Iran’s Geometry Problems (1).pdfthe quadrilateral and Xa point collinear with Oand E0such that XA...

Iran’s Geometry Problems

Problems and Solutions from Contests

2014-2015

This booklet is prepared by Hirad Aalipanah, Iman Maghsoudi.With special thanks to Morteza Saghafian, Mahdi Etesami Fard,

Davood Vakili, Erfan Salavati.Copyright c©Young Scholars Club 2014-2015. All rights reserved.

Ministry of education, Islamic Republic of Iran.www.ysc.ac.ir - www.igo-official.ir

The first Iranian Geometry Olympiad was held simultaneously in Tehran and Is-fahan on September 4th, 2014 with over 300 participants. This competition had twolevels, junior and senior which each level had 5 problems. The contestants solvedproblems in 4 hours and 30 minutes.In the end, the highest ranked participants in each level awarded with gold ruler,silver ruler or bronze ruler respectively.This booklet have the problems of this competition plus other geometry problemsused in other Iranian mathematical competition since summer of 2014 till spring of2015.This year the second Iranian Geometry Olympiad will be held in Tehran on Septem-ber 3th, 2015. We tend to provide online presence for those who are interested fromother countries. Those who wish to participate can contact Mr. Salavati for moreinformation at [email protected]

Iranian Geometry Olympiads website: www.igo-official.ir

Problems

Problems 4

1.(Geometry Olympiad(Junior and Senior level)) In a right triangle ABCwe have ∠A = 90◦ ,∠C = 30◦. Denot by C the circle passing through A which istangent to BC at the midpoint.Assume that C intersects AC and the circumcircle ofABC at N and M respectively. Prove that MN⊥BC.

Proposed by Mahdi Etesami Fard

2.(Geometry Olympiad(Junior Level)) The inscribed circle of4ABC touchesBC, AC and AB at D, E and F respectively. Denote the perpendicular foots fromF , E to BC by K, L respectively. Let the second intersection of these perpendicularswith the incircle be M , N respectively. Show that

S4BMD

S4CND= DK

DL


3.(Geometry Olympiad (Junior Level)) Each of Mahdi and Morteza hasdrawn an inscribed 93-gon. Denote the first one by A1A2...A93 and the second byB1B2...B93. It is known that AiAi+1 ‖ BiBi+1 for 1 6 i 6 93 (A93 = A1, B93 = B1).Show that AiAi+1

BiBi+1is a constant number independent of i.

Proposed by Morteza Saghafian

4.(Geometry Olympiad (Junior Level)) In a triangle ABC we have ∠C =∠A + 90◦. The point D on the continuation of BC is given such that AC = AD. Apoint E in the side of BC in which A doesnt lie is chosen such that

∠EBC = ∠A,∠EDC =1

2∠A

Prove that ∠CED = ∠ABC.


5.(Geometry Olympiad (Junior Level)) Two points X, Y lie on the arc BC ofthe circumcircle of 4ABC (this arc does not contain A) such that ∠BAX = ∠CAY .Let M denotes the midpoint of the chord AX . Show that BM + CM > AY

Proposed by Mahan Tajrobekar

Problems 5

6.(Geometry Olympiad(Senior level)) In a quadrilateral ABCD we have∠B = ∠D = 60◦. Consider the line whice is drawn from M , the midpoint of AD,parallel to CD. Assume this line intersects BC at P . A point X lies on CD suchthat BX = CX. Prove that AB = BP ⇔ ∠MXB = 60◦

Proposed by Davood Vakili

7.(Geometry Olympiad(Senior level)) An acute-angled triangleABC is given.The circle with diameter BC intersects AB, AC at E, F respectively. Let M be themidpoint of BC and P the intersection point of AM and EF . X is a point on the arcEF and Y the second intersection point of XP with circle mentioned above. Showthat ∠XAY = ∠XYM .

Proposed by Ali Zooelm

8.(Geometry Olympiad(Senior level)) The tangent line to circumcircle of theacute-angled triangle ABC (AC > AB) at A intersects the continuation of BC at P .We denote by O the circumcenter of ABC. X is a point OP such that ∠AXP = 90◦.Two points E, F respectively on AB, AC at the same side of OP are chosen suchthat

∠EXP = ∠ACX, ∠FXO = ∠ABX

If K, L denote the intersection points of EF with the circumcircle of 4ABC, showthat OP is tangent to the circumcircle of 4KLX.


9.(Geometry Olympiad(Senior level)) Two points P , Q lie on the side BCof triangle ABC and have the same distance to the midpoint. The pependicularsfromP , Q tp BC intesects AC, AB at E, F respectively. LEt M be the intersectionpoint of PFand EQ. If H1 and H2 denote the orthocenter of 4BFP and 4CEQrecpectively, show that AM ⊥ H1H2.


10.(IGO Short list)Suppose that I is incenter of 4ABC and CI inresects ABat D.In circumcircle of4ABC, T is midpoint of arc BAC and BI intersect this circleat M . If MD intersects AT at N , prove that: BM ‖ CN .


Problems 6

11.(IGO Short list) Consider two parallelogram ABCD and A′B′C ′D′ such thatAB ‖ A′B′ and BC ‖ B′C ′. Let P intersection of BB′ and DD′. Prove that P,A,C ′

are collinear if only if P,C,A′ are collinear.


12.(IGO Short list) Soppose that ABCD is a cyclic quadrilateral. Let M mid-point of arc AB and P intersection of AC and BD. If MP⊥CD and ∠DAC =∠ADC + ∠DCB prove that ABCD is trapezoid.


13.(IGO Short list)let ABCP be a Tetrahedron and ω be a sphere which con-tains incircles of 4PBC, 4PCD, 4PBD. prove that it contains incircle of 4BCA.


14.(IGO Short list)in triangle 4ABC, let S be the intersection of BC and tan-gent from A to the circumcircle of 4ABC.Let L, K be points lay on line SO suchthat ∠LAC = ∠KAB = 90◦. Show that OL = OK.

Proposed by Ali Zamani

15.(IGO Short list)Suppose that BE, CD are the altitudes of the4ABC whichintersects at H. Let B′, C ′ lay on BE, CD such that BE = B′E, CD = C ′D. Sup-pose that O’ is the circumcenter of 4B′HC ′. Let K be the midpoint of AH. O′Kmeets BC at T . Prove that ∠OTC = ∠O′TB.


16.(IGO Short list)In acute-triangle ABC, H is orthocenter and X, Y lie onAB,AC respectively such that AX = 2XB and AY = 2Y C. If 3BC2 = AC2 +BC2,prove that circumcircle of 4AXY tangent to circumcirle of 4BHC.


Problems 7

17.(IGO Short list)Suppose AD is the angle bisector of the 4ABC and D′ isthe midpoint of it. We draw a line from A which is parallel to BC. This line meetsthe circumcircle of 4ABC at A1. Let T ′ be the midpoint of the smaller arc AA1.T ′D′ meets the circumcircle at A2. We define B1, B2, C1, C2 in the same way. Provethat A1A2, B1B2, C1C2 are concurrent.


18.(National Math Olympiad (Second Round)) The arbitrary circle whichpasses through B and C from4ABC intersects AC and AB in D and E respectively.Suppose CE intersects BD in P and H be the foot of the perpendicular drawn fromP to AC .If M and N be the midpoints of BC and AP respectively. Prove that4MNH and 4CAE are similar.

Proposed by Amir Saeedi

19.(National Math Olympiad (Second Round)) Let ABCD be a quadrilat-eral such that AC is the angle bisector of ∠DAB and ∠ADC = ∠ACB. Let X andY be the foot of the perpendiculars drawn from A to BC and CD respectively. provethat BD passes through the orthocenter of 4AXY .


20.(Third Round) Let A′ on circumcircle of 4ABC such that AA′ be diameter.Suppose that Line I and I ′ are parallel to extenal bisector of ∠BAC and intenal bi-sector of ∠BAC respectively, such that passing through A′. Line I intersects BC,ACat C2, C1 respectively and Line I ′ intersects BC,AB at B2, B1 respectively. Provethat circumcircles of 4ABC, 4CC1C2 and 4BB1B2 are concur.

21.(Third Round)Tow points P,Q lay in isosceles triangle ABC (AB = AC).Suppose Q lay in ∠PAC and ∠PAQ = ∠BAC

2. X is intersection of AP,BQ and Y is

intersection of AQ,CP . If CQ = PQ = BP prove that quadrilateral PQY X is cyclic.


Problems 8

22.(Third Round)Distinct points B,B′, C, C ′ lie on an arbitrary line `. A is apoint not lying on `. A line passing through B and parallel to AB′ intersects withAC in E and a line passing through C and parallel to AC ′ intersects with AB in F .Let X be the intersection point of the circumcircles of 4ABC and 4AB′C ′(A 6= X).Prove that EF ‖ AX.

Proposed by Ja’far Naamdaar

23.(Third Round)D is an arbitrary point lying on side BC of 4ABC. Circleω1 is tangent to segments AD , BD and the circumcircle of 4ABC and circle ω2 istangent to segments AD , CD and the circumcircle of 4ABC. Let X and Y be theintersection points of ω1 and ω2 with BC respectively and take M as the midpoint ofXY . Let T be the midpoint of arc BC which does not contain A. If I is the incenterof 4ABC, prove that TM goes through the midpoint of ID.

24.(Third Round) Let X, Y on line BC from triangle ABC such that ∠XAY =90◦. Suppose that H is orthocenter of 4ABC. If AX intersects BH at X ′ and AYintersects CH at Y ′, prove that intersection of circumcircle of 4BXX ′ and circum-circle of 4CY Y ′ lay on X ′Y ′.

25.(Team Selection Test)Point Ib is the B-excenter of triangle ABC. If wedenote by M the midpoint of arc BC of the circumcircle of triangle ABC (the onethat does not contain vertex A), and MIb intersects the circumcircle of triangle ABCat T , prove that TI2

b = TB × TC.


26.(Team Selection Test)Quadrilateral ABCD is both inscribed and circum-scribed. Let E be the intersection point ofAD andBC, F the intersection point ofABand CD, S the intersection point of AC and BD and O the circumcenter of quadri-lateral ABCD. E ′ and F ′ are selected on AB and AD such that ∠BEE ′ = ∠AEE ′

and ∠AFF ′ = ∠DFF ′. Let M be the midpoint of arc BAD of the circumcircle ofthe quadrilateral and X a point collinear with O and E ′ such that XA

XB= EA

EB. Also

let Y be a point collinear with O and F ′ such that Y AY D

= FAFD

. Prove that the circlewith diameter OS, the circumcircle of triangle OAM and the circumcircle of triangleOXY are co-axis.


Problems 9

27.(Team Selection Test)Point D is the intersection point of the angle bisectorof vertex A with side BC of triangle ABC, and point E is the tangency point of theinscribed circle of triangle ABC with side BC. A1 is a point on the circumcircle oftriangle ABC such that AA1||BC. If we denote by T the second intersection pointof line EA1 with the circumcircle of triangle AED and by I the incenter of triangleABC, prove that IT = IA.


28.(Team Selection Test) Let ABC be an acute-angled triangle. Point Z onthe altitude of vertex A and points X and Y on the extensions of the altitudes ofvertices B and C are selected such that,

∠AY B = ∠BZC = ∠CXA = 90◦.

Prove that X, Y and Z are collinear if and only if the length of the tangent fromvertex A to the nine-point circle of the triangle is equal to the sum of the lengths oftangents from vertices B and C to this circle.

Proposed by Mohammad Javad Shabani

29.(Team Selection Test)From a point A outside circle ω, tangents AS and ATare drawn to the circle. Points X and Y are the midpoints of segments AT and AS,respectively. Tangent XR is drawn from point X to the circle and P and Q are themidpoints of segments XT and XR, respectively. If XY and PQ intersect each otherat K, and SX and TK intersect at L, prove that KRLQ is an inscribed quadrilateral.

Proposed by Farhad Seifollahi

30.(Team Selection Test)H is the foot of the altitude of vertex A of triangleABC and H ′ is the reflection of H with respect to the midpoint of BC. If tangentsto the circumcircle of triangle ABC at points B and C intersect each other at X andthe perpendicular to XH ′ at H ′ intersects lines AB and AC at Y and Z, respectively,prove that ∠Y XB = ∠ZXC.


Solutions

Solutions 11

1.(Geometry Olympiad(Junior and Senior Level)) In a right triangle ABCwe have ∠A = 90◦ ,∠C = 30◦. Denot by C the circle passing through A which istangent to BC at the midpoint.Assume that C intersects AC and the circumcircle ofABC at N and M respectively. Prove that MN⊥BC.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.Let K midpoint of side BC. Therefore:

AK = KC ⇒ ∠KAC = ∠NKC = 30◦

∠ANK = ∠NKC + ∠ACB = 60◦

A,K,N,M lie on circle (C). Therefore:

∠KAN = ∠KMN = 30◦,∠AMK = 60◦

We know that K is the circumcenter of4ABC. So we can say KM = KC = AK.Therefore 4AKM is equilateral.( because of ∠AMK = 60◦ ). So ∠AKM = 60◦. Weknow that ∠AKB = 60◦, so we have ∠MKC = 60◦. On the other hand:

∠KMN = 30◦ ⇒MN⊥BC

Solutions 12

2.(Geometry Olympiad(Junior Level)) The inscribed circle of4ABC touchesBC, AC and AB at D, E and F respectively. Denote the perpendicular foots fromF , E to BC by K, L respectively. Let the second intersection of these perpendicularswith the incircle be M , N respectively. Show that

S4BMD

S4CND= DK

DL


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.Let I be the incenter of 4ABC. We know that

∠BFK = 90◦ − ∠B

∠BFD = 90◦ − 12∠B

}⇒ ∠DFM =

1

2∠B

But ∠DFM = ∠MDK. Therefore

∠MDK =1

2∠B

Hense 4MDK and 4BID are similar (same angles) and MKDK

= rBD

. In the sameway we have NL

DL= r

CD. Therefore

r =MK ·BD

DK=NL · CDDL

⇒ area of 4BMD

area of 4CND=MK ·BDNL · CD

=DK

DL

Solutions 13

3.(Geometry Olympiad (Junior Level)) Each of Mahdi and Morteza hasdrawn an inscribed 93-gon. Denote the first one by A1A2...A93 and the second byB1B2...B93. It is known that AiAi+1 ‖ BiBi+1 for 1 6 i 6 93 (A93 = A1, B93 = B1).Show that AiAi+1

BiBi+1is a constant number independent of i.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.We draw a 93-gon similar with the second 93-gon in the circumcircle of the first

93-gon (so the sides of the second 93-gon would be multiplying by a constant numberc). Now we have two 93-gons witch are inscribed in the same circle and apply theproblem’s conditions. We name this 93-gons A1A2...A93 and C1C2...C93.

We know that A1A2 ‖ C1C2. Therefore_A1C1=

_A2C2 but they lie on the opposite

side of each other. In fact,_AiCi=

_Ai+1Ci+1 and they lie on the opposite side of each

other for all 1 6 i 6 93 (_

A94C94=_A1C1). Therefore

_A1C1 and

_A1C1 lie on the opposite

side of each other. So_A1C1= 0◦ or 180◦. This means that the 93-gons are coincident

or reflections of each other across the center. So AiAi+1 = CiCi+1 for 1 6 i 6 93.Therefore, AiAi+1

BiBi+1= c.

