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 erfan.salavati@gmail.com

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

Problems

Problems 4

1.(Geometry Olympiad(Junior and Senior level)) In a right triangle 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

Proposed by Mahdi Etesami Fard

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.

Proposed by Morteza Saghafian

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.

Proposed by Mahdi Etesami Fard

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.

Proposed by Mahdi Etesami Fard

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 .

Proposed by Ali Zooelm

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.

Proposed by Mahdi Etesami Fard

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

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

Proposed by Mahdi Etesami Fard

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

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.

Proposed by Morteza Saghafian

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

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

93-gon (so the sides of the second 93-gon would be multiplying by a constant 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.

Proposed by Morteza Saghafian

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

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

CD. AM intersects DE and BE at P,Q respectively. Therefore, 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◦

Proposed by Davood Vakili

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

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 .

Proposed by Ali Zooelm

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

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.

Proposed by Mahdi Etesami Fard

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−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.

Proposed by Mahdi Etesami Fard

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

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.