Olympiad problems

2005 Turkey MO (2nd round), 2

Tags: geometry , parallelogram , modular arithmetic , circumcircle , power of a point , radical axis , geometry proposed

In a triangle $ABC$ with $AB<AC<BC$, the perpendicular bisectors of $AC$ and $BC$ intersect $BC$ and $AC$ at $K$ and $L$, respectively. Let $O$, $O_1$, and $O_2$ be the circumcentres of triangles $ABC$, $CKL$, and $OAB$, respectively. Prove that $OCO_1O_2$ is a parallelogram.

2006 Cono Sur Olympiad, 1

Tags: geometry , parallelogram , geometry proposed

Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BC$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram.

2002 Poland - Second Round, 2

Tags: geometry , 3D geometry , pyramid , geometry proposed

Triangle $ABC$ with $\angle BAC=90^{\circ}$ is the base of the pyramid $ABCD$. Moreover, $AD=BD$ and $AB=CD$. Prove that $\angle ACD\ge 30^{\circ}$.

2007 ITAMO, 1

Tags: geometry , inequalities , circumcircle , triangle inequality , geometry proposed

It is given a regular hexagon in the plane. Let P be a point of the plane. Define s(P) as the sum of the distances from P to each side of the hexagon, and v(P) as the sum of the distances from P to each vertex. a) Find the locus of points P that minimize s(P) b) Find the locus of points P that minimize v(P)

2009 Indonesia TST, 3

Tags: geometry , circumcircle , geometry proposed

Let $ ABC$ be an isoceles triangle with $ AC\equal{}BC$. A point $ P$ lies inside $ ABC$ such that \[ \angle PAB \equal{} \angle PBC, \angle PAC \equal{} \angle PCB.\] Let $ M$ be the midpoint of $ AB$ and $ K$ be the intersection of $ BP$ and $ AC$. Prove that $ AP$ and $ PK$ trisect $ \angle MPC$.

2012 Singapore MO Open, 1

Tags: geometry , trigonometry , geometry proposed

The incircle with centre $I$ of the triangle $ABC$ touches the sides $BC, CA$ and $AB$ at $D, E, F$ respectively. The line $ID$ intersects the segment $EF$ at $K$. Proof that $A, K, M$ collinear, where $M$ is the midpoint of $BC$.

2012 Pan African, 3

Tags: trigonometry , geometry proposed , geometry

(i) Find the angles of $\triangle ABC$ if the length of the altitude through $B$ is equal to the length of the median through $C$ and the length of the altitude through $C$ is equal to the length of the median through $B$. (ii) Find all possible values of $\angle ABC$ of $\triangle ABC$ if the length of the altitude through $A$ is equal to the length of the median through $C$ and the length of the altitude through $C$ is equal to the length of the median through $B$.

2004 China Western Mathematical Olympiad, 2

Tags: geometry , incenter , exterior angle , geometry proposed

Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.

2013 JBMO TST - Turkey, 1

Tags: geometry , incenter , circumcircle , geometry proposed

Let $D$ be a point on the side $BC$ of an equilateral triangle $ABC$ where $D$ is different than the vertices. Let $I$ be the excenter of the triangle $ABD$ opposite to the side $AB$ and $J$ be the excenter of the triangle $ACD$ opposite to the side $AC$. Let $E$ be the second intersection point of the circumcircles of triangles $AIB$ and $AJC$. Prove that $A$ is the incenter of the triangle $IEJ$.

2010 Sharygin Geometry Olympiad, 12

Tags: geometry , geometric transformation , reflection , geometry proposed

Let $AC$ be the greatest leg of a right triangle $ABC,$ and $CH$ be the altitude to its hypotenuse. The circle of radius $CH$ centered at $H$ intersects $AC$ in point $M.$ Let a point $B'$ be the reflection of $B$ with respect to the point $H.$ The perpendicular to $AB$ erected at $B'$ meets the circle in a point $K$. Prove that [b]a)[/b] $B'M \parallel BC$ [b]b)[/b] $AK$ is tangent to the circle.

2010 Baltic Way, 15

Tags: geometry , circumcircle , angle bisector , geometry proposed

The points $M$ and $N$ are chosen on the angle bisector $AL$ of a triangle $ABC$ such that $\angle ABM=\angle ACN=23^{\circ}$. $X$ is a point inside the triangle such that $BX=CX$ and $\angle BXC=2\angle BML$. Find $\angle MXN$.

2010 Poland - Second Round, 2

Tags: geometry , 3D geometry , tetrahedron , circumcircle , incenter , geometry proposed

The orthogonal projections of the vertices $A, B, C$ of the tetrahedron $ABCD$ on the opposite faces are denoted by $A', B', C'$ respectively. Suppose that point $A'$ is the circumcenter of the triangle $BCD$, point $B'$ is the incenter of the triangle $ACD$ and $C'$ is the centroid of the triangle $ABD$. Prove that tetrahedron $ABCD$ is regular.

1997 Pre-Preparation Course Examination, 5

Tags: inequalities , geometry , circumcircle , trigonometry , geometry proposed

Let $ABC$ be an acute angled triangle, $O$ be the circumcenter of $ABC$, and $R$ be the cicumradius. $AO$ meets the circumcircle of $BOC$ at $A'$, $BO$ meets the circumcircle of $COA$, and $CO$ meets the circumcircle of $AOB$ at $C'$. Prove that \[OA' \cdot OB' \cdot OC' \geq 8R^3.\] When does inequality occur?

2005 Greece JBMO TST, 3

Tags: geometry , geometry proposed

I have a very good solution of this but I want to see others. Let the midpoint$ M$ of the side$ AB$ of an inscribed quardiletar, $ABCD$.Let$ P $the point of intersection of $MC$ with $BD$. Let the parallel from the point $C$ to the$ AP$ which intersects the $BD$ at$ S$. If $CAD$ angