Length Bashing in Olympiad Geometry

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. Anglechasing is simple enough for a lot of us. But sometimes finding the lengths is aboring and hard job to do; especially if you don’t like calculations in geometry.Here are some important theorems, lemmas and important triangle lengths thatwill appear a lot in bashing geometry problems. In the last section I’ve provideda bunch of computational problems for you. At the end I have to thank mydear friend Matin Yousefi for helping me by gathering many of those wonderfulproblems and Amir Hossein Parvardi who helped me a lot in writing this articleand 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 =a+ b+ c

2

1

Length Bashing in Olympiad Geometry Sina Ghaseminejad

3 Useful Trigonometry Formulas

In this section, there are some useful trigonometry formulas that can help youwhile bashing. Sum and difference formulas are the most important ones:

sin (α+ β) = sinα cosβ + sinβ cosα

sin (α− β) = sinα cosβ − cosα sinβ

cos (α+ β) = cosα cosβ − sinα sinβ

cos (α− β) = cosα cosβ + sinα sinβ

tan (α+ β) =tanα+ tanβ

1− tanα tanβ

tan (α− β) =tanα− tanβ

1 + tanα tanβ.

Using the equations above, you will have double and half angle formulas:

sin (2α) = 2 sinα cosα

cos (2α) = cos2 α− sin2 α

cos (2α) = 2 cos2 α− 1

cos (2α) = 1− 2 sin2 α

tan (2α) =2 tanα

1− tan2 α

sin(α

2

)= ±

√1− cosα

2

cos(α

2

)= ±

√1 + cosα

2

tan(α

2

)=

1− cosα

sinα

tan(α

2

)=

sinα

1 + cosα

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Factorization may help you during the bashing as well:

cosα cosβ =1

2(cos (α+ β) + cos (α− β))

sinα sinβ =1

2(cos (α− β)− cos (α+ β))

sinα cosβ =1

2(sin (α+ β) + sin (α− β))

sinα+ sinβ = 2 sin

(α+ β

2

)cos

(α− β

2

)sinα− sinβ = 2 sin

(α− β

2

)cos

(α+ β

2

)cosα− cosβ = −2 sin

(α+ β

2

)sin

(α− β

2

)cosα+ cosβ = 2 sin

(α+ β

2

)sin

(α− β

2

)

4 Some Important Theorems

Here are some theorems which are the bases of computational geometry andwill be used in this article a lot.

Theorem 4.1. (Law of Sines) In the triangle 4ABC we have:

a

sinA=

b

sinB=

c

sinC= 2R.

Theorem 4.2. (Law of Cosines) In the triangle 4ABC we have:

a2 = b2 + c2 − 2bc cosA.

We have the exact same formula for b and c.

Corollary 4.3. We can calculate the cosine of an angle in a triangle directlyusing the Law of Cosines:

cosA =b2 + c2 − a2

2bc

cosB and cosC can be written in the exact same way.

As an exercise, try to prove that the equations below are true:

cos

(A

2

)=

√p(p− a)

bc

sin

(A

2

)=

√(p− b)(p− c)

bc

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Theorem 4.4. (Stewart’s Theorem) In the triangle 4ABC, if a line passingthrough A meets BC at X, we have:

b2 ·BX + c2 · CX = a(AX2 +BX · CX).

Theorem 4.5. (Angle Bisector Theorem) In the triangle 4ABC, we have:

b

c=CDa

BDa=

sinB

sinC

If and only if ADa is an angle bisector.

Theorem 4.6. (Medians Theorem) In the triangle 4ABC, we have:

b

c=

sinB

sinC=

sin (BAMa)

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 ver-tices of triangle meeting the opposite side, a Cevian. In triangle 4ABC, threeCevians AA′, BB′, and CC ′ are concurrent if and only if we have:

AC ′

C ′B· BA

′

A′C· CB

′

B′A= 1.

Theorem 4.9. (Trigonometric Ceva) In triangle 4ABC, three Cevians AA′,BB′, and CC ′ are concurrent if and only if we have:

sinCAA′

sinA′AB· sinABB′

sinB′BC· sinBCC ′

sinC ′CA= 1.

Theorem 4.10. (Menelaus’s Theorem) Given a triangle4ABC, and a transver-sal line ` that meets BC, AC, and AB at points D, E, and F respectively, thenwe will have:

AF

FB· BDDC· CEEA

= 1.

The inverse of this theorem is also true.

