and
EF meet ` at K and J. Show that: ∠JIK < 90o .
47. The incircle of triangle 4ABC with center I touches AB, BC and CA
at D, E and F . ` is a line passing through A parallel to BC. Lines ED and
EF meet ` at K and J. Show that: ∠JIK < 90o .
48. Let ha , hb and hc be the altitudes of triangle 4ABC with circumradius R
and area S, Prove that:
r
R
1X 1
3
.
≥
3
ha
2S 2
49. In the triangle 4ABC, Point A0 is symmetry of A with respect to BC.
B 0 and C 0 are defined in the same way. Prove that: [A0 B 0 C 0 ] ≤ 4[ABC]
. (Here [ABC] represents the area of triangle 4ABC).
50. If G is the centroid of triangle 4ABC and angles ∠BAG and ∠ACG
are equal. Prove that:
2
sin CAG + sin CBG ≤ √ .
3
51. Triangle 4ABC has circumcenter O and circumradius R. Circumcenter
of triangle 4BOC is Ωa . The line AO meets Ωa for the second time at A0 . B 0
and C 0 are defined the same way. Prove that:
OA0 · OB 0 · OC 0 ≥ 8R2 .
52. Let ha , hb and hc be the altitudes of triangle 4ABC with semiperimeter P . Prove that:
X
√
ha ≤ 3P.
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53. Let ha , hb , hc , be the altitudes of 4ABC from A, B, C and ra , rb
, rc be inradii of excircles relative to vertices A, B, C. Prove that:
X ra
≥ 3.
ha
54. Triangle 4ABC with circumradius R and inradius r is given. Prove that:
X A
B
5
r
· sin
≤ +
.
sin
2
2
8 4R
55. Circumcircle of triangle 4ABC is Ω. Let the angle bisector of ∠A meet BC
and Ω at D and T . We choose an arbitrary point X on BT . The intersection
point of XD and T C is Y . Show that the angle ∠XAY is greater that ∠BAC.
56. Circle ω with radius R is inscribed inside trapezoid ABCD with AB k CD.
Show that: AB · CD ≥ 4R2 .
57. Triangle 4ABC has integer-length sides and AC = 2007. The Internal
bisector of ∠A meets BC at D. Given that AB = CD. Determine AB and BC.
58. (Casey’s Theorem) Assume the line ` is the radical axis of two circle C1 and
C2 with centers O1 and O2 . For an arbitrary point A on the plane, H is the
projection of A on `. Prove that:
|PCK1 − PCK2 | = 2AH · O1 O2 .
59. A triangle 4ABC is given with AB > AC. Line ` is tangent to the
circumcircle of 4ABC at A. A circle centered in A with radius |AC| cuts AB
in the point D and the line ` in points E, F (such that C and E are in the
same halfplane with respect to AB). Prove that the line DE passes through
the incenter of 4ABC.
60. Let P be a moving point on side BC of triangle 4ABC. Line `1 is drawn
from P such that `1 and P B make a specific angle α. Line `2 is also drawn from
P the same way such that `2 and P C make a specific angle β. Lines `1 and `2
meet AB and AC at K and L. Prove that the circumcircle of triangle 4AKL
passes trough a fixed point.
61. Triangle 4ABC with circumcircle Γ is given. Let M be the midpoint
>
>
of arc ABC and N be the midpoint of arc BAC. Prove that the incenter of
4ABC lies on line M N if and only if we have: AC + BC = 3AB.
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62. Circle ω is internally tangent to circle Ω at K. Let A be a point on Ω
and chords AB and AC of Ω be tangent to ω at M and N . AK meets M N at
S. Prove that the angles ∠M SB and ∠N SC are equal.
63. ABCD is a convex quadrilateral with ∠BAC = 30◦ , ∠CAD = 20◦ ,
∠ABD = 50◦ , ∠DBC = 30◦ . If the diagonals intersect at P , show that:
P C = P D.
64. Let H be the orthocenter of an acute-angled triangle 4ABC. Show that
the triangles 4ABH, 4BCH and 4CAH have the same perimeter if and only
if the triangle 4ABC is equilateral.
