olymmath

IGO 2018

Iranian Geometry Olympiad 2018.

Elementary

Intermediate

I1

  • Intermediate

There are three rectangles in the following figure. The lengths of some segments are shown. Find the length of the segment XYXY . Image Proposed by Hirad Aalipanah

Answer, solutions and source

I2

  • Intermediate

In convex quadrilateral ABCDABCD , the diagonals ACAC and BDBD meet at the point PP . We know that ∠DAC=90o\angle DAC = 90^o and 2∠ADB=∠ACB2 \angle ADB = \angle ACB . If we have ∠DBC+2∠ADC=180o\angle DBC + 2 \angle ADC = 180^o prove that 2AP=BP2AP = BP .

Proposed by Iman Maghsoudi

Answer, solutions and source

I3

  • Intermediate

Let ω1,ω2\omega_1,\omega_2 be two circles with centers O1O_1 and O2O_2 , respectively. These two circles intersect each other at points AA and BB . Line O1BO_1B intersects ω2\omega_2 for the second time at point CC , and line O2AO_2A intersects ω1\omega_1 for the second time at point DD . Let XX be the second intersection of ACAC and ω1\omega_1 . Also YY is the second intersection point of BDBD and ω2\omega_2 . Prove that CX=DYCX = DY .

Proposed by Alireza Dadgarnia

Answer, solutions and source

I4

  • Intermediate

We have a polyhedron all faces of which are triangle. Let PP be an arbitrary point on one of the edges of this polyhedron such that PP is not the midpoint or endpoint of this edge. Assume that P0=PP_0 = P . In each step, connect PiP_i to the centroid of one of the faces containing it. This line meets the perimeter of this face again at point Pi+1P_{i+1} . Continue this process with Pi+1P_{i+1} and the other face containing Pi+1P_{i+1} . Prove that by continuing this process, we cannot pass through all the faces. (The centroid of a triangle is the point of intersection of its medians.)

Proposed by Mahdi Etesamifard - Morteza Saghafian

Answer, solutions and source

I5

  • Intermediate

Suppose that ABCDABCD is a parallelogram such that ∠DAC=90o\angle DAC = 90^o . Let HH be the foot of perpendicular from AA to DCDC , also let PP be a point along the line ACAC such that the line PDPD is tangent to the circumcircle of the triangle ABDABD . Prove that ∠PBA=∠DBH\angle PBA = \angle DBH .

Proposed by Iman Maghsoudi

Answer, solutions and source

Advanced

A1

  • Advanced

Two circles ω1,ω2\omega_1,\omega_2 intersect each other at points A,BA,B . Let PQPQ be a common tangent line of these two circles with P∈ω1P \in \omega_1 and Q∈ω2Q \in \omega_2 . An arbitrary point XX lies on ω1\omega_1 . Line AXAX intersects ω2\omega_2 for the second time at YY . Point Y′≠YY'\ne Y lies on ω2\omega_2 such that QY=QY′QY = QY' . Line Y′BY'B intersects ω1\omega_1 for the second time at X′X' . Prove that PX=PX′PX = PX' .

Proposed by Morteza Saghafian

Answer, solutions and source

A2

  • Advanced

In acute triangle ABC,∠A=45oABC, \angle A = 45^o . Points O,HO,H are the circumcenter and the orthocenter of ABCABC , respectively. DD is the foot of altitude from BB . Point XX is the midpoint of arc AHAH of the circumcircle of triangle ADHADH that contains DD . Prove that DX=DODX = DO .

Proposed by Fatemeh Sajadi

Answer, solutions and source

A3

  • Advanced

Find all possible values of integer n>3n > 3 such that there is a convex nn -gon in which, each diagonal is the perpendicular bisector of at least one other diagonal.

Proposed by Mahdi Etesamifard

Answer, solutions and source

A4

  • Advanced

Quadrilateral ABCDABCD is circumscribed around a circle. Diagonals AC,BDAC,BD are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments AB,BC,CDAB,BC,CD and DADA at points K,L,MK,L,M and NN . Given that KLMNKLMN is cyclic, prove that so is ABCDABCD .

Proposed by Nikolai Beluhov (Bulgaria)

Answer, solutions and source

A5

  • Advanced

ABCDABCD is a cyclic quadrilateral. A circle passing through A,BA,B is tangent to segment CDCD at point EE . Another circle passing through C,DC,D is tangent to ABAB at point FF . Point GG is the intersection point of AE,DFAE,DF , and point HH is the intersection point of BEBE , CFCF . Prove that the incenters of triangles AGFAGF , BHFBHF , CHECHE , DGEDGE lie on a circle.

Proposed by Le Viet An (Vietnam)

Answer, solutions and source