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Theorem 4.11. (Area of the Triangle) The area of a triangle can be calculatedusing following formulas which can be proved easily.Let ha be the altitude from vertex A, then the area of 4ABC can be written asthese equations:

[ABC] =a · ha

2

[ABC] =bc · sinA

2

Let R, r and p be the circumradius, inradius and semiperimeter of 4ABCrespectively, we will have the following equations as well:

[ABC] = pr

[ABC] =abc

4R

[ABC] = 2R2 sinA sinB sinC

[ABC] =√p(p− a)(p− b)(p− c)

Theorem 4.12. (Euler Line) In the triangle 4ABC, let O be the circumcen-ter, G be the centroid, N9 be the center of Nine-Point Circle, and H be theorthocenter.Then, O, G, N9 and H are colinear and they share the ratio:

OG : GN9 : N9H = 2 : 1 : 3.

Theorem 4.13. (Nagel Line) In the triangle 4ABC, let I be the incenter, Gbe 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 triangle4ABC, with orthocenter H and circumcenterO, 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 mid-point 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 andin many situations you need to mention and prove them) that will be useful forreal Olympiad-based problem solving. Using these lemmas you can tackle manyhard problems.

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Lemma 5.1. (Lemma of Ratio) In the triangle 4ABC, if a line passing throughA meet BC at X, we have:

BX

CX=c · sin (BAX)

b · sin (CAX).

Corollary 5.2. In the quadrilateral ABCD, we have:

sinBAD

sinCAD=

sinB

sinC· BDCD

.

Lemma 5.3. (The Lemma of Perpendicularly) For a line BC, Point A on theplane and point H on BC are assumed. For a point P that lies on AH, Wehave:

AB2 −AC2 = PB2 − PC2.

If and only if AH⊥BC.

Lemma 5.4. For four angles α , α′ , β , β′, if we know that:

sinα

sinα′=

sinβ

sinβ′and α+ α′ = β + β′

Then we can conclude that: α = β and α′ = β′.

Lemma 5.5. For four angles α , α′ , β , β′, if we know that:

sinα

sinα′=

sinβ

sinβ′and α− α′ = β − β′

Then we can conclude that: α, β = α′, β′.

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 andγ are supplementary (their sum is equal to 180o).

Lemma 5.6. (Trigonometric Ptolemy) Let A, B, C, D, be four points on theplane. ABCD is a cyclic quadrilateral of this order, if and only if we have:

AB · sinBAC +AD · sinCAD = AC · sinBAD.

Lemma 5.7. (Trigonometric Colinearity) Let A, B, C, D, be four points onthe plane. B, C, D are colinear if and only if we have:

sinBAC

AD+

sinCAD

AB=

sinBAD

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 ∠PAB = ∠CAQ, we

will have:AP ·AQ = bc.

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Proof. Since we know that ∠PAB = ∠CAQ, it’s obvious that ∠ABP = ∠AQC =∠B, so we get:

4ABP ∼ 4AQC ⇒ b

AP=AQ

c

=⇒ AP ·AQ = bc.

Corollary 5.9. Let I be the incenter of 4ABC and Ia be the center of theexcircle relative to the vertex A, then this lemma can be written for I and Ia aswell.

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+C2 . So we have:

4ABIa ∼ 4AIC ⇒c

AIa=AI

b

=⇒ AI ·AIa = bc.

6 Geometric Calculations

6.1 Median


• 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 = c2 +

(a2

)2− 2c

(a2

)cosB = c2 +

a2

4− ac

(a2 + c2 − b2

2ac

)

m2a = c2 +

a2

4− a2

2− c2

2+b2

2

m2a =

c2

2+b2

2− a2

4

=⇒ ma =

√2b2 + 2c2 − a2

4.

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6.2 Altitude


• 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:

sinB

ha=

sin 90

c

=⇒ ha = c · sinB.ii) Using the formulas of the area, we will get:

a · ha2

=bc · sinA

2

⇒ ha =bc · sinA

a.

By the Law of Sines we know that asinA = 2R , using that, we get:

=⇒ ha =bc

2R.

6.2.2 Length of BHa

In the triangle 4ABHa , we have:

sin (90−B)

BHa=

sinB

c.

Since we know that sin (90−B) = cosB, 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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6.2.3 Length of AH

We know that HHbCHa is a cyclic quadrilateral; So:

AH ·AHa = AHb ·AC.

Knowing that AHa = b · sinC and AHb = c · cosA, we get:

AH · b · sinC = c · cosA · b

=⇒ AH = a · cotA.

Using asinA = 2R, it can be written like:

=⇒ AH = 2R · cosA.

6.2.4 Length of HHa

Since we know that HHbCHa is a cyclic quadrilateral; So we get ∠BHHa = ∠C.Using the law of sines in triangle 4BHHa and BH = b · cotB, we have:

=⇒ HHa = b · cotB · cosC.

Which can be transformed as:

=⇒ HHa = 2R · cosB · cosC.

6.2.5 Length of HbHc

With some angle chasing we will know that ∠AHbHc = ∠ABC and ∠AHcHb =∠ACB, this will lead us to the similarity 4AHbHc ∼ 4ABC; so:

HbHc

a=AHb

c= cosA

=⇒ HbHc = a · cosA.

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6.3 Internal and Exterior Angle Bisectors


• 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.

• d′a , d′b , d′c as the exterior angle bisectors of ∠A, ∠B, ∠C, respectively.

• D′a , D′b , D′c are where d′a , d′b , d′c meet the extension of BC, AC, AB,respectively.

6.3.1 Length of da

Using the Stewart’s Theorem, we will have:

a

(d2a +

a2bc

(b+ c)2

)+ b2

(ac

b+ c

)+ c2

(ab

b+ c

)⇒ d2a = bc−BDa · CDa

d2a = bc

(1− a2

(b+ c)2

)=

2b2c2

(b+ c)2

((b2 + c2 − a2) + 2bc

2bc

)d2a =

4b2c2

(b+ c)2· cos2

(A

2

)=⇒ da =

2bc

b+ c· cos

(A

2

).

6.3.2 d′a and D′a

We can calculate the length of d′a , just like da with the Stewart’s Theorem.The results are quite similar:

⇒ (d′a)2 = BD′a · CD′a − bc

=⇒ d′a =2bc

|b− c|· sin

(A

2

)

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6.3.3 Length of BDa

Using the Angle Bisector Theorem, we have:

BDa

CDa=c

b

⇒ BDa

BDa + CDa=

c

b+ c

BDa

a=

c

b+ c

=⇒ BDa =a · cb+ c

.

6.3.4 Length of AI

i) Using the Angle Bisector Theorem in the triangle 4ABDa , we will get:

AI

IDa=

c

BD

AI

IDa=

cacb+c

=b+ c

a

AI

AI + IDa=

b+ c

a+ b+ c

⇒ AI

ADa=b+ c

2p

AI = ADa ·b+ c

2p=

2bc

b+ c· cos

(A

2

)· b+ c

2p

=⇒ AI =bc

p· cos

(A

2

).

ii) If we connect I to Ec , it’s obvious that IEc⊥AB and IEc = r; in the triangle4AIEc , we will get:

⇒ AI =r

sin(A2

) .iii) Having Ec and the incircle again, it’s easy to prove that AEc = p − a; inthe triangle 4AIEc, we will get:

=⇒ AI =p− a

cos(A2

) .

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6.4 Mixtilinear Incircles

A Mixtilinear Incircle is a circle tangent to two of a triangle’s segments, and thecircumcircle of that triangle. There are three Mixtilinear Incircles in a triangle,we show Mixtilinear Incircle opposite to vertex A (the circle that is tangent toAB, 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 lieson 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− bc

p

=⇒ BE =c(p− b)

p

CF = b− bc

p

=⇒ CF =b(p− c)

p

6.5 T and T ′

We defined T and T ′ as the points where the internal and exterior bisectorsof ∠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 ′ 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.


• d′a , d′b , d′c as the exterior angle bisectors of ∠A, ∠B, ∠C, respectively.

• D′a , D′b , D′c are where d′a , d′b , d′c meet the extension of BC, AC, AB,respectively.

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6.5.1 Length of AT

An special case of the Isogonal Lines Lemma that we proved, is when A, P , Qare collinear; And that is when P ≡ Da and Q ≡ T .Using the lemma, we get:

ADa ·AT = bc.

We know that ADa = 2bcb+c · cos

(A2

), by putting that in our equation we have:

AT · 2bc

b+ c· cos

(A

2

)= bc

=⇒ AT =b+ c

2 cos (A2 ).

