problems of 3rd iranian geometry olympiad 2016...

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Problems of 3rd Iranian Geometry Olympiad 2016 (Elementary)

1. Ali wants to move from point A to point B. He cannot walk inside theblack areas but he is free to move in any direction inside the white areas (notonly the grid lines but the whole plane). Help Ali to find the shortest pathbetween A and B. Only draw the path and write its length.

Proposed by Morteza Saghafian

2. Let ω be the circumcircle of triangle ABC with AC > AB. Let X be apoint on AC and Y be a point on the circle ω, such that CX = CY = AB.(The points A and Y lie on different sides of the line BC). The line XY inter-sects ω for the second time in point P . Show that PB = PC.

Proposed by Iman Maghsoudi

3. Suppose that ABCD is a convex quadrilateral with no parallel sides. Makea parallelogram on each two consecutive sides. Show that among these 4 newpoints, there is only one point inside the quadrilateral ABCD.


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4. In a right-angled triangle ABC (∠A = 90◦), the perpendicular bisectorof BC intersects the line AC in K and the perpendicular bisector of BK inter-sects the line AB in L. If the line CL be the internal bisector of angle C, findall possible values for angles B and C.

Proposed by Mahdi Etesami Fard

5. Let ABCD be a convex quadrilateral with these properties:∠ADC = 135◦ and ∠ADB−∠ABD = 2∠DAB = 4∠CBD. IfBC =

√2CD

prove that AB = BC +AD.


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Problems of 3rd Iranian Geometry Olympiad 2016 (Medium)

1. In trapezoid ABCD with AB ‖ CD, ω1 and ω2 are two circles with di-ameters AD and BC, respectively. Let X and Y be two arbitrary points on ω1

and ω2, respectively. Show that the length of segment XY is not more thanhalf of the perimeter of ABCD.


2. Let two circles C1 and C2 intersect in points A and B. The tangent toC1 at A intersects C2 in P and the line PB intersects C1 for the second timein Q (suppose that Q is outside C2). The tangent to C2 from Q intersects C1

and C2 in C and D, respectively (The points A and D lie on different sides ofthe line PQ). Show that AD is bisector of the angle CAP .


3. Find all positive integers N such that there exists a triangle which canbe dissected into N similar quadrilaterals.

Proposed by Nikolai Beluhov (Bulgaria) and Morteza Saghafian

4. Let ω be the circumcircle of right-angled triangle ABC (∠A = 90◦). Tan-gent to ω at point A intersects the line BC in point P . Suppose that M is themidpoint of (the smaller) arc AB, and PM intersects ω for the second time inQ. Tangent to ω at point Q intersects AC in K. Prove that ∠PKC = 90◦.

Proposed by Davood V akili

5. Let the circles ω and ω′ intersect in points A and B. Tangent to circleω at A intersects ω′ in C and tangent to circle ω′ at A intersects ω in D.Suppose that the internal bisector of ∠CAD intersects ω and ω′ at E and F ,respectively, and the external bisector of ∠CAD intersects ω and ω′ in X andY , respectively. Prove that the perpendicular bisector of XY is tangent to thecircumcircle of triangle BEF .


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Problems of 3rd Iranian Geometry Olympiad 2016 (Advanced)

1. Let the circles ω and ω′ intersect in A and B. Tangent to circle ω at Aintersects ω′ in C and tangent to circle ω′ at A intersects ω in D. Suppose thatthe segment CD intersects ω and ω′ in E and F , respectively (assume that Eis between F and C). The perpendicular to AC from E intersects ω′ in pointP and perpendicular to AD from F intersects ω in point Q (The points A, Pand Q lie on the same side of the line CD). Prove that the points A, P and Qare collinear.


2. In acute-angled triangle ABC, altitude of A meets BC at D, and M ismidpoint of AC. Suppose that X is a point such that ∠AXB = ∠DXM = 90◦

(assume that X and C lie on opposite sides of the line BM). Show that∠XMB = 2∠MBC.


3. Let P be the intersection point of sides AD and BC of a convex qualri-lateral ABCD. Suppose that I1 and I2 are the incenters of triangles PAB andPDC, respectively. Let O be the circumcenter of PAB, and H the orthocenterof PDC. Show that the circumcircles of triangles AI1B and DHC are tangenttogether if and only if the circumcircles of triangles AOB and DI2C are tangenttogether.

