Navvye/Olympiad-Problems-CSV huggingface dataset:
2019 Iranian Geometry Olympiad
Quadrilateral $ABCD$ is given such that

$$\angle DAC = \angle CAB = 60^\circ,$$and

$$AB = BD - AC.$$Lines $AB$ and $CD$ intersect each other at point $E$. Prove that $$

    \angle ADB = 2\angle BEC.

    $$

Proposed by Iman Maghsoudi
2019 Iranian Geometry Olympiad
For a convex polygon (i.e. all angles less than $180^\circ$) call a diagonal bisector if its bisects both area and perimeter of the polygon. What is the maximum number of bisector diagonals for a convex pentagon?



Proposed by Morteza Saghafian
2019 Iranian Geometry Olympiad
Two circles $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ respectively intersect each other at points $A$ and $B$, and point $O_1$ lies on $\omega_2$. Let $P$ be an arbitrary point lying on $\omega_1$. Lines $BP, AP$ and $O_1O_2$ cut $\omega_2$ for the second time at points $X$, $Y$ and $C$, respectively. Prove that quadrilateral $XPYC$ is a parallelogram.



Proposed by Iman Maghsoudi
2019 Iranian Geometry Olympiad
Find all quadrilaterals $ABCD$ such that all four triangles $DAB$, $CDA$, $BCD$ and $ABC$ are similar to one-another.



Proposed by Morteza Saghafian
2019 Iranian Geometry Olympiad
Three circles $\omega_1$, $\omega_2$ and $\omega_3$ pass through one common point, say $P$. The tangent line to $\omega_1$ at $P$ intersects $\omega_2$ and $\omega_3$ for the second time at points $P_{1,2}$ and $P_{1,3}$, respectively. Points $P_{2,1}$, $P_{2,3}$, $P_{3,1}$ and $P_{3,2}$ are similarly defined. Prove that the perpendicular bisector of segments $P_{1,2}P_{1,3}$, $P_{2,1}P_{2,3}$ and $P_{3,1}P_{3,2}$ are concurrent.



Proposed by Mahdi Etesamifard
2019 Iranian Geometry Olympiad
Let $ABCD$ be a parallelogram and let $K$ be a point on line $AD$ such that $BK=AB$. Suppose that $P$ is an arbitrary point on $AB$, and the perpendicular bisector of $PC$ intersects the circumcircle of triangle $APD$ at points $X$, $Y$. Prove that the circumcircle of triangle $ABK$ passes through the orthocenter of triangle $AXY$.



Proposed by Iman Maghsoudi
2019 Iranian Geometry Olympiad
Let $ABC$ be a triangle with $\angle A = 60^\circ$. Points $E$ and $F$ are the foot of angle bisectors of vertices $B$ and $C$ respectively. Points $P$ and $Q$ are considered such that quadrilaterals $BFPE$ and $CEQF$ are parallelograms. Prove that $\angle PAQ > 150^\circ$. (Consider the angle $PAQ$ that does not contain side $AB$ of the triangle.)



Proposed by Alireza Dadgarnia
2019 Iranian Geometry Olympiad
Circles $\omega_1$ and $\omega_2$ intersect each other at points $A$ and $B$. Point $C$ lies on the tangent line from $A$ to $\omega_1$ such that

$\angle ABC = 90^\circ$. Arbitrary line $\ell$ passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut $\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot of altitude from $A$ to $\ell$. Prove that points $X, Y$ and $Z$ are collinear.



Proposed by Iman Maghsoudi
2019 Iranian Geometry Olympiad
Is it true that in any convex $n$-gon with $n > 3$, there exists a vertex and a diagonal passing through this vertex such that the angles of this diagonal with both sides adjacent to this vertex are acute?



Proposed by Boris Frenkin - Russia
2019 Iranian Geometry Olympiad
Circles $\omega_1$ and $\omega_2$ have centres $O_1$ and $O_2$, respectively. These two circles intersect at points $X$ and $Y$. $AB$ is common tangent line of these two circles such that $A$ lies on $\omega_1$ and $B$ lies on $\omega_2$. Let tangents to $\omega_1$ and $\omega_2$ at $X$ intersect $O_1O_2$ at points $K$ and $L$, respectively. Suppose that line $BL$ intersects $\omega_2$ for the second time at $M$ and line $AK$ intersects $\omega_1$ for the second time at $N$. Prove that lines $AM, BN$ and $O_1O_2$ concur.