6.5.2 Length of AT ′

Here, we need to use D′a again. it’s obvious that T ′, A and D′a are collinear;with angle chasing we know that ∠ABD′a = ∠AT ′C and ∠D′aAB = ∠CAT ′,so:

4ABD′a ∼ 4AT ′C ⇒c

AT ′=d′ab

Just like the length of AT , by putting d′a = 2bc|b−c| · sin

(A2

)in the equation, we

get:

=⇒ AT ′ =|b− c|

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 andIT are all equal. with some angle chasing, it’s simple to show that ∠TBI =∠TIB = ∠A+B

2 and ∠TCI = ∠TIC = ∠A+C2 ; so BT = CT = IT (and it’s not

hard to prove they are all equal to TIa as well).Adding TMa to the structure, where Ma is the midpoint of side BC, we willhave:

TB · cos

(A

2

)=a

2

=⇒ TB = TC = TI =a

2 cos(A2

) .

Page 13


6.5.4 T and T’ images on AC

In this part, without loss of generality, we assume that AC > AB.S and S′ will be images of T and T ′ on the segment AC.By the definition, T ′S′⊥AC and TS⊥AC, using the Law of Sines, we will havethe following results:

=⇒ AS = CS′ =b+ c

2

=⇒ AS′ = CS =|b− c|

2

6.6 Excircles and Excenters

Just like Mixtilinear Incircles, there are three Excircles in a triangle. We definethe Excircle opposite to vertex A with ωa.ωb and ωc are defined in the same way. In this section, in triangle 4ABC, wehave:

• 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.


• I as the incenter of the triangle.

6.6.1 Length of AIa

i) Using the Isogonal Lines Lemma again, and having AI = bcp · cos

(A2

), we get:

AIa ·bc

p· cos

(A

2

)= bc.

With some calculations, we have:

=⇒ AIa =bc

p− a· cos

(A

2

).

ii) Let ωa touch AB at Zc.It’s obvious that and IaZc⊥AZc and IaZc = ra ; using that in the triangle4AZcIa , we get:

=⇒ AIa =ra

sin(A2

) .

Page 14


iii) Having Zc again; it’s easy to prove that AZc = p; using that in the triangle4AZcIa , we get:

=⇒ AIa =p

cos(A2

) .

6.6.2 Length of AIb

If we connect Ib to Xb , it’s obvious that IbXb⊥AC and IbXb = rb ; using that,we will have:

AIb =rb

sin(90− A

2

)=⇒ AIb =

rb

cos(A2

) .6.6.3 Length of ra

i) Calculating the length of ra is quite like proving r = [ABC]p ; we use the area

of 4ABC to calculate it.Consider ra ; we know that:

[ABC] = [AIaB] + [AIaC]− [CIaB].

⇒ [ABC] =p · ra

2+p · ra

2− a · ra

2With 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 IaZc⊥AZc ; If we connect Ito Ec , we know that IEc⊥AB as well. It is well known that A, I and Ia arecolinear, so we can prove that 4AIEc ∼ 4AIaZc.Since in triangle 4AIEc we have AEc = p−a , IEc = r and in triangle 4AIaZc

we have AZc = p , IaZc = ra, by using the similarity:

AEc

AZc=IEc

IZc

p− ap

=r

ra

By putting r = [ABC]p , we get the same result:

=⇒ ra =[ABC]

p− a

Page 15


7 Training Problems

Here are some problems gathered from Mathematics Olympiads all over theworld that can be solved by length bashing for some exercise. You may solvesome of these in much easier ways; But please don’t! There’s much more benefitin solving them by calculations than other ways.

1. (Blanchet’s Theorem) In an acute triangle 4ABC, F is the foot of theperpendicular from C to AB and P is a point on CF. Let the lines AP and BPmeet BC and AC at D and E, respectively. Show that the angles ∠EFC and∠DFC 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 fromP 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:

BY

XP+CY

XQ=AY

AX.

3. Let B1 , C1 be the midpoints of sides AC, AB of a triangle 4ABC. Therays CC1 , BB1 meet the tangents to the circumcircle at B and C at K and Lrespectively. Prove that the angles ∠BAK and ∠CAL are equal.

4. Equal circles ω1 and ω2 are passing through each others centers. The triangle4ABC is inscribed into the circle ω1 such that lines AC and BC are tangentto the circle ω2. Prove that: cosA+ cosB = 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 ∠BPC = 90o, Prove that:AE +AP = PD.