Proposed by Hooman Fattahimoghaddam

4. In a convex quadrilateral ABCD, the lines AB and CD meet at pointE and the lines AD and BC meet at point F . Let P be the intersection pointof diagonals AC and BD. Suppose that ω1 is a circle passing through D andtangent to AC at P . Also suppose that ω2 is a circle passing through C andtangent to BD at P . Let X be the intersection point of ω1 and AD, and Ybe the intersection point of ω2 and BC. Suppose that the circles ω1 and ω2

intersect each other in Q for the second time. Prove that the perpendicularfrom P to the line EF passes through the circumcenter of triangle XQY .


5. Do there exist six points X1, X2, Y1, Y2, Z1, Z2 in the plane such that allof the triangles XiYjZk are similar for 1 ≤ i, j, k ≤ 2?


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Solutions of 3nd Iranian Geometry Olympiad 2016 (Elementary)

1. Ali wants to move from point A to point B. He cannot walk inside theblack areas but he is free to move in any direction inside the white areas (notonly the grid lines but the whole plane). Help Ali to find the shortest pathbetween A and B. Only draw the path and write its length.


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Solution.

According to Pythagorean theorem, the length of the path AB is equal to:√32 + 32 +

√32 + 42 + 1 +

√22 + 22 + 1 = 7 + 5

√2

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2.Let ω be the circumcircle of triangle ABC with AC > AB. Let X be a pointon AC and Y be a point on the circle ω, such that CX = CY = AB. (Thepoints A and Y lie on different sides of the line BC). The line XY intersects ωfor the second time in point P . Show that PB = PC.


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Solution.We know that CX = CY therefore:

∠Y XC = ∠XY C ⇒_AP +

_CY=

_PC

Also we have AB = CY therefore_AP +

_CY=

_AP +

_AB=

_PB, so PB = PC.

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3. Suppose that ABCD is a convex quadrilateral with no parallel sides. Makea parallelogram on each two consecutive sides. Show that among these 4 newpoints, there is only one point inside the quadrilateral ABCD.


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Solution.It’s clear that the ray from B parallel to AD passes through the quadrilateralif and only if ∠DAB + ∠ABC > 180◦.

We have to find a parallelogram such that both of it’s rays pass thoroughABCD. Among A,B and C,D there is exactly one set with sum of anglesgreater than 180◦. Also among A,D and B,D there is exactly one set with sumof angles greater than 180◦. These two good sets have a vertex in common, sayA. So both of the rays from B parallel to AD, and from D parallel to AB, areinside the quadlirateral.

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4. In a right-angled triangle ABC (∠A = 90◦), the perpendicular bisector ofBC intersects the line AC in K and the perpendicular bisector of BK intersectsthe line AB in L. If the line CL be the internal bisector of angle C, find allpossible values for angles B and C.


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Solution.We have three cases:

Case i. AC > AB. We know that:

∠LBK = ∠LKB = α ⇒ ∠KLA = 2α ⇒ ∠LKA = 90◦ − 2α

BK = CK ⇒ ∠KBC = ∠KCB =∠BKA

2= 45◦ − α

2

Let T be a point on BC such that LT⊥BC. We know that the line CL is theinternal bisector of angle C, so LT = LA also we have LB = LK therefore twotriangles BTL and KAL are congruent.

⇒ ∠LBT = ∠LKA ⇒ 45◦ +α

2= 90◦ − 2α ⇒ α = 18◦

Therefore ∠B = 45◦ + α2 = 54◦ and ∠C = 36◦

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Case ii. AC < AB. We know that:

∠LBK = ∠LKB = α ⇒ ∠KLA = 2α ⇒ ∠LKA = 90◦ − 2α

Let T be a point on BC such that LT⊥BC. We know that the line CL is theinternal bisector of angle C, so LT = LA also we have LB = LK therefore twotriangles BTL and KAL are equal.

⇒ ∠LBT = ∠LKA = 90◦ − 2α ⇒ ∠CBK = ∠BKC = 90◦ − α

On the other hand we have:

BK = CK ⇒ ∠CBK = ∠BKC = 60◦ ⇒ α = 30◦

Therefore ∠B = 90◦ − 2α = 30◦ and ∠C = 60◦

Case iii. AC = AB. In this case, K ≡ A and L is the midpoint of AB. Let Tbe a point on BC such that LT⊥BC. We know that the line CL is the internalbisector of angle C, so LT = LA = LB which is impossible.

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5. Let ABCD be a convex quadrilateral with these properties:∠ADC = 135◦ and ∠ADB−∠ABD = 2∠DAB = 4∠CBD. IfBC =

√2CD

prove that AB = BC +AD.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Solution.Suppose that ∠CBD = α, so ∠DAB = 2α, therefore:

∠ADB − ∠ABD = 4α , ∠ADB + ∠ABD = 180◦ − 2α

⇒ ∠ADB = 90◦ + α , ∠ABD = 90◦ − 3α ⇒ ∠DAB + ∠CBA = 90◦

Let P be intersection point of AD and BC. So we have ∠APB = 90◦. On the

other hand we know that ∠PDC = 45◦, therefore PD =√22 CD = BC

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Let the point Q be the reflection of point D in point P , Thus QD = 2PD = BC.We know that two triangles DPB and QPB are congruent. So ∠CBD =∠CBQ = α, therefore ∠ABQ = 90◦ − α. On the other hand ∠DAB = 2α, sothe triangle ABQ is isosceles.

⇒ AB = AQ ⇒ AB = DQ+AD = BC +AD

.

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Solutions of 3nd Iranian Geometry Olympiad 2016 (Medium)

1. In trapezoid ABCD with AB ‖ CD, ω1 and ω2 are two circles with di-ameters AD and BC, respectively. Let X and Y be two arbitrary points on ω1

and ω2, respectively. Show that the length of segment XY is not more thanhalf of the perimeter of ABCD.


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First solution.Let O1 and O2 be the centers of circles ω1 and ω2, respectively. It’s clear thatO1 and O2 are the midpoints of AD and BC, respectively.

XO1 =AD

2, Y O2 =

BC

2, O1O2 =

AB + CD

2

⇒ XY ≤ XO1 +O1O2 + Y O2 =AB +BC + CD +DA

2

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Second solution.The farthest points of two circles lie on their center line.

And it’s clear in the figure that:

XO1 =AD

2, O1O2 =

AB + CD

2, Y O2 =

BC

2

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2. Let two circles C1 and C2 intersect in points A and B. The tangent to C1

at A intersects C2 in P and the line PB intersects C1 for the second time in Q(suppose that Q is outside C2). The tangent to C2 from Q intersects C1 andC2 in C and D, respectively (The points A and D lie on different sides of theline PQ). Show that AD is bisector of the angle CAP .


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Solution.We know that:

∠CAB = ∠CQB , ∠DAB = ∠BDQ

⇒ ∠CAD = ∠CAB + ∠DAB = ∠CQB + ∠BDQ = ∠PBD = ∠PAD

Therefore AD is the bisector of ∠CAP .

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3. Find all positive integers N such that there exists a triangle which can bedissected into N similar quadrilaterals.

Proposed by Nikolai Beluhov (Bulgaria) and Morteza Saghafian

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Solution.For N = 1 it’s clear that this is impossible. Also for N = 2 this dissection isimpossible too, because one of the two quadrilaterals is convex and the other isconcave. For N ≥ 3 we can do this kind of dissection in equilateral triangle.

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4. Let ω be the circumcircle of right-angled triangle ABC (∠A = 90◦). Tangentto ω at point A intersects the line BC in point P . Suppose that M is themidpoint of (the smaller) arc AB, and PM intersects ω for the second time inQ. Tangent to ω at point Q intersects AC in K. Prove that ∠PKC = 90◦.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Solution.Suppose that AB < AC. It’s enough to show that PK ‖ AB.

4PMA ∼ 4PAQ ⇒ AQ

MA=PQ

PA, 4PMB ∼ 4PCQ ⇒ MB

QC=PB

PQ

4PBA ∼ 4PAC ⇒ AC

BA=PA

PBWe know that MA = MB, so according to above three equations we can saythat:

AQ

QC=BA

AC(1)

4KAQ ∼ 4KQC ⇒ KA

KQ=KQ

KC=AQ

QC⇒ KA

KC= (

AQ

QC)2 (2)

4PBA ∼ 4PAC ⇒ PB

PA=PA

PC=BA

AC⇒ PB

PC= (

BA

AC)2 (3)

(1), (2), (3) ⇒ KA

KC=PB

PC⇒ PK ‖ AB

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The solution is the same in case of AB > AC.

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5. Let the circles ω and ω′ intersect in points A and B. Tangent to circle ω at Aintersects ω′ in C and tangent to circle ω′ at A intersects ω in D. Suppose thatthe internal bisector of ∠CAD intersects ω and ω′ at E and F , respectively, andthe external bisector of ∠CAD intersects ω and ω′ in X and Y , respectively.Prove that the perpendicular bisector of XY is tangent to the circumcircle oftriangle BEF .


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Solution.Suppose that P is the intersection point of XE and Y F . We know that:

∠EXA = ∠EAC = ∠EAD = ∠FY A = α ⇒ PX = PY

∠ABE = ∠EXA = α , ∠ABF = 180◦ − ∠FY A = 180◦ − α⇒ ∠EBF = ∠XPY = 180◦ − 2α ⇒ PEBF : cyclic

EF⊥XY ⇒ ∠PEF = ∠AEX = ∠AFY ⇒ PE = PF

We proved that PE = PF and the quadrilateral PEBF is cyclic. Therefore, Pis the midpoint of arc EF in the circumcircle of triangle BEF . Also we knowthat the perpendicular bisector of XY is parallel to EF and passes through P .So the perpendicular bisector of XY is tangent to the circumcircle of triangleBEF at P .

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Solutions of 3nd Iranian Geometry Olympiad 2016 (Advanced)

1. Let the circles ω and ω′ intersect in A and B. Tangent to circle ω at Aintersects ω′ in C and tangent to circle ω′ at A intersects ω in D. Suppose thatCD intersects ω and ω′ in E and F , respectively (assume that E is between Fand C). The perpendicular to AC from E intersects ω′ in point P and perpen-dicular to AD from F intersects ω in point Q (The points A, P and Q lie onthe same side of the line CD). Prove that the points A, P and Q are collinear.


−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Solution.We know that:

∠AFC = ∠AED = 180◦ − ∠CAD , ∠AEF = 180◦ − ∠AQD

⇒ ∠AFD = ∠AQD

So the point Q is the reflection of the point F in the line AD. Similarly we cansay the point P is the reflection of the point E in the line AC. Therefore:

∠DAQ = ∠DAF = ∠ACD , ∠CAP = ∠CAE = ∠CDA

⇒ ∠DAQ+ ∠CAD + ∠CAP = ∠ACD + ∠CAD + ∠CDA = 180◦

So the points A, P and Q are collinear.

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2. In acute-angled triangle ABC, altitude of A meets BC at D, and M ismidpoint of AC. Suppose that X is a point such that ∠AXB = ∠DXM =90◦ (assume that X and C lie on opposite sides of the line BM). Show that∠XMB = 2∠MBC.


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First solution.Let N be the midpoint of side AB. So MN ‖ BC and ∠MBC = ∠NMB.Therefore it’s enough to show that the line MN is the bisector of ∠XMB.

∠ADB = ∠AXB = 90◦ ⇒ AXDB : cyclic

⇒ ∠BXD = ∠BAD = 90◦−∠ABC ⇒ ∠BXM = 180◦−∠ABC = ∠BNM

⇒ BNXM ; cyclic , AN = NX = BN ⇒ ∠BMN = ∠XMN

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Second solution.Let P be the intersection point of XM and BC. Suppose that Q is the pointsuch that the quadrilateral ADBQ be a rectangle. We know that:

∠DXP = ∠ADP = 90◦ ⇒ ∠ADX = ∠XPD

Also we know that AXDBQ is cyclic, so:

∠ADX = ∠AQX ⇒ ∠AQX = ∠XPD

So Q, X and P are collinear because AQ ‖ BP .