Solutions 14

4.(Geometry Olympiad (Junior Level)) In a triangle ABC we have ∠C =∠A + 90◦. The point D on the continuation of BC is given such that AC = AD. Apoint E in the side of BC in which A doesnt lie is chosen such that

∠EBC = ∠A,∠EDC =1

2∠A

Prove that ∠CED = ∠ABC.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.Suppose M is the midpoint of CD. Hense AM is the perpendicular bisector of

CD. AM intersects DE and BE at P,Q respectively. Therefore, PC = PD. Wehave

∠EBA+ ∠CAB = ∠A+ ∠B + ∠A = 180◦ − ∠C + ∠A = 90◦

Hense AC ⊥ BE. Thus in 4ABQ, BC,AC are altitudes. This means C is theorthocenter of this triangle and

∠CQE = ∠CQB = ∠A =1

2∠A+

1

2∠A = ∠PDC + ∠PCD = ∠CPE

Hense CPQE is cyclic. Therefore

∠CED = ∠CEP = ∠CQP = ∠CQA = ∠CBA = ∠B.

Solutions 15

5.(Geometry Olympiad (Junior Level)) Two points X, Y lie on the arc BC ofthe circumcircle of 4ABC (this arc does not contain A) such that ∠BAX = ∠CAY .Let M denotes the midpoint of the chord AX . Show that

BM + CM > AY

Proposed by Mahan Tajrobekar

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.O is the circumcenter of 4ABC, so OM ⊥ AX. We draw a perpendicular line

from B to OM . This line intersects with the circumcircle at Z. Since OM ⊥ BZ,OM is the perpendicular bisector of BZ. This means MZ = MB. By using triangleinequality we have

BM +MC = ZM +MC > CZ

But BZ ‖ AX, thus

_AZ =

_BX =

_CY ⇒

_ZAC =

_Y CA ⇒ CZ = AY

Hense BM + CM > AY.

Solutions 16

6.(Geometry Olympiad(Senior level)) In a quadrilateral ABCD we have∠B = ∠D = 60◦. Consider the line whice is drawn from M , the midpoint of AD,parallel to CD. Assume this line intersects BC at P . A point X lies on CD suchthat BX = CX. Prove that:

AB = BP ⇔ ∠MXB = 60◦


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.Suppose X ′ is a point such that 4MBX ′ is equilateral.(X ′ and X lie on the same

side of MB) It’s enough to show that:

AB = BP ⇔ X ′ ≡ X

We want to prove that if AB = BP then ∠MXB = 60◦.AB = BP therefore 4ABP is equilateral. We know that ∠ABP = ∠MBX ′ = 60◦,Therefore ∠ABM = ∠PBX ′. On the other hand AB = BP,BM = BX ′ therefore4BAM and 4BPX ′ are equal.

∠X ′PM = 360◦ − ∠MPB − ∠BPX ′ = 360◦ − ∠DCB − ∠BAM ′ = 120◦

Solutions 17

MP ‖ DC, so we can say ∠PMD = 120◦. If we draw the line passing throughX ′ such that be parallel with CD and this line intersects AD in D′, then quadri-lateral MPX ′D′ is isosceles trapezoid. Therefore PX ′ = MD′. In the other handPX ′ = AM = MD ( becauese 4BAM and 4BPX ′ are equal.) According to thestatements we can say MD′ = MD. In other words, D′ ≡ D and X ′ lie on CD.Therefore both of X and X ′ lie on intersection of DC and perpendicular bisector ofMB, so X ′ ≡ X.

Now we prove if ∠MXB = 60◦ then AB = BP .Let P ′ such that 4MP ′X be equilateral.(P ′ and X be on the same side of AB) It’senough to show that P ′ ≡ P .

Draw the line passing through P ′ such that be parallel with CD. Suppose that thisline intersects AD in M ′.

∠XP ′M ′ = 360◦ − ∠M ′P ′B − ∠BP ′X = 360◦ − ∠DCA− ∠BAM = 120◦

Also ∠P ′M ′D = 120◦. Therefore quadrilateral XP ′M ′D is isosceles trapezoid andDM ′ = P ′X = AM = DM . So we can say M ′ ≡M ⇒ P ′ ≡ P .

Solutions 18

7.(Geometry Olympiad(Senior level)) An acute-angled triangleABC is given.The circle with diameter BC intersects AB, AC at E, F respectively. Let M be themidpoint of BC and P the intersection point of AM and EF . X is a point on the arcEF and Y the second intersection point of XP with circle mentioned above. Showthat ∠XAY = ∠XYM .


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.Suppose point K is intersection AM and circumcircle of 4AEF . MF tangent tocircumcircle of 4AEF at F .( because of ∠MFC = ∠MCF = ∠AEF ). Therefore MF 2 = MK.MA . In theother hand, MY = MF so MY 2 = MK.MA. It means

∠MYK = ∠Y AM (1)

Also AP.PK = PE.PF = PX.PY therefore AXKY is(...??) .Therefore

∠XAY = ∠XYK (2)

According to equation 1 and 2 we can say ∠XAY = ∠XYM .

Solutions 19

8.(Geometry Olympiad(Senior level)) The tangent line to circumcircle of theacute-angled triangle ABC (AC > AB) at A intersects the continuation of BC at P .We denote by O the circumcenter of ABC. X is a point OP such that ∠AXP = 90◦.Two points E, F respectively on AB, AC at the same side of OP are chosen suchthat

∠EXP = ∠ACX, ∠FXO = ∠ABX

If K, L denote the intersection points of EF with the circumcircle of 4ABC, showthat OP is tangent to the circumcircle of 4KLX.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−solution.

Let M and N on continuation of XF and XE such that M,L,X,N,K lie onsame circle. We have to prove ∠AMX = ∠ACX. In other hand, ∠ACX = ∠NXPso we have to prove ∠ACX = ∠NMX.

We know that XF.FM = FL.FK = AF.FC. Therefore AMCX is cyclic and∠AMX = ∠ACX. similarly we can say ANBX is cyclic. Now it’s enough to showthat ∠AMX = ∠NMX. In other words, we have to show that A, N , M lie on sameline. we know that ANBX is cyclic therefore:

∠NAM = ∠NAE + ∠A+ ∠FAM = ∠EXB + ∠A+ ∠CXF

= ∠A+ 180◦ − ∠BXC + ∠ABX + ∠ACX

= ∠A+ 180◦ − ∠BXC + ∠BXC − ∠A = 180◦

Solutions 20

9.(Geometry Olympiad(Senior level)) Two points P , Q lie on the side BCof triangle ABC and have the same distance to the midpoint. The pependicularsfromP , Q tp BC intesects AC, AB at E, F respectively. LEt M be the intersectionpoint of PFand EQ. If H1 and H2 denote the orthocenter of 4BFP and 4CEQrecpectively, show that AM ⊥ H1H2.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solution.