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. Fori ∈ a, b, c, let wi be the circle with MiTi as diameter. Let pi be the commonexternal 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 thelines 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

Page 16


8. Bisector of side BC intersects circumcircle of triangle4ABC in points P andQ. Points A and P lie on the same side of line BC. Point R is an orthogonalprojection of point P on line AC. Point S is middle of line segment AQ. Showthat 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 triangle4ABC such that BD = CE = AF and ∠BDF = ∠CED = ∠AFE. Show that4ABC 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. Provethat:

AB ·AC +BC · CA+ CA · CBAH ·AD +BH ·BE + CH · CF

≤ 2.

11. Let4ABC be an acute angled triangle. LetD,E, F be points onBC,CA,ABsuch that AD is the median, BE is the internal bisector and CF is the alti-tude. Suppose that ∠FDE = ∠C,∠DEF = ∠A and ∠EFD = ∠B. Show that4ABC 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 possiblemeasures of angle ∠BAC?

13. Let ABCD be a cyclic quadrilateral. Let P , Q, R be the feet of the perpen-diculars from D to the lines BC, CA, AB, respectively. Show that PQ = QRif and only if the bisectors of ∠ABC and ∠ADC are concurrent with AC.

14. Bisector of angle ∠BAC of triangle 4ABC intersects circumcircle of thistriangle in point D 6= A. Points K and L are orthogonal projections on line ADof 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 perpen-dicular bisectors of AN and NC intersect AB and BC respectively in K andM . Prove that the circumcenter O of 4ABC lies on the circumcircle of triangle4KBM

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.

Page 17


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. Sup-

pose that AC ·BD = EG·FH. Show that AC, BD, EG, FH are all concurrent.

18. In a scalene triangle 4ABC the altitudes AA1 and CC1 intersect at H,Ois the circumcenter, and B0 the midpoint of side AC. The line BO intersectsside AC at P , while the lines BH and A1C1 meet at Q. Prove that the linesHB0 and PQ are parallel.

19. Triangle 4ABC has integer-length sides. Line ` is tangent to the cir-cumcircle of 4ABC at A. Let S be the intersection of ` and BC. if AS is aninteger as well, Prove that: gcd(AB,AC) > 1.

20. Let AD be an angle bisector of triangle 4ABC. Perpendiculars fromD to the other sides of 4ABC, meet AB at X and AC at Y , Let P be theintersection 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, re-spectively. Similarly, let Let A2 , B2 , and C2 be the points where L2 intersectsthose side lines. Prove that the midpoints of the three segments A1A2, B1B2,and C1C2 are collinear.

23. Let ABCD be a trapezoid with AB ‖ CD. Lines AC and BD meet atP . H1 and H2 are the orthocenters of triangles 4APD and 4BPC respec-tively. Let M be the midpoint of H1H2 . Prove that: MP⊥CD.

24. Let P , Q be two points on the side BC of triangle 4ABC such that2PQ = 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 andH2 be the orthocenters of 4ABP and 4CDP respectively. Let G1 and G2 bethe centroids of 4ADP and 4BCP respectively. Prove that: H1H2⊥G1G2.

26. Let ω be the incircle of triangle 4ABC with center I. Circle ω touches ABand AC at E and F respectively. Line BI and CI intersect EF at X and Yrespectively. If M is the midpoint of BC, Prove that 4MXY is equilateral ifand only if ∠A = 60o.

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27. Let Ω be the circumcircle of triangle 4ABC and G be its centroid. LinesAG, BG and CG meet Ω at A′, B′ and C ′ respectively. Prove that:∑

A′G ≥∑

AG.

28. 4ABC is a triangle with incenter I, Midpoints of AB and AC are Mand N . Line CI meets MN at K. We call the line perpendicular to MN at K,`1 and the line parallel to BI passing through N , `2 . Let P be the intersectionof `1 and `2 . Prove that: PI⊥AC.

29. If I is the incenter of triangle 4ABC and a circle passing through A andI meets AB and AC at E and F , And M is the midpoint of EF , Show that:∠BMC > 90o

30. The incircle of triangle 4ABC touches the sides AB, BC and CA atD, 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′ and CC ′ are two altitudes of triangle 4ABC, Angle bisector of ∠Ameets B′C ′ and BC at E and F . Let `1 be the perpendicular line to B′C ′ atE and `2 be the perpendicular line to BC at F . The intersection of `1 and `2is 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:∑

PA2 =1

3

∑AB2 + 3PG.