AM = MC and AQ ‖ BP ⇒ QM = MP

Now we know that ∠QBC = 90◦, thus:

QM = BM = MP ⇒ ∠XMB = 2∠MBC

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3. In a convex qualrilateral ABCD, let P be the intersection point of AC andBD. Suppose that I1 and I2 are the incenters of triangles PAB and PDCrespectively. Let O be the circumcenter of PAB, and H the orthocenter ofPDC. Show that the circumcircles of triangles AI1B and DHC are tangenttogether if and only if the circumcircles of triangles AOB and DI2C are tangenttogether.

Proposed by Hooman Fattahimoghaddam

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Solution.Suppose that the circumcircles of triangles AI1B and DHC is tangent togetherat point K. Let Q be the second intersection point of circumcircles of trianglesAKD and BKC. we know that:

∠DHC = ∠DKC = 180◦ − ∠P

∠P + ∠PDK + ∠PCK = ∠DKC ⇒ ∠PDK + ∠PCK = 180◦ − 2∠P

AQKD : cyclic ⇒ ∠AQK = 180◦ − ∠PDK

BQKC : cyclic ⇒ ∠BQK = 180◦ − ∠PCK

⇒ ∠AQB = ∠PDK+∠PCK = 180◦−2∠P = 180◦−∠AOB ⇒ AOBQ : cyclic

Also we have ∠AKD = ∠AQD , ∠BKC = ∠BQC and ∠AQB = ∠DKC−∠P .So ∠CQD = ∠AKB + ∠P = 180◦ − ∠AI1B + ∠P = 90◦ + ∠P

2 = ∠CI2D.

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So the qudrilateral CDQI2 is cyclic. So we have to show that circumcircles oftriangles AOB and DI2C is tangent together at the point Q. It’s enough toshow that:

∠ABQ+ ∠DCQ = ∠AQD

We know that the circumcircles of triangles AI1B and DHC are tangent to-gether at the point K, so we have:

∠ABK + ∠DCK = ∠AKD

⇒ (∠ABQ+ ∠KBQ) + (∠DCQ− ∠KCQ) = ∠AKD

We know that ∠KBQ = ∠KCQ and ∠AKD = ∠AQD, So:

∠ABQ+ ∠DCQ = ∠AQD

Therefore the circumcircles of triangles AOB and DI2C are tangent togetherat point Q.On the other side of the problem, Suppose that the circumcircles of trianglesCI2D and AOB are tangent together at point Q. Let the point K be the secondintersection of circumcircles of triangles AQD and BQC. Similarly we can showthat the circumcircles of triangles AI1B and DHC are tangent together at thepoint K.

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Comment.Also there is another solution using inversion with respect to a circle withMichel’s point of the quadrilateral as its center.

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4. In a convex quadrilateral ABCD, the lines AB and CD meet at point Eand the lines AD and BC meet at point F . Let P be the intersection pointof diagonals AC and BD. Suppose that ω1 is a circle passing through D andtangent to AC at P . Also suppose that ω2 is a circle passing through C andtangent to BD at P . Let X be the intersection point of ω1 and AD, and Ybe the intersection point of ω2 and BC. Suppose that the circles ω1 and ω2

intersect each other in Q for the second time. Prove that the perpendicularfrom P to the line EF passes through the circumcenter of triangle XQY .


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First solution.

Lemma 1. In the convex quadrilateral ABCD, the lines AB and CD meetat point E and the lines AD and BC meet at point F . Let point P be theintersection of AC and BD. Suppose that X and Y be two arbitrary points onAD and BC, respectively. If BC ∩ PX = U and AD ∩ PY = V , then the linesXY , UV and EF are concurrent.proof.Let point Z be the intersection of XY and UV . Suppose that PF ∩ UV = Land PF ∩ CD = K. We know that:

(Z,L, V, U) = −1 , (E,K,D,C) = −1

If ZF intersects CD in E′, so we can say that (E′,K,D,C) = −1. ThereforeE ≡ E′, so the lines XY , UV and EF are concurrent.

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Lemma 2. In the cyclic quadrilateral ABCD with circumcenter O, the linesAB and CD meet at point E and the lines AD and BC meet at point F . Ifpoint P be the intersection of AC and BD, then PO⊥EF .proof.Let ω be the circumcircle of quadrilateral ABCD. Suppose that point R is theintersection of tangents to circle ω at A and C, and point S is the intersectionof tangents to circle ω at B and D.According to Pascal’s theorem in Hexagonal AABCCD and ABBCDD, we cansay that points R and S lie on line EF .