First we show that if we move P and Q, the line AM doesn’t move. To show thatwe calculate sin∠A1

sin∠A2. By the law of sines in 4AFM and 4AEM we have

sin∠A1

sin∠A2

=sin∠F1

sin∠E1

· FMEM

(3)

also, for 4FBP and 4CEQ we have

sin∠F1 = BPPF· sin∠B

sin∠E1 = CQEQ· sin∠C

}⇒ sin∠F1

sin∠E1

=sin∠Bsin∠C

· EQFP

(4)

from (3) and (4) we have

sin∠A1

sin∠A2

=sin∠Bsin∠C

· EQFP· FMEM

(5)

4FMQ and 4EMP are similar, thus

FM

FP=

FQ

FQ+ EP,EQ

EM=FQ+ EP

EP

with putting this into (5) we have

Solutions 21

sin∠A1

sin∠A2

=sin∠Bsin∠C

· FQEP

(6)

on the other handtan∠B = FQ

BQ

tan∠C = EPCP

BQ = CP

⇒ FQ

EP=

tan∠Btan∠C

if we put this in (6) we have

sin∠A1

sin∠A2

=sin∠Bsin∠C

· tan∠Btan∠C

wich is constant.now we show that H1H2s are parallel. consider α the angle between H1H2 and

BC. Hense we have

tanα =H2P −H1Q

QP(7)

H1 and H2 are the orthometers of 4BFP and 4CQE respectively. Thus we have

QF ·H1Q = BQ ·QP ⇒ H1Q =BQ ·QPFQ

EP ·H2P = CP · PQ⇒ H2P =CP · PQEP

but CP = BQ. Thus

H2P −H1Q =PQ ·BQ · (FQ− EP )

EP · FQ

by putting this in (7) :

tanα =BQ · (FQ− EP )

EP · FQ=BQ

EP− BQ

FQ=CP

EP− BQ

FQ

⇒ tanα = cot∠B − cot∠C (8)

Solutions 22

hense tanα is constant, thus H1H2s are parallel.Soppuse θ is the angle between AM and BC. we have to show

tanα · tan θ = 1

let AM intersects with BC at X. We have

BX

CX=

sin∠A1

sin∠A2

· sin∠Csin∠B

⇒ BX

CX=

tan∠Btan∠C

let D be the foot of the altitude drawn from A. We have

BX

CX=

tan∠Btan∠C

=ADBDADCD

=CD

BD⇒ BD = CX

tan θ =AD

DX=

AD

CD − CX=

AD

CD −BD=

1CDAD− BD

AD

=1

cot∠B − cot∠C

this equality and (8) implies that AM ⊥ H1H2.

Solutions 23

10.(IGO Short list)Suppose that I is incenter of 4ABC and CI inresects ABat D.In circumcircle of4ABC, T is midpoint of arc BAC and BI intersect this circleat M . If MD intersects AT at N , prove that: BM ‖ CN .


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−solution.

Let X = AI ∩ DM , Y = AC ∩ DM and Z = CX ∩ BM , K = AC ∩ BM . Weknow that (D, Y,X,N) = −1 (because of ∠XAN = 90◦ and ∠DAX = ∠XAY ). Byprojecting D, Y,X,N on BM from C, we have to prove IZ = KZ.

According to sine from of Ceva’s theorem in 4ADC:

sin(∠DCX)

sin(∠XCA).sin(∠XAC)

sin(∠XAD).sin(∠XDA)

sin(∠XDC)= 1

We know that ∠XAC = ∠XAD, therefore:

sin(∠XDC)

sin(∠XDA)=sin(∠DCX)

sin(∠XCA)(1)

According to low of sine in 4AMD and 4DMC:

MC

DM=sin(∠CDX)

sin(∠DCM),

AM

DM=sin(∠XDA)

sin(∠DAM)

because of AM = MC we can say:

sin(∠XDC)

sin(∠DCM)=sin(∠XDA)

sin(∠DAM)⇒ sin(∠XDC)

sin(∠XDA)=sin(∠DCM)

sin(∠DAM)(2)

Solutions 24

∠DCM = ∠KIC = ∠C2

+ ∠B2

and ∠DAM = ∠IKC = ∠A + ∠B2

, according toequation 1 and 2:

sin(∠DCX)

sin(∠XCA)=sin(∠KIC)

sin(∠IKC)

According to Median’s theorem in 4IKC we show that IZ = KZ.

Solutions 25

11.(IGO Short list) Consider two parallelogram ABCD and A′B′C ′D′ such thatAB ‖ A′B′ and BC ‖ B′C ′. Let P intersection of BB′ and DD′. Prove that P,A,C ′

are collinear if only if P,C,A′ are collinear.


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solution.Soppose that P,A,C ′ are collinear, now we show that P,C,A′ are collinear.Let C ′′ = PC ∩ B′C ′, A′′ = PA ∩ A′B′, X = PC ′ ∩ A′D′, Y = PD ∩ A′B′. Drawthe line passing through A′′ such that be parallel with AD. Suppos that this lineintersects PD′ at D′′. We know that two parallelogram ABCD and A′′B′C ′′D′′ arehomogeneouse. We have to show C ′′, A′, P are collinear.

According to Menelaus’s theorem in 4AYD′ for collinearity X,A′′, P :

A′X

XD′.PD′

PY.Y A′′

A′′A′= 1 (1)

Also according to Tales’s theorem:

A′′A′ ‖ D′C ′ ⇒ A′X

XD′=A′′X

XC ′, A′X ‖ B′C ′ ⇒ A′′A′

A′B′=A′′X

XC ′

Solutions 26

⇒ A′X

XD′=A′′A′

A′B′(2)

D′′A′′ ‖ A′D′ ⇒ Y A′′

A′′A′=

Y D′′

D′D′′, Y B′ ‖ D′′C ′′ ‖ D′C ′ ⇒ B′C ′′

C ′′C ′=

Y D′′

D′D′′

⇒ Y A′′

A′′A′=B′C ′′

C ′′C ′(3)

A′′Y ‖ D′C ′ ⇒ PD′

PY=PC ′

PA′′(4)

According to equation 1, 2, 3, 4 we can say:

B′C ′′

C ′′C ′.PC ′

PA′′.A′′A′

A′B′= 1

Therefore C ′′, A′, P are collinear.(according to Menelaus’s theorem in 4A′′B′C)Similarly if P,C,A′ be collinear, we can prove P,A,C ′ are collinear.

Solutions 27

12.(IGO Short list) Soppose that ABCD is a cyclic quadrilateral. Let M mid-point of arc AB and P intersection of AC and BD. If MP⊥CD and ∠DAC =∠ADC + ∠DCB prove that ABCD is trapezoid.


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solution.Let ∠BDC = α, ∠DCA = θ, ∠ADB = ∠ACB = 2x therefore:

∠ADM = ∠MDB = ∠ACM = ∠MCB = x

(because M is midpoint of arc AB). Suppose that CM intersects AD at E and DMintersects BC at F .Also soppose that DE intersects BC at K.

we know that ∠DAC + ∠ADC + ∠DCA = 180◦. therefore:

∠DAC = 180◦ − α− θ − 2x = α + θ + 4x ⇒ α + θ + 3x = 90◦

Also ∠DEC = 180◦ − α− θ − 3x = 90◦. therefore CE⊥KD and similary DF⊥KC.therefore M is orthocenter of 4KDC and KM⊥CD. In the other hand MP⊥CD,therefore K,M,P are collinear.

According to sine from of Ceva’s theorem in 4KDC:

sin(∠DKP )

sin(∠PKC).sin(∠KCP )

sin(∠PCD).sin(∠PDC)

sin(∠PDK)= 1 ⇒ sin(∠DKM)

sin(∠MKC).sin(α)

sin(θ)= 1 (1)

Solutions 28

In the other hand M is orthocenter of 4KDC therefore:

∠DKM = ∠MCD = θ + x , ∠MKC = ∠MDC = α + x

According to equation 1:

sin(θ + x).sin(α) = sin(α + x).sin(θ)

⇒ cos(α + θ + x)− cos(x+ θ − α) = cos(α + θ + x)− cos(x− θ + α)

⇒ cos(x+ θ − α) = cos(x− θ + α)

we know that:

(x+ θ − α) + (x− θ + α) = 2x < 180◦ ⇒ x+ θ − α = x+ θ + α ⇒ α = θ

⇒ ∠BAC = ∠DCA ⇒ AB ‖ CD

Therefore ABCD is trapezoid.

Solutions 29

13.(IGO Short list)let ABCP be a Tetrahedron and ω be a sphere which con-tains incircles of 4PBC, 4PCD, 4PBD. prove that it contains incircle of 4BCA.