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 ∠PBA = ∠QBC,∠PCB = ∠QCA and ∠PAB = ∠QAC. Prove that:

AP · sinBPC = AQ · sinBQC.

35. AA′, BB′ and CC ′ a altitudes of triangle 4ABC. We extent lines AA′,CC ′ from A′, C ′ and extend BB′ from B. Let X, Y and Z be points on linesAA′, BB′ and CC ′ such that AA′ = A′X, CC ′ = C ′Z and BY = 3BB′. Provethat angles ∠ABC and ∠XY Z are equal.

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36. Let M be the midpoint of side BC in triangle 4ABC. Points X, Yare outside the triangle and ∠AXB = ∠AY C = 90o and ∠XBA = ∠Y CA.a) Prove that: XM = YM .b) Prove that: ∠XMB = 2∠XBA.

37. Let r be the radius of the circle inscribed in triangle 4ABC and Ωa bethe excircle relative to vertex A and Ia be its center. Ωa touches lines BC, ABand AC at D, E and F respectively. EF meets IaB and IaC at X and Y . Theintersection of XC and Y B is K. Prove that the distance of K and BC is equalto r.

38. A circle is inscribed in Trapezoid ABCD with AB ‖ CD. Lines AC andBD meet at P . Prove that: ∠CPD > 90o.

39. Line ` is passing through triangle 4ABC. AX, BY and CZ are per-pendiculars to `. Prove that the centroid of 4ABC lies on ` if and only if wehave: AX = BY + CZ.

40. A Line passing through the centroid of triangle 4ABC, meets AB atX and AC at Y . Prove that:

AX

BX· AYCY≥ 4.

41. In triangle4ABC, PointsX, 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 MN isparallel 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 medianthrough A in triangle 4ABC respectively. If we know that DH < DM , Provethat: BC < 2AD

44. ABCD is a quadrilateral and points E, F are on BC, AD such thatBCBE = AD

AF = n for some real number n. Prove that:

[ABEF ] ≤ AD ·BC + n(n− 1) ·AB2 + n ·AB ·DC2n2

.

(Here [ABCD] represents the area of quadrilateral ABCD).

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45. In the triangle 4ABC, points X, X ′ are on BC, Y , Y ′ are on CA andZ, Z ′ are on AB such that BX = X ′C, CY = Y ′A and AZ = Z ′B. Provethat: [XY Z] = [X ′Y ′Z ′].(Here [ABC] represents the area of triangle 4ABC).

46. The incircle of triangle 4ABC with center I touches AB, BC and CAat D, E and F . ` is a line passing through A parallel to DF . Lines ED andEF meet ` at K and J . Show that: ∠JIK < 90o.

47. The incircle of triangle 4ABC with center I touches AB, BC and CAat D, E and F . ` is a line passing through A parallel to BC. Lines ED andEF meet ` at K and J . Show that: ∠JIK < 90o.

48. Let ha, hb and hc be the altitudes of triangle 4ABC with circumradius Rand area S, Prove that:

1

3

∑ 1

ha≥ 3

√R

2S2.

49. In the triangle 4ABC, Point A′ is symmetry of A with respect to BC.B′ and C ′ are defined in the same way. Prove that: [A′B′C ′] ≤ 4[ABC]. (Here [ABC] represents the area of triangle 4ABC).

50. If G is the centroid of triangle 4ABC and angles ∠BAG and ∠ACGare equal. Prove that:

sinCAG+ sinCBG ≤ 2√3.

51. Triangle 4ABC has circumcenter O and circumradius R. Circumcenterof triangle 4BOC is Ωa. The line AO meets Ωa for the second time at A′. B′

and C ′ are defined the same way. Prove that:

OA′ ·OB′ ·OC ′ ≥ 8R2.

52. Let ha, hb and hc be the altitudes of triangle 4ABC with semiperime-ter P . Prove that: ∑

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:∑ ra

ha≥ 3.

54. Triangle 4ABC with circumradius R and inradius r is given. Prove that:∑(sin

(A

2

)· sin

(B

2

))≤ 5

8+

r

4R.

55. Circumcircle of triangle 4ABC is Ω. Let the angle bisector of ∠A meet BCand Ω at D and T . We choose an arbitrary point X on BT . The intersectionpoint of XD and TC is Y . Show that the angle ∠XAY is greater that ∠BAC.