Proposed by Dominik Burek - Poland
2019 Iranian Geometry Olympiad
Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of arc $BC$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX\parallel CY\parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN\parallel AM$, $\omega_b$ be a circle that passes through $B$, $X$ and tangents to $BC$ and $\omega_c$ be a circle that passes through $C$, $Y$ and tangents to $BC$. Prove that circle with center $K$ and radius $KP$ is tangent to 3 circles $\omega_b$, $\omega_c$ and $\Gamma$.



Proposed by Tran Quan - Vietnam
2019 Iranian Geometry Olympiad
Let points $A, B$ and $C$ lie on the parabola $\Delta$ such that the point $H$, orthocenter of triangle $ABC$, coincides

with the focus of parabola $\Delta$. Prove that by changing the position of points $A, B$ and $C$ on $\Delta$ so that the orthocenter remain at $H$, inradius of triangle $ABC$ remains unchanged.



Proposed by Mahdi Etesamifard
2018 Iranian Geometry Olympiad
As shown below, there is a $40\times30$ paper with a filled $10\times5$ rectangle inside of it. We want to cut out the filled rectangle from the paper using four straight cuts. Each straight cut is a straight line that divides the paper into two pieces, and we keep the piece containing the filled rectangle. The goal is to minimize the total length of the straight cuts. How to achieve this goal, and what is that minimized length? Show the correct cuts and write the final answer. There is no need to prove the answer.



Proposed by Morteza Saghafian
2018 Iranian Geometry Olympiad
Convex hexagon $A_1A_2A_3A_4A_5A_6$ lies in the interior of convex hexagon $B_1B_2B_3B_4B_5B_6$ such that $A_1A_2 \parallel B_1B_2$, $A_2A_3 \parallel B_2B_3$,..., $A_6A_1 \parallel B_6B_1$. Prove that the areas of simple hexagons $A_1B_2A_3B_4A_5B_6$ and $B_1A_2B_3A_4B_5A_6$ are equal. (A simple hexagon is a hexagon which does not intersect itself.)



Proposed by Hirad Aalipanah - Mahdi Etesamifard
2018 Iranian Geometry Olympiad
In the given figure, $ABCD$ is a parallelogram. We know that $\angle D = 60^\circ$, $AD = 2$ and $AB = \sqrt3 + 1$. Point $M$ is the midpoint of $AD$. Segment $CK$ is the angle bisector of $C$. Find the angle $CKB$.



Proposed by Mahdi Etesamifard
2018 Iranian Geometry Olympiad
There are two circles with centers $O_1,O_2$ lie inside of circle $\omega$ and are tangent to it. Chord $AB$ of $\omega$ is tangent to these two circles such that they lie on opposite sides of this chord. Prove that $\angle O_1AO_2 + \angle O_1BO_2 > 90^\circ$.



Proposed by Iman Maghsoudi
2018 Iranian Geometry Olympiad
There are three rectangles in the following figure. The lengths of some segments are shown.

Find the length of the segment $XY$ .

//cdn.artofproblemsolving.com/images/6/f/f/6ff334abf1755ad7fa792de3941109208965d5f8.png

Proposed by Hirad Aalipanah
2018 Iranian Geometry Olympiad
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that  $\angle DAC = 90^o$ and  $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.



Proposed by Iman Maghsoudi
2018 Iranian Geometry Olympiad
Let $\omega_1,\omega_2$ be two circles with centers $O_1$ and $O_2$, respectively. These two circles intersect each other at points $A$ and $B$. Line $O_1B$ intersects $\omega_2$ for the second time at point $C$, and line $O_2A$ intersects $\omega_1$ for the second time at point $D$ . Let $X$ be the second intersection of $AC$ and $\omega_1$. Also $Y$ is the second intersection point of $BD$ and $\omega_2$. Prove that $CX = DY$ .



Proposed by Alireza Dadgarnia
2018 Iranian Geometry Olympiad
We have a polyhedron all faces of which are triangle. Let $P$ be an arbitrary point on one of the edges of this polyhedron such that $P$ is not the midpoint or endpoint of this edge. Assume that $P_0 = P$. In each step, connect $P_i$ to the centroid of one of the faces containing it. This line meets the perimeter of this face again at point $P_{i+1}$. Continue this process with $P_{i+1}$ and the other face containing $P_{i+1}$. 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
2018 Iranian Geometry Olympiad
Suppose that $ABCD$ is a parallelogram such that $\angle DAC = 90^o$. Let $H$ be the foot of perpendicular from $A$ to $DC$, also let $P$ be a point along the line $AC$ such that the line $PD$ is tangent to the circumcircle of the triangle $ABD$. Prove that $\angle PBA = \angle DBH$.