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solution.first we show that X and Y lay on each other. if X and Y are diffrent then APCplane and ω intersect at a circle and a point which is wrong. Therefore, X ≡ Y

Solutions 30

we have AX = AY = AE, CX = CZ = CF , BD = BZ = BY . Thus the incircleof 4ABC passes throught X, Y , Z.

assume that O is the center of ω. We have :

OX = OY = OZ = OE = OF = OD

Plain ABC and ω intersect at a circle which X, Y , Z lay on it.So this circle is theincircle of 4ABC.

Solutions 31

14.(IGO Short list)in triangle 4ABC, let S be the intersection of BC and tan-gent from A to the circumcircle of 4ABC.Let L, K be points lay on line SO suchthat ∠LAC = ∠KAB = 90◦. Show that OL = OK.


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solution.

Let B′, C ′ be the intersecion of AK and AL with the circumcircle (other than A)respectively. BB′, CC ′ are the diameters of the circumcircle.So we can change theproblem and redefine K, L.

Solutions 32

AK ′A′L′ is a parallelogram and the diameter K ′L′ passes throught O so OK ′ =OL′.

Now we have to show that K ≡ K ′, L ≡ L′. Let S be the intersection of BC andthe tangent from A to the circumcircle. From Pascal’s theorem we know :1) AC ′CBA′ → L′, O, S are colinear2) AB′BCA′ → K ′, O, S are colinear

so K ′ ≡ K, L′ ≡ L. Therefore OK = OL.

Solutions 33

15.(IGO Short list)Suppose that BE, CD are the altitudes of the4ABC whichintersects at H. Let B′, C ′ lay on BE, CD such that BE = B′E, CD = C ′D. Sup-pose that O’ is the circumcenter of 4B′HC ′. Let K be the midpoint of AH. O′Kmeets BC at T . Prove that ∠OTC = ∠O′TB.


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solution.AHBC ′ and AHCB′ are cyclic. Therefore their circumradius are equal to R(thecircumradius of 4ABC). Let their circumcenters be P and Q respectively. We knowthat PQ ⊥ AH, PO′ ⊥ C ′H, QO′ ⊥ B′H. Therefore 4PO′Q ∼= 4ABC and O′K isthe median drawn from O′. So O′K ‖ AM . O′′ is the reflection of O about BC. ThusAKO′′M is a parallelogram. This means KO′′ ‖ AM . So O′, K, O′′ are colinear andthis is equal to what we want to prove.

Solutions 34

16.(IGO Short list)In acute-triangle ABC, H is orthocenter and X, Y lie onAB,AC respectively such that AX = 2XB and AY = 2Y C. If 3BC2 = AC2 +BC2,prove that circumcircle of 4AXY tangent to circumcirle of 4BHC.


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Let M midpoint of BC. Soppose that AM intersects circumcircle of4ABC at A′′.LetA′ on AM such that A′M = A′′M .

MA′′.MA = MB2 = MC2 = MA′.MA ⇒ MC

MA′=MA

MC

Also ∠A′MC = ∠CMA, therefore:

4A′MC ∼ 4CMA ⇒ ∠A′CM = ∠MAC ⇒ ∠MA′C = ∠C

Similarly we can say:

4A′MB ∼ 4BMA ⇒ ∠A′BM = ∠MAB ⇒ ∠MA′B = ∠B

⇒ ∠BA′C = ∠MA′B + ∠MA′C = ∠B + ∠C = 180◦ − ∠A = ∠BHC

Therefore BHA′C is cyclic quadrilateral.

On the other hand we know:

MA′ = MA′′ =BC2

4AM=

a2

4ma

, ma2 =

b2 + c2

2− a2

4

Solutions 35

⇒ AA′

AA′′=ma − a2

ama

ma + a2

ma

=ma

2 − a2

4

ma2 + a2

4

=b2 + c2 − a2

b2 + c2=

2

3⇒ AA′

AA”=AX

XB=AY

Y C

According torevers of Tales’s theorem we can say: XY ‖ BC, XA′ ‖ BA”, A′Y ‖A”C. Therefore ∠XY A = ∠C = ∠BA′′A′ = ∠XA′A, therefore AXA′Y is cyclicquadrilateral.

Now we show that circumcircle of 4AXY tangent to circumcirle of 4BHC at A′.Letline L such that be tangent to circumcircle 4BA′C at A′. Suppose that L intersectsAB, AC at E,F respectively. We have to show L is tangent to circumcircle 4AXYat A′. To prove this we have to show:

∠EA′A = ∠XY A+ ∠XAA′

We know that:

∠XY A = ∠C = ∠MA′C , ∠XAA′ = ∠A′BC = ∠CA′F

Also:∠EA′A = ∠MA′F = ∠MA′C + ∠CA′F

Therefore:∠EA′A = ∠XY A+ ∠XAA′

Solutions 36

17.(IGO Short list)Suppose AD is the angle bisector of the 4ABC and D′ isthe midpoint of it. We draw a line from A which is parallel to BC. This line meetsthe circumcircle of 4ABC at A1. Let T ′ be the midpoint of the smaller arc AA1.T ′D′ meets the circumcircle at A2. We define B1, B2, C1, C2 in the same way. Provethat A1A2, B1B2, C1C2 are concurrent.


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solution.Let M , N be the midpoints of AB, AC. With an inversion with the center A and theratio AB·AC

2and a reflection about AD, B goes to N , C goes to M , the circumcircle

goes to MN so T ′ goes to T ′1(the intersection of external bisector and MN), D′ goesto T so A1 goes to the intersection of the circumcircle of 4AT ′1T and MN or A′1(theintersection of perpendicular bisector and MN), A2 goes to A′2(the intersection oftangent from A to the circumcircle of 4ABC and MN).

Because OA′1AA2 is cyclic so A1A2 passes throught the inverse of O (which is thefoot of altitude drawn from A). so the following problem should be solved :

Solutions 37

We have to show that A2A3, B2B3, C2C3 are concurrent. Because of AA2 = 2·A3Mthese lines passes throught the centroid of 4ABC.

Solutions 38

18.(National Math Olympiad (Second Round)) The arbitrary circle whichpasses through B and C from4ABC intersects AC and AB in D and E respectively.Suppose CE intersects BD in P and H be the foot of the perpendicular drawn fromP to AC .If M and N be the midpoints of BC and AP respectively. Prove that4MNH and 4CAE are similar.

Proposed by Amir Saeedi

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solution.Let T and K be reflections of piont P about H and M respectively. We know

that 4MNH and 4KAT are similar.(because PN = NA, PH = HT , PN = NK. Therefore, sides of this two trianglesare parallel due to Thales’ Theorem). On the other hand, 4ABD and 4ACE aresimilar because of ∠EBD = ∠ECD. So we have to prove that 4AKT and 4ABDare similar.To prove that we have to prove ∠BAD = ∠KAT and AB

AD= AK

AT.

Notice that this is equal to the similarity of4ABK and4ABT (because ∠BAD =∠KAT if and only if ∠BAK = ∠DAT ).We know that T is reflection of point P about AC. Therefore, 4ADP and 4ADTare equal. so we have to show that 4ABK and 4ADP are similar.

In quadrilateral BKCP , BM = MC and PM = MK. Therefore this quadrilat-

Solutions 39

eral is a parallelogram. This leades to :

∠ABK = ∠AEC = 180◦ − ∠BEC = 180◦ − ∠BDC = ∠ADP

On the other hand, BK = CP . so we have to show that ADDP

= ABBK

= ABCP

Because of the Law of Sines in 4ABD and 4PDC we have to show :

sin(∠ABD)

sin(∠ADB)=sin(∠DCP )

sin(∠CDP )

We know that :

∠ABD = ∠DCP ⇒ sin(∠ABD) = sin(∠DCP )

∠ADB = 180◦ − ∠CDP ⇒ sin(∠ADB) = sin(∠CDP )

So the equality holds.