65. In the rectangle ABCD, Points X, Y are chosen on sides DC, BC such
that CX 6= CY . Let P be the intersection of DY and BX. If AP ⊥XY , show
that DX = BY .
66. Let P , Q and R be the midpoints of sides AB, AC and BC in triangle
4ABC. Points X and Y are on P R and QR such that BX⊥P R and CY ⊥QR.
if O0 is the circumcenter of P QR, prove that RO0 bisects XY .
67. In the isosceles triangle 4ABC with AB = AC, Point K is on segment
BC such that BK = 2KC and point P is chosen on AK such that angles
∠BP Kand ∠BAC are equal. Prove that: 2∠CP K = ∠BAC.
68. Circle ω is inscribed in triangle 4ABC and touches AC at B1 . Points
E and F are chosen on ω such that ∠AEB1 = 90◦ and ∠CF B1 = 90◦ . Let M
>
be the midpoint of AC. Prove that lines DM , CF and AE are concurrent.
69. In an acute-angled triangle 4ABC with AB 6= AC and A = 60◦ , O and
H are circumcircle and orthocenter respectively. if OH meets AB and AC (not
their extension) at P and Q and we show the area of triangle 4AP Q with S
S
.
and area of quadrilateral BP QC with R, find all possible values of R
70. Circle ω is internally tangent to circle Ω at S. Let A be a point on Ω
and chords AB and AC of Ω be tangent to ω at E and F . Let T be the mid>
point of arc BC. Show that lines BC, EF and T S are concurrent.
71. Let 4ABC be a triangle with incenter I, incircle ω and circumcircle Γ.
Let M, N, P be the midpoints of sides BC, CA, AB and let E, F be the tangency points of ω with CA and AB, respectively. Let U, V be the intersections
of line EF with line M N and line M P , respectively, and let X be the midpoint
>
of arc BAC of Γ.
a) Prove that I lies on ray CV .
b) Prove that line XI bisects U V .
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Length Bashing in Olympiad Geometry
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72. Circle ω is tangent to line ` at M . Let A be the point on ω such that AM is
the diameter. Points X and Y are moving on ` such that always M X · M Y = k
for some real k and M is between X, Y . Lines AX and AY meet ω at E and
F . Prove that by Moving X and Y , EF passes trough a fixed point.
73. In an equilateral triangle 4ABC, circle ω meets side AB at D, E; side
BC at F , G; and side CA at H, I internally (D, E, F , G, H and I are on ω in
order). Show that:
AD + BF + CH = BE + GH + IA.
74. Let 4ABC be an acute-angled triangle with AB 6= AC. Let H be the
orthocenter of triangle 4ABC, and let M be the midpoint of the side BC. Let
D be a point on the side AB and E a point on the side AC such that AE = AD
and the points D, H, E are on the same line. Prove that the line HM is perpendicular to the common chord of the circumscribed circles of triangle 4ABC
and triangle 4ADE.
75. Chord AB of circle Γ is assumed. P is an arbitrary point on AB. Circles ω1 with radius r1 and ω2 with radius r2 are on two sides of AB such that
they’re tangent to AB at P and internally tangent to Γ. Show that with changing the position of P , the ratio rr21 won’t change.
76. The lengths of the sides of a convex hexagon ABCDEF satisfy AB = BC,
CD = DE, EF = F A. Prove that:
DE
FA
3
BC
+
+
≥ .
BE
DA F C
2
77. Given a triangle 4ABC satisfying AC + BC = 3AB. The incircle of
triangle 4ABC has center I and touches the sides BC and CA at the points
D and E, respectively. Let K and L be the reflections of the points D and E
with respect to I. Prove that the points A, B, K, L lie on one circle.
78. In an acute-angled and not isosceles triangle 4ABC. we draw the median AM and the height AH. Points Q and P are marked on the lines AB
and AC, respectively, so that the QM ⊥ AC and P M ⊥ AB. The circumcircle of 4P M Q intersects the line BC for second time at point X. Prove that
BH = CX.
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