Proposed by Iman Maghsoudi
2018 Iranian Geometry Olympiad
Two circles $\omega_1,\omega_2$ intersect each other at points $A,B$. Let $PQ$ be a common tangent line of these two circles with $P \in \omega_1$ and $Q  \in  \omega_2$. An arbitrary point $X$ lies on $\omega_1$. Line $AX$ intersects $ \omega_2$ for the second time at $Y$ . Point $Y'\ne Y$ lies on $\omega_2$ such that $QY = QY'$. Line $Y'B$ intersects $ \omega_1$ for the second time at $X'$. Prove that $PX = PX'$.



Proposed by Morteza Saghafian
2018 Iranian Geometry Olympiad
In acute triangle $ABC, \angle A = 45^o$. Points $O,H$ are the circumcenter and the orthocenter of $ABC$, respectively. $D$ is the foot of altitude from $B$. Point $X$ is the midpoint of arc $AH$ of the circumcircle of triangle $ADH$ that contains $D$. Prove that $DX = DO$.



Proposed by Fatemeh Sajadi
2018 Iranian Geometry Olympiad
Find all possible values of integer $n > 3$ such that there is a convex $n$-gon in which, each diagonal is the perpendicular bisector of at least one other diagonal.



Proposed by Mahdi Etesamifard
2018 Iranian Geometry Olympiad
Quadrilateral $ABCD$ is circumscribed around a circle. Diagonals $AC,BD$ are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments $AB,BC,CD$ and $DA$ at points $K,L,M$ and $N$. Given that $KLMN$ is cyclic, prove that so is $ABCD$.



Proposed by Nikolai Beluhov (Bulgaria)
2018 Iranian Geometry Olympiad
$ABCD$ is a cyclic quadrilateral. A circle passing through $A,B$ is tangent to segment $CD$ at point $E$. Another circle passing through $C,D$ is tangent to $AB$ at point $F$. Point $G$ is the intersection point of $AE,DF$, and point $H$ is the intersection point of $BE,CF$. Prove that the incenters of triangles $AGF,BHF,CHE,DGE$ lie on a circle.



Proposed by Le Viet An (Vietnam)
2017 Iranian Geometry Olympiad
Each side of square $ABCD$ with side length of $4$ is divided into equal parts by three points. Choose one of the three points from each side, and connect the points consecutively to obtain a quadrilateral. Which numbers can be the area of this quadrilateral? Just write the numbers without proof.

[asy]

import graph; size(4.662701220158751cm); 

real labelscalefactor = 0.5; /* changes label-to-point distance */

pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 

pen dotstyle = black; /* point style */ 

real xmin = -4.337693339683693, xmax = 2.323308403400238, ymin = -0.39518100382374105, ymax = 4.44811779880594;  /* image dimensions */

pen yfyfyf = rgb(0.5607843137254902,0.5607843137254902,0.5607843137254902); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); 



filldraw((-3.,4.)--(1.,4.)--(1.,0.)--(-3.,0.)--cycle, white, linewidth(1.6)); 

filldraw((0.,4.)--(-3.,2.)--(-2.,0.)--(1.,1.)--cycle, yfyfyf, linewidth(2.) + sqsqsq); 

 /* draw figures */

draw((-3.,4.)--(1.,4.), linewidth(1.6)); 

draw((1.,4.)--(1.,0.), linewidth(1.6)); 

draw((1.,0.)--(-3.,0.), linewidth(1.6)); 

draw((-3.,0.)--(-3.,4.), linewidth(1.6)); 

draw((0.,4.)--(-3.,2.), linewidth(2.) + sqsqsq); 

draw((-3.,2.)--(-2.,0.), linewidth(2.) + sqsqsq); 

draw((-2.,0.)--(1.,1.), linewidth(2.) + sqsqsq); 

draw((1.,1.)--(0.,4.), linewidth(2.) + sqsqsq); 

label("$A$",(-3.434705427329005,4.459844914550807),SE*labelscalefactor,fontsize(14)); 

label("$B$",(1.056779902954702,4.424663567316209),SE*labelscalefactor,fontsize(14)); 

label("$C$",(1.0450527872098359,0.07390362597090126),SE*labelscalefactor,fontsize(14)); 

label("$D$",(-3.551976584777666,0.12081208895036549),SE*labelscalefactor,fontsize(14)); 