Solutions 40

19.(National Math Olympiad (Second Round)) Let ABCD be a quadrilat-eral such that AC is the angle bisector of ∠DAB and ∠ADC = ∠ACB. Let X andY be the foot of the perpendiculars drawn from A to BC and CD respectively. provethat BD passes through the orthocenter of 4AXY .


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solution.Let E the foot of the perpendicular drawn from Y to AX. Suppose Y E intersets

BD in P . It’s enough to show that XP ⊥ AY . This equals to XP ‖ CD. We knowthat:

Y E ‖ CB ⇒ DY

Y C=DP

PB

On the other hand:

4ADC ∼ 4ACB ⇒ DY

Y C=CX

XB

Therefore:DP

PB=CX

XB

So because of reverse of Thales’ Theorem XP and CD are parallel.

Solutions 41

20.(Third Round) Let A′ on circumcircle of 4ABC such that AA′ be diameter.Suppose that Line I and I ′ are parallel to extenal bisector of ∠BAC and intenal bi-sector of ∠BAC respectively, such that passing through A′. Line I intersects BC,ACat C2, C1 respectively and Line I ′ intersects BC,AB at B2, B1 respectively. Provethat circumcircles of 4ABC, 4CC1C2 and 4BB1B2 are concur.

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solution.Suppose that circumcircle of 4ABC intesects circumcircle of 4CC1C2 at P and I ′

intesects AC at K. We have to show quadrilateral BB1PB2 is cyclic. Suppose that:∠ABC = 2α , ∠ABC = θ. We know that:

∠A′KC = α , ∠KA′C1 = ∠A′CK = 90◦ ⇒ ∠CA′C2 = α

∠CC2A′ + ∠CA′C2 = ∠BCA′ = 2α + θ − 90◦ ⇒ ∠CC2A

′ = α + θ − 90◦

On the other hand quadrilaterals PCC1C2 and ABCP are cyclic, therefore:

∠CPC1 = ∠CC2C1 = α + θ − 90◦ , ∠A′PC = 90◦ − θ ⇒ ∠A′PC1 = α

therefore quadrilateral KPC1A′ is cyclic.

⇒ ∠KA′P = ∠KC1P = ∠B2C2P ⇒ PB2A′C2 : cyclic

⇒ ∠B2PA′ = ∠B2C2A

′ = α + θ − 90◦ , ∠BPA′ = 2α + θ − 90◦

⇒ ∠BPB2 = ∠BB1B2 = α

therefore quadrilateral BB1PB2 is cyclic.

Solutions 42

21.(Third Round)Tow points P,Q lay in isosceles triangle ABC (AB = AC).Suppose Q lay in ∠PAC and ∠PAQ = ∠BAC

2. X is intersection of AP,BQ and Y is

intersection of AQ,CP . If CQ = PQ = BP prove that quadrilateral PQY X is cyclic.


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solution.Let Z be a point such that: AZ = AB = AC and ∠ZAP = ∠PAB.We know that:

AZ = AB , AX = AX , ∠ZAP = ∠PAB ⇒ 4BAP = 4ZAP

Also we know that ∠PAQ = ∠BAC2

, therefore ∠ZAQ = ∠QAC. Similarly We cansay: 4CAQ = 4ZAQ.

Also we know that: 4BPX = 4ZPX and 4ZQY = 4QY C. Therfore:

∠PBX = ∠PZX , ∠QZY = ∠Y CQ

BP = PQ ⇒ ∠PBQ = ∠PQB , QC = PQ ⇒ ∠QPC = ∠PCQ

⇒ ∠PZX = ∠XQP , ∠QZY = ∠QPY

We can say quadrilaterals PXZQ and PZY Q are cyclic. Therefore quadrilateralsPQY X is cyclic.

Solutions 43

22.(Third Round)Distinct points B,B′, C, C ′ lie on an arbitrary line `. A is apoint not lying on `. A line passing through B and parallel to AB′ intersects withAC in E and a line passing through C and parallel to AC ′ intersects with AB in F .Let X be the intersection point of the circumcircles of 4ABC and 4AB′C ′(A 6= X).Prove that EF ‖ AX.

Proposed by Ja’far Naamdaar

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Because of the sine form of Ceva’s theorem in 4ACX we have :

sin∠XAB′

sin∠B′CX=

sin∠AXB′

sin∠B′XC· sin∠B′AC

sin∠B′CAWe know that ∠XAB′ = ∠XC ′C, ∠B′XC = 180◦ − ∠BAC ′, ∠AXB′ = ∠BC ′A.Therefore :

sin∠XC ′Csin∠XCC ′

=sin∠B′CAsin∠BAC ′

· sin∠B′ACsin∠B′CA

Because of the Law of sines we have XCXC′ = sin∠XC′C

sin∠XCC′ ,BABC′ = sin∠B′CA

sin∠BAC′ ,B′CB′A

= sin∠B′ACsin∠B′CA

.Thus we have :

XC

XC ′=BA ·B′CB′A ·BC ′

From Thales’ theorem we have BEB′A

= BCB′C

, BFBA

= BCBC′ . So we have :

XC

XC ′=BF

BE

Clearly we have ∠CXC ′ = ∠B′AB = ∠FBE. So 4XCC ′ ∼ 4BEF . This leads to∠BEF = ∠CC ′X = ∠B′AX. Therefore EF ‖ AX.

Solutions 44

23.(Third Round)D is an arbitrary point lying on side BC of 4ABC. Circleω1 is tangent to segments AD , BD and the circumcircle of 4ABC and circle ω2 istangent to segments AD , CD and the circumcircle of 4ABC. Let X and Y be theintersection points of ω1 and ω2 with BC respectively and take M as the midpoint ofXY . Let T be the midpoint of arc BC which does not contain A. If I is the incenterof 4ABC, prove that TM goes through the midpoint of ID.

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solution.First we prove this lemma :

lemma(1):D is an arbitrary point lying on side BC of 4ABC. Circle ω is tan-gent to segments AD , BD and the circumcircle of 4ABC. Let F and E be theintersection points of AD and BD with ω. prove that EF passes throught the incen-ter of 4ABC.

proof. Suppose that T is the midpoint of arc BC. We know that KE passesthrought T . Let X be the intersection of AT , EF . We need to show that X is theincenter of 4ABC. KF passes throught the midpiont of arc AD′. Suppose that∠KAD = 2α, ∠ACB = 2θ, ∠DAT = x. Thus we have :

Solutions 45

∠FKA = ∠FKD′ = 90◦ − α− θ∠D′KT = x

}⇒ ∠EKF = 90◦ − α− θ − x

⇒ ∠EFD = ∠XFA = 90◦ − α− θ − x⇒ ∠FXT = 90◦ − α− θ

On the other hand we have ∠FKA = 90◦ − α − θ. So KFXA is cyclic and wehave :

∠XKF = x⇒ ∠XKE = 90◦ − α− θ = ∠FXT ⇒4TXE ∼ 4TKX

⇒ TX2 = TE · TK = TB2 = TC2

So TX = TB = TC and we proved the statement.

Now according to the lemma(1), I is the intersection of XX ′ and Y Y ′ and∠XIY = 90◦. Let N be the midpoint of X ′Y ′. Points T , M , N have equal powerto both ω1 and ω2. Therefore T , M , N are colinear. So we only have to prove thatMN passes throught the midpoint of ID. Because of IM = IY and DY = DY ′ wehave IM ‖ DN . On the other hand we have IM = DN = XD+DY

2. So IMDN is a

parallelogram. Therefore MN passes throught midpoint of ID.