We know that polar of the point R with respect to circle ω passes through P .So polar of the point P with respect to circle ω passes through R. Similarly,we can say that polar of the point P with respect to circle ω passes through S.Therefore polar of the point P with respect to circle ω is EF . So PO⊥EF .

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Suppose that PX intersects BC in point U , and PY intersects AD in point V .

∠XQP = α ⇒ ∠XDP = ∠XPA = ∠UPC = α

∠Y QP = θ ⇒ ∠Y CP = ∠Y PB = ∠V PD = θ

⇒ ∠XV Y = ∠XQY = ∠XUY = α+ θ ⇒ QVXY U : cyclic

Let point O be the circumcenter of QVXY U . According to lemma 1, we cansay that XY , UV and EF are concurrent at point Z. Now according to lamme2, we can say that PO⊥EF . So the perpendicular from P to EF passes throughthe circumcircle of triangle XQY .

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Second solution.Suppose that point O is the circumcenter of triangle XQY . The inversion withrespect to a circle with center P trasnforms the problem into this figure. Supposethat X ′ is the inversion of point X wrt P . We have to show that the line PO′

is the diameter of circumcircle of triangle E′PF ′. Let O′′ be the circumcenterof triangle X ′Q′Y ′. We know that the points P , O′ and O′′ are collinear. Sowe have to show that he line PO′′ passes through the circumcenter of triangleE′PF ′.

Suppose that O1, O2, O3 and O4 are the centers of circles in the above figureand K be the intersection point of O1O3 and O2O4. We know that point Klies on perpendicular bisector of PE′ and PF ′, thus K is the circumcenter oftriangle PE′F ′. So we have to show that P , K and O′′ are collinear. On theother hand, we know that the quadrilateral D′B′Y ′Q′ is isosceles trapezoid. Sothe point O′′ lies on perpendicular bisector of B′D′. Similarly, the point O′′ lieson perpendicular bisector of A′C ′. Therefore, the point O′′ is the intersectionof A′C ′ and B′D′.

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Suppose that:A′C ′ ∩O1O2 = M , A′C ′ ∩O3O4 = T

B′D′ ∩O2O3 = N , B′D′ ∩O1O4 = L

Let points U and V be on A′C ′ such that KU⊥A′C ′ and O′′V⊥A′C ′. Also letpoints R and S be on B′D′ such that KR⊥B′D′ and O′′S⊥B′D′.

We know that O1O2 and O3O4 are perpendicular to A′C ′. So O1O2 ‖ O3O4

Similarly O2O3 ‖ O1O4, therefore the quadrilateral O1O2O3O4 is a parallelo-gram. It means that the point K lies on the midpoint of the segments O1O3

and O2O4. So UM = UT . Also we have A′M = PM and C ′T = PT

⇒ PV = A′V −A′P = (PM + PT )− 2PM = PT − PM

⇒ TV = PT − PV = PM ⇒ UP = UV

Similarly, we can show that RP = RS, so point K lies on the perpendicularbisector of PV and PS. It means that K is the citcumcenter of triangle PSV .Therefore the points P , K and O′′ are collinear.

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5. Do there exist six points X1, X2, Y1, Y2, Z1, Z2 in the plane such that all ofthe triangles XiYjZk are similar for 1 ≤ i, j, k ≤ 2.


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Solution. (by Ilya Bogdanov from Russia)Suppose a triangle XY Z, in such a way that XY = 1, Y Z = t2, ZX = t3 and∠Z = ∠X + 2∠Y .

Such a triangle exists, because for the minimum possible value of t, we have∠Z > ∠X + 2∠Y and for t = 1 we have ∠Z < ∠X + 2∠Y . So there exists atriangle with the above properties. Now consider the following 6 points, thesepoints have the properties of the problem.

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So there exist the points X1, X2, Y1, Y2, Z1, Z2 in the plane such that XiYjZkbe the similar triangles for all of 1 ≤ i, j, k ≤ 2

problems of 3rd iranian geometry olympiad 2016...

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