 /* dots and labels */

dot((-2.,4.),linewidth(4.pt) + dotstyle); 

dot((-1.,4.),linewidth(4.pt) + dotstyle); 

dot((0.,4.),linewidth(4.pt) + dotstyle); 

dot((1.,3.),linewidth(4.pt) + dotstyle); 

dot((1.,2.),linewidth(4.pt) + dotstyle); 

dot((1.,1.),linewidth(4.pt) + dotstyle); 

dot((0.,0.),linewidth(4.pt) + dotstyle); 

dot((-1.,0.),linewidth(4.pt) + dotstyle); 

dot((-3.,1.),linewidth(4.pt) + dotstyle); 

dot((-3.,2.),linewidth(4.pt) + dotstyle); 

dot((-3.,3.),linewidth(4.pt) + dotstyle); 

dot((-2.,0.),linewidth(4.pt) + dotstyle); 

clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 

 /* end of picture */

[/asy]

Proposed by Hirad Aalipanah
2017 Iranian Geometry Olympiad
Find the angles of triangle $ABC$.

[asy]

import graph; size(9.115122858763474cm); 

real labelscalefactor = 0.5; /* changes label-to-point distance */

pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 

pen dotstyle = black; /* point style */ 

real xmin = -7.9216637359954705, xmax = 10.308581981531479, ymin = -6.062398124651168, ymax = 9.377503860601273;  /* image dimensions */



 /* draw figures */

draw((1.1862495478417192,2.0592342833377844)--(3.0842,-3.6348), linewidth(1.6)); 

draw((2.412546402365528,-0.6953452852662508)--(1.8579031454761883,-0.8802204313959623), linewidth(1.6)); 

draw((-3.696094000229639,5.5502174997511595)--(1.1862495478417192,2.0592342833377844), linewidth(1.6)); 

draw((-1.0848977580797177,4.042515022414228)--(-1.4249466943082025,3.5669367606747184), linewidth(1.6)); 

draw((1.1862495478417192,2.0592342833377844)--(9.086047353374928,-3.589295214974483), linewidth(1.6)); 

draw((9.086047353374928,-3.589295214974483)--(3.0842,-3.6348), linewidth(1.6)); 

draw((6.087339936804166,-3.904360930324946)--(6.082907416570757,-3.319734284649538), linewidth(1.6)); 

draw((3.0842,-3.6348)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6)); 

draw((0.08549258342923806,-3.9498657153504615)--(0.08106006319583221,-3.3652390696750536), linewidth(1.6)); 

draw((-2.9176473533749285,-3.6803047850255166)--(-6.62699301304923,-3.7084282888220432), linewidth(1.6)); 

draw((-6.62699301304923,-3.7084282888220432)--(-4.815597805533209,2.0137294983122676), linewidth(1.6)); 

draw((-5.999986761815922,-0.759127399441624)--(-5.442604056766517,-0.9355713910681529), linewidth(1.6)); 

draw((-4.815597805533209,2.0137294983122676)--(-3.696094000229639,5.5502174997511595), linewidth(1.6)); 

draw((-4.815597805533209,2.0137294983122676)--(1.1862495478417192,2.0592342833377844), linewidth(1.6)); 

draw((-1.8168903889624484,2.3287952136627297)--(-1.8124578687290425,1.744168567987322), linewidth(1.6)); 

draw((-4.815597805533209,2.0137294983122676)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6)); 

draw((-3.5893009510093994,-0.7408500702917692)--(-4.1439442078987385,-0.9257252164214806), linewidth(1.6)); 

label("$A$",(-4.440377746205339,7.118654172569505),SE*labelscalefactor,fontsize(14)); 

label("$B$",(-7.868514331571194,-3.218904987952353),SE*labelscalefactor,fontsize(14)); 

label("$C$",(9.165869786409527,-3.0594567746795223),SE*labelscalefactor,fontsize(14)); 