Solutions 46

24.(Third Round) Let X, Y on line BC from triangle ABC such that ∠XAY =90◦. Suppose that H is orthocenter of 4ABC. If AX intersects BH at X ′ andAY intersects CH at Y ′, prove that intersection of circumcircle of 4BXX ′ andcircumcircle of 4CY Y ′ lay on X ′Y ′.

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solution.Suppose that CH intersects AB at E and BH intersects AC at F . Let P intersectionof circumcircle of 4CY Y ′ and circumcircle of BEFC. therefore: ∠BPC = 90◦.

∠XAY = ∠AFX ′ = 90◦ ⇒ ∠Y ′AF = ∠AX ′F , ∠EBF = ∠ECF

⇒ 4ABX ′ ∼ 4Y ′CA ⇒ Y ′A

AX ′=AB

Y ′C(1)

We know that: ∠AY ′P = ∠Y CP = ∠AEP , therefore quadrilateral AEY ′P iscyclic.Therefore ∠EAP = ∠PY ′C = ∠PY C. So we can say quadrilateral ABY P iscyclic. Therefore: ∠PY ′C = ∠PY C = ∠PAB.

Also:

∠Y ′CP = ∠Y ′Y P = ∠ABP ⇒ 4PY ′C ∼ 4PAB ⇒ AB

Y ′C=BP

PC(2)

According to equation 1 and 2 we can say:

Y ′A

AX ′=BP

PC, ∠Y ′AX ′ = ∠BPC = 90◦ ⇒ 4Y ′AX ′ ∼ 4CPB

Solutions 47

Therefore ∠AY ′P = ∠BCP = ∠AY ′X ′, so we can say Y ′, P,X ′ are collinear. Nowwe show that quadrilateral XX ′PB is cyclic.

ABY P : cyclic ⇒ ∠BAY = ∠BPY , ∠BAH = ∠ECB = ∠Y ′PY

In other hand quadrilateralABY P is cyclic, therefore ∠BAY = ∠BPY . So ∠HAY ′ =∠BPY ′ Also:

∠XAY = 90◦ , AH⊥BC ⇒ ∠BXA = ∠HAY ′ ⇒ ∠BXA = ∠BPY ′

therefore quadrilateral XX ′PB is cyclic.

Solutions 48

25.(Team Selection Test)Point Ib is the B-excenter of triangle ABC. If wedenote by M the midpoint of arc BC of the circumcircle of triangle ABC (the onethat does not contain vertex A), and MIb intersects the circumcircle of triangle ABCat T , prove that TI2

b = TB × TC.


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solution.Firstly, we prove two lammas.

lemma(1).In each triangle ABC with altitude AH and circumradius R, we haveAC.AB = 2R.AH.

proof.The proof is easy.

lemma(2). In triangle ABC, let N be the second intersection point of IbIc andω (other than A), where Ib and Ic are excenters of ABC and ω is its circumcircle.Then, the points Ib, Ic, C and B lie on a common circle with center N .

proof.Since CIc and CIb are internal and external angle bisectors of ∠C, we have∠IcCIb = 90◦. Similarly, ∠IbBIc = 90◦, and so quadrilateral IcBCIb is cyclic. Nowwe have

∠NIcB = ∠AIcB = 180◦ − ∠IcAB − ∠IcBA = 90◦ − ∠C2

Now since ∠IcNB = ∠ANB = ∠C, we conclude that triangle IcNB is isoscelesand so N lies on perpendicular bisector of IcB. By simialar arguments, N lies on theperpendicular bisector of IbC, too. Therefore, N is the center this circle.

Solutions 49

We keep using notations in lemma(2). Furthermore, dentoe by Γ the circumcirleof IbCBIc and let T1 be the second intersection point of MIb and Γ. Since M andN lie on the internal and external bisectors of ∠A, respectively, then MN must be adiameter of ω and so ∠NTM = 90◦. Now T must be the midpoint of IbT1, becauseNT ⊥ IbT1 and N is the center of circle passing through T1 and Ib. So NT is theperpendicular bisector of IbT1 and T1T = TIb. If T ′ is the foot of perpendicualr fromT to line BC(radical axis of ω and Γ) then by Casey’s theorem for point T and circlesω and Γ we get

P TΓ − P T

ω = 2ON.TT ′ ⇒ TI2b = TIb.TT1 = 2R.TT ′ (1)

According to the lemma 1 for triangle TBC, we deduce 2R.TT ′ = TB.TC (2).(1) and (2) imply TI2

b = TB.TC, as desired.

Solutions 50

26.(Team Selection Test)Quadrilateral ABCD is both inscribed and circum-scribed. Let E be the intersection point ofAD andBC, F the intersection point ofABand CD, S the intersection point of AC and BD and O the circumcenter of quadri-lateral ABCD. E ′ and F ′ are selected on AB and AD such that ∠BEE ′ = ∠AEE ′

and ∠AFF ′ = ∠DFF ′. Let M be the midpoint of arc BAD of the circumcircle ofthe quadrilateral and X a point collinear with O and E ′ such that XA

XB= EA

EB. Also

let Y be a point collinear with O and F ′ such that Y AY D

= FAFD

. Prove that the circlewith diameter OS, the circumcircle of triangle OAM and the circumcircle of triangleOXY are co-axis.


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solution.

An inversion with center O and constant r2 (r is the radius of the circumcircle ofquadrilateral), takes the circle with diameter OS to the line EF and the circumcircleof the triangle OAM to the line AM . Furthermore, since XO is the bisector of ∠AXBand OA = OB, we get the quadrilateral AXBO is cyclic and so OE ′.OX = r2.Simillarly, OF ′.OY = r2. Therefore, under this inversion the circumcircle of triangleOXY maps to the line E ′F ′. So it is enough to prove that the line EF , E ′F ′ andAM are concurrent.Denote by ω, ωE and ωF the incircles of quadrilateral ABCD, triangle EAB andtriangle FAD, respectively. Now E and E ′ are the external and internal homotheticcenters of ωE and ω, respectively. In the same manner F and F ′ are the external andinternal homothetic centers of ωF and ω, respectively. Therefore, EF and E ′F ′ meetat the internal homothetic center of ωF and ωE, say R. On the other hand, AM isthe bisector of ∠EAB = ∠FAD and so is the line of centers of ωF and ωE. Thus,R the internal homothetic center of ωE and ωF lies on AM . So we have proved thatthe lines EF , E ′F ′ and AM meet at K, which completes the proof. or equivalentlytheir poles with respect to the incircle of ABCD are collinear. Let K, G, H and Pbe the tangency points of this circle and sides AB, BC, CD and DA, respectively.Let L be the intersection point of KH and GP . Now the pole of EF is the pointL and the pole of AM is the midpoint of the segment KP . And the pole of E ′F ′ isthe intersection point of parallel line to PG from K and to KH from P ,say N . Nowsince NKLP is a parallelogram, evidently we have N , L and the midpoint of KP areon a common line and the proof is complete.

Solutions 51

27.(Team Selection Test)Point D is the intersection point of the angle bisectorof vertex A with side BC of triangle ABC, and point E is the tangency point of theinscribed circle of triangle ABC with side BC. A1 is a point on the circumcircle oftriangle ABC such that AA1||BC. If we denote by T the second intersection pointof line EA1 with the circumcircle of triangle AED and by I the incenter of triangleABC, prove that IT = IA.


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solution.

Let E1 be the reflection of E with respect to the midpoint of BC and X theintersection point of AE1 and EA1. We claim that IX ‖ BC. For this reason, wehave (Suppose that R is the radius of circumcircle of ABC and ∠B ≥ ∠C).