56. Circle ω with radius R is inscribed inside trapezoid ABCD with AB ‖ CD.Show that: AB · CD ≥ 4R2.

57. Triangle 4ABC has integer-length sides and AC = 2007. The Internalbisector 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 andC2 with centers O1 and O2. For an arbitrary point A on the plane, H is theprojection of A on `. Prove that:

|PKC1− PK

C2| = 2AH ·O1O2.

59. A triangle 4ABC is given with AB > AC. Line ` is tangent to thecircumcircle of 4ABC at A. A circle centered in A with radius |AC| cuts ABin the point D and the line ` in points E,F (such that C and E are in thesame halfplane with respect to AB). Prove that the line DE passes throughthe incenter of 4ABC.

60. Let P be a moving point on side BC of triangle 4ABC. Line `1 is drawnfrom P such that `1 and PB make a specific angle α. Line `2 is also drawn fromP the same way such that `2 and PC make a specific angle β. Lines `1 and `2meet AB and AC at K and L. Prove that the circumcircle of triangle 4AKLpasses 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 MN 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 Ωand chords AB and AC of Ω be tangent to ω at M and N . AK meets MN atS. Prove that the angles ∠MSB and ∠NSC 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:PC = PD.

64. Let H be the orthocenter of an acute-angled triangle 4ABC. Show thatthe triangles 4ABH,4BCH and 4CAH have the same perimeter if and onlyif the triangle 4ABC is equilateral.

65. In the rectangle ABCD, Points X, Y are chosen on sides DC, BC suchthat CX 6= CY . Let P be the intersection of DY and BX. If AP⊥XY , showthat DX = BY .

66. Let P , Q and R be the midpoints of sides AB, AC and BC in triangle4ABC. Points X and Y are on PR and QR such that BX⊥PR and CY⊥QR.if O′ is the circumcenter of PQR, prove that RO′ bisects XY .

67. In the isosceles triangle 4ABC with AB = AC, Point K is on segmentBC such that BK = 2KC and point P is chosen on AK such that angles∠BPKand ∠BAC are equal. Prove that: 2∠CPK = ∠BAC.

68. Circle ω is inscribed in triangle 4ABC and touches AC at B1 . PointsE and F are chosen on ω such that ∠AEB1 = 90 and ∠CFB1 = 90. Let Mbe 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 andH are circumcircle and orthocenter respectively. if OH meets AB and AC (nottheir extension) at P and Q and we show the area of triangle 4APQ with Sand area of quadrilateral BPQC with R, find all possible values of S

R .

70. Circle ω is internally tangent to circle Ω at S. Let A be a point on Ωand 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 TS 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 tan-gency points of ω with CA and AB, respectively. Let U, V be the intersectionsof line EF with line MN and line MP , 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 UV .

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72. Circle ω is tangent to line ` at M . Let A be the point on ω such that AM isthe diameter. Points X and Y are moving on ` such that always MX ·MY = kfor some real k and M is between X, Y . Lines AX and AY meet ω at E andF . 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; sideBC at F , G; and side CA at H, I internally (D, E, F , G, H and I are on ω inorder). Show that:

AD +BF + CH = BE +GH + IA.

74. Let 4ABC be an acute-angled triangle with AB 6= AC. Let H be theorthocenter of triangle 4ABC, and let M be the midpoint of the side BC. LetD be a point on the side AB and E a point on the side AC such that AE = ADand the points D, H, E are on the same line. Prove that the line HM is per-pendicular to the common chord of the circumscribed circles of triangle 4ABCand triangle 4ADE.

75. Chord AB of circle Γ is assumed. P is an arbitrary point on AB. Cir-cles ω1 with radius r1 and ω2 with radius r2 are on two sides of AB such thatthey’re tangent to AB at P and internally tangent to Γ. Show that with chang-ing the position of P , the ratio r1

r2won’t change.

76. The lengths of the sides of a convex hexagon ABCDEF satisfy AB = BC,CD = DE, EF = FA. Prove that:

BC

BE+DE

DA+FA

FC≥ 3

2.

77. Given a triangle 4ABC satisfying AC + BC = 3AB. The incircle oftriangle 4ABC has center I and touches the sides BC and CA at the pointsD and E, respectively. Let K and L be the reflections of the points D and Ewith 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 me-dian AM and the height AH. Points Q and P are marked on the lines ABand AC, respectively, so that the QM ⊥ AC and PM ⊥ AB. The circumcir-cle of 4PMQ intersects the line BC for second time at point X. Prove thatBH = CX.

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Length Bashing in Olympiad Geometry