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dot((3.0842,-3.6348),linewidth(3.pt) + dotstyle); 

dot((9.086047353374928,-3.589295214974483),linewidth(3.pt) + dotstyle); 

dot((1.1862495478417192,2.0592342833377844),linewidth(3.pt) + dotstyle); 

dot((-2.9176473533749285,-3.6803047850255166),linewidth(3.pt) + dotstyle); 

dot((-4.815597805533209,2.0137294983122676),linewidth(3.pt) + dotstyle); 

dot((-6.62699301304923,-3.7084282888220432),linewidth(3.pt) + dotstyle); 

dot((-3.696094000229639,5.5502174997511595),linewidth(3.pt) + dotstyle); 

clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 

 /* end of picture */

[/asy]

Proposed by Morteza Saghafian
2017 Iranian Geometry Olympiad
In the regular pentagon $ABCDE$, the perpendicular at $C$ to $CD$ meets $AB$ at $F$. Prove that $AE+AF=BE$.



Proposed by Alireza Cheraghi
2017 Iranian Geometry Olympiad
$P_1,P_2,\ldots,P_{100}$ are $100$ points on the plane, no three of them are collinear. For each three points, call their triangle clockwise if the increasing order of them is in clockwise order. Can the number of clockwise triangles be exactly $2017$?



Proposed by Morteza Saghafian
2017 Iranian Geometry Olympiad
Let $ABC$ be an acute-angled triangle with $A=60^{\circ}$. Let $E,F$ be the feet of altitudes through $B,C$ respectively. Prove that $CE-BF=\tfrac{3}{2}(AC-AB)$



Proposed by Fatemeh Sajadi
2017 Iranian Geometry Olympiad
Two circles $\omega_1,\omega_2$ intersect at $A,B$. An arbitrary line through $B$ meets $\omega_1,\omega_2$ at $C,D$ respectively. The points $E,F$ are chosen on $\omega_1,\omega_2$ respectively so that $CE=CB,\ BD=DF$. Suppose that $BF$ meets $\omega_1$ at $P$, and $BE$ meets $\omega_2$ at $Q$. Prove that $A,P,Q$ are collinear.



Proposed by Iman Maghsoudi
2017 Iranian Geometry Olympiad
On the plane, $n$ points are given ($n>2$). No three of them are collinear. Through each two of them the line is drawn, and among the other given points, the one nearest to this line is marked (in each case this point occurred to be unique). What is the maximal possible number of marked points for each given $n$?



Proposed by Boris Frenkin (Russia)
2017 Iranian Geometry Olympiad
In the isosceles triangle $ABC$ ($AB=AC$), let $l$ be a line parallel to $BC$ through $A$. Let $D$ be an arbitrary point on $l$. Let $E,F$ be the feet of perpendiculars through $A$ to $BD,CD$ respectively. Suppose that $P,Q$ are the images of $E,F$ on $l$. Prove that $AP+AQ\le AB$



Proposed by Morteza Saghafian
2017 Iranian Geometry Olympiad
Let $X,Y$ be two points on the side $BC$ of triangle $ABC$ such that $2XY=BC$ ($X$ is between $B,Y$). Let $AA'$ be the diameter of the circumcirle of triangle $AXY$. Let $P$ be the point where $AX$ meets the perpendicular from $B$ to $BC$, and $Q$ be the point where $AY$ meets the perpendicular from $C$ to $BC$. Prove that the tangent line from $A'$ to the circumcircle of $AXY$ passes through the circumcenter of triangle $APQ$.



Proposed by Iman Maghsoudi
2017 Iranian Geometry Olympiad
In triangle $ABC$, the incircle, with center $I$, touches the sides $BC$ at point $D$. Line $DI$ meets $AC$ at $X$. The tangent line from $X$ to the incircle (different from $AC$) intersects $AB$ at $Y$. If $YI$ and $BC$ intersect at point $Z$, prove that $AB=BZ$.



Proposed by Hooman Fattahimoghaddam
2017 Iranian Geometry Olympiad
We have six pairwise non-intersecting circles that the radius of each is at least one (no circle lies in the interior of any other circle). Prove that the radius of any circle intersecting all the six circles, is at least one.



Proposed by Mohammad Ali Abam - Morteza Saghafian
2017 Iranian Geometry Olympiad
Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic.



Proposed by Ali Daeinabi - Hamid Pardazi
2017 Iranian Geometry Olympiad
Three circles $\omega_1,\omega_2,\omega_3$ are tangent to line $l$ at points $A,B,C$ ($B$ lies between $A,C$) and $\omega_2$ is externally tangent to the other two. Let $X,Y$ be the intersection points of $\omega_2$ with the other common external tangent of $\omega_1,\omega_3$. The perpendicular line through $B$ to $l$ meets $\omega_2$ again at $Z$. Prove that the circle with diameter $AC$ touches $ZX,ZY$.



Proposed by Iman Maghsoudi - Siamak Ahmadpour
2017 Iranian Geometry Olympiad
Sphere $S$ touches a plane. Let $A,B,C,D$ be four points on the plane such that no three of them are collinear. Consider the point $A'$ such that $S$ in tangent to the faces of tetrahedron $A'BCD$. Points $B',C',D'$ are defined similarly. Prove that $A',B',C',D'$ are coplanar and the plane $A'B'C'D'$ touches $S$.



Proposed by Alexey Zaslavsky (Russia)
2016 Iranian Geometry Olympiad
Ali wants to move from point $A$ to point $B$. He cannot walk inside the black areas but he is free to move in any direction inside the white areas (not only the grid lines but the whole plane). Help Ali to find the shortest path between $A$ and $B$. Only draw the path and write its length.

//cdn.artofproblemsolving.com/images/1/8/c/18c5684618fc23f1c3ee07eaa8ef313a3a90fd03.png

by Morteza Saghafian
2016 Iranian Geometry Olympiad
Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$, such that $CX = CY = AB$. (The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $\omega$ for the second time in point $P$. Show that $PB = PC$.



by Iman Maghsoudi
2016 Iranian Geometry Olympiad
Suppose that $ABCD$ is a convex quadrilateral with no parallel sides. Make a parallelogram on each two consecutive sides. Show that among these $4$ new points, there is only one point inside the quadrilateral $ABCD$.



by Morteza Saghafian
2016 Iranian Geometry Olympiad
In a right-angled triangle $ABC$ ($\angle A = 90^o$), the perpendicular bisector of $BC$ intersects the line $AC$ in $K$ and the perpendicular bisector of $BK$ intersects the line $AB$ in $L$. If the line $CL$ be the internal bisector of angle $C$, find all possible values for angles $B$ and $C$.



by Mahdi Etesami Fard
2016 Iranian Geometry Olympiad
Let $ABCD$ be a convex quadrilateral with these properties:

$\angle ADC = 135^o$ and $\angle ADB - \angle ABD = 2\angle DAB = 4\angle CBD$.

If $BC = \sqrt2  CD$ , prove that $AB = BC + AD$.



by Mahdi Etesami Fard
2016 Iranian Geometry Olympiad
In trapezoid $ABCD$ with $AB || CD$, $\omega_1$ and $\omega_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $\omega_1$ and $\omega_2$, respectively. Show that the length of segment $XY$ is not more than half the perimeter of $ABCD$.



Proposed by Mahdi Etesami Fard
2016 Iranian Geometry Olympiad
Let two circles $C_1$ and $C_2$ intersect in points $A$ and $B$. The tangent to $C_1$ at $A$ intersects $C_2$ in $P$ and the line $PB$ intersects $C_1$ for the second time in $Q$ (suppose that $Q$ is outside $C_2$). The tangent to $C_2$ from $Q$ intersects $C_1$ and $C_2$ in $C$ and $D$, respectively. (The points $A$ and $D$ lie on different sides of the line $PQ$.) Show that $AD$ is the bisector of $\angle CAP$.



Proposed by Iman Maghsoudi
2016 Iranian Geometry Olympiad
Find all positive integers $N$ such that there exists a triangle which can be dissected into $N$ similar quadrilaterals.



Proposed by Nikolai Beluhov (Bulgaria) and Morteza Saghafian
2016 Iranian Geometry Olympiad
Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$.



Proposed by Davood Vakili
2016 Iranian Geometry Olympiad
Let the circles $\omega$ and $\omega'$ intersect in points $A$ and $B$. The tangent to circle $\omega$ at $A$ intersects $\omega'$ at $C$ and the tangent to circle $\omega'$ at $A$ intersects $\omega$ at $D$. Suppose that the internal bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $E$ and $F$, respectively, and the external bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $X$ and $Y$, respectively. Prove that the perpendicular bisector of $XY$ is tangent to the circumcircle of triangle $BEF$.



Proposed by Mahdi Etesami Fard