AX

XE1

=AA1

EE1

=AA1

AC − ABAI

IE=AB

BD=AC

CD=AB + AC

BC

So referring to the Thales’ Theorem, we must prove AA1.BC = AC2 − AB2.We have AA1 = BC − 2BH = BC − 2AB. cos(∠B), where H is the foot of per-pendicular from A to BC and on the other hand, by The Law of Cosines we haveAC2 − AB2 = BC2 − 2AB.BC cos(∠B). Therefore, the claim is proved.Now since the quadrilateral DEAT is cyclic and AA1 ‖ IX, we get that the quadri-lateral IATX is cyclic. Also, since pairs (E,E1) and (A,A1) are symmetric withrespect to the perpendicular bisector of the side BC, we have XE = XE1 and so

∠ATI = ∠AXI = ∠XE1E = ∠XEE1 = ∠IXE = ∠TAI

Thus, IT = IA.

Solutions 52

28.(Team Selection Test) Let ABC be an acute-angled triangle. Point Z onthe altitude of vertex A and points X and Y on the extensions of the altitudes ofvertices B and C are selected such that,

∠AY B = ∠BZC = ∠CXA = 90◦.

Prove that X, Y and Z are collinear if and only if the length of the tangent fromvertex A to the nine-point circle of the triangle is equal to the sum of the lengths oftangents from vertices B and C to this circle.

Proposed by Mohammad Javad Shabani

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solution.We denote the feet of altitudes of vertices A, B and C by D, E and F , respectively.Firstly, we prove that if X, Y and Z are collinear then the length of the tangent fromA to the nine-point circle of triangle ABC (which we denote by ω) is equal to thesum of the lengths of tangents from vertices B and C to this circle.

PAω =

1

2AE.AC =

1

2AF.AB =

1

2AX2 =

1

2AY 2

So the length of the tangent from A to ω is√

22AX =

√2

2AY . Similarly, the lengths of

tangents from vertices B and C are equal to BY and CX, respectively. If X, Y andZ lie on a common line we have Y Z + ZX = Y X and ∠XZC + ∠Y ZB = 90◦ (1).Since CX2 = CE.CA = CD.CB = CZ2, we get CX = CZ. In the same manner, weinfer AX = AY and BY = BZ. Using (1) we can deduce ∠XCZ + ∠ZBY = 180◦.The quadrilateral FZCB is cyclic and so ∠ABZ = ∠FCZ. Therefore,

∠Y BA+ ∠FCX = 90◦ ⇒ ∠Y BA+ (90◦ − ∠A) + ∠ACX = 90◦

Thus, ∠Y AX = 90◦ and so ∠AXY = 45◦. From these results, we get ∠ZBY =∠ZCX = 90◦. Hence,

√2Y B +

√2CX =

√2AX and so Y B + CX = AX, which is

what we wanted to show.For the converse we present the following lemma.

lemma.Let C1 and C2 be two perpendicular circles meeting each other at twopoints A and B. Then, There are exactly two points like T on the line AB satisfyingthe following property. If TY and TX are tangents from T to C1 and C2, respesc-tively, suth Y and X are in two different sides of AB. We have A, Y and X are on acommon line. Furthermore, these two points are symmetric with respect to the lineof centers of C1 and C2 and power of them with respect to C1 and C2 are both equalto (R1 +R2)2, where Ri is the radius of Ci.

Solutions 53

proof.First note that if the point T has this property, we have

1

2

_AY= ∠TY A = ∠TY X = ∠TXY = ∠TXA =

1

2

_AX

And since ∠O1AO2 = 90◦ (Oi is the center of Ci), we get

_AY=

_AX=

1

2

_AY +

1

2

_AX= ∠XAO2 + ∠Y AO1 = 180◦ − ∠O1AO2 = 90◦

So we must have_AY=

_AX= 90◦.

Now for the existence of such T , consider points Y and X on C1 and C2, respectively,

such that_AY=

_AX= 90◦. Note that we can choose such points in a uniqe way in

one side of O1O2. It is easy to see Y , A and X are collinear. Therefore, if T is theintersection point of the tangent to C1 at Y and the tangent to C2 at X, we infer

∠TY X =_Y A=

_XA= ∠TXY

So TY = TX. Thus, T lies on the radical axis of C1 and C2.On the other hand, in the hexagon TXO2AO1Y we have

∠TY O1 = ∠Y O1A = ∠O1AO2 = ∠AO2X = ∠O2XT = 90◦

Hence, ∠Y TX = 90◦, and obviously TX = AO2 + Y O1 = R2 +R1, as desired.

Now for the main problem note that circle ω1 with center B and radius BZ isperpendicular to the circle ω2 with center C and radius CZ, because ∠BZX = 90◦.Furthermore, since ∠AY B = ∠AXC = 90◦, AY is tangent to ω1 at Y and AX istangent to ω2 at X. on the other hand AY = AX = BY + CX, so according to thelemma the point having this property is unique in one side of BC. Therefore, againby lemma X, Y and Z are collinear.

Solutions 54

29.(Team Selection Test)From a point A outside circle ω, tangents AS and ATare drawn to the circle. Points X and Y are the midpoints of segments AT and AS,respectively. Tangent XR is drawn from point X to the circle and P and Q are themidpoints of segments XT and XR, respectively. If XY and PQ intersect each otherat K, and SX and TK intersect at L, prove that KRLQ is an inscribed quadrilateral.

Proposed by Farhad Seifollahi

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solution.If we consider A and X as circles with radius zero, then K is the radical centerof A, X and ω. Therefore, K lies on the perpendicular bisector of AX, and so∠STA = ∠KXA = ∠KAT . Let U be the intersection point of AK and TS.

By Thales’s Theorem K is the midpoint of AU . On the othe hand ∠ATS = ∠AST .Thus, triangles AUT and ATS are similar. Since TK and SX are medians of thesetriangles, we infer that ∠XSA = ∠STK. Hence, L lies on ω and also AXLK is acyclic quadrilateral. Now we have

RT ‖ PK ⇒ ∠RTK = ∠QKL

On the other hand, since XR is tangent to ω, we have ∠XRL = ∠RTK.Therefore,∠XRL = ∠QKL and so four points K, R, L and Q lie on a common circle.

Solutions 55

30.(Team Selection Test)H is the foot of the altitude of vertex A of triangleABC and H ′ is the reflection of H with respect to the midpoint of BC. If tangentsto the circumcircle of triangle ABC at points B and C intersect each other at X andthe perpendicular to XH ′ at H ′ intersects lines AB and AC at Y and Z, respectively,prove that ∠Y XB = ∠ZXC.


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solution.Let P , Q and M be the feet of perpendicular lines from X to AB, AC and BC,respectively. Obviousely, M is the midpoint of BC. We have

∠ZXY = ∠ZXH ′ + ∠H ′XY = ∠AQH ′ + ∠APH ′ = ∠PH ′Q− ∠A

Since we know ∠BXC = 180◦ − ∠2A, it suffices to prove ∠PH ′Q = 180◦ − ∠A.Note that ∠APM = ∠BXM = 90◦ − ∠A and so PM ⊥ AQ. Similarly, QM ⊥ AP .So M is the orthocenter of triangle APQ, and as a consequence AM ⊥ PQ.

Let R be the foot of the perpendicular line from A to XM . Thus, ARH ′M is aparallelogram. Now since AM ⊥ PQ and RH ′ ‖ AM , we infer PH ′ ⊥ PQ (1). On theother hand, since ∠ARX = ∠APX = ∠AQX, R lies on the circumcircle of APQ andso ∠PRQ = ∠A. Therefore, RH ′ = AM = 2r cos(∠A) = 2r cos(∠PRQ) (2), wherer is the radius of circle passing through A, R, P and Q. From (1) and (2), we deducethat H ′ is the orthocenter of triangle PRQ. This implies that ∠PH ′Q = 180◦ − ∠Aand this completes the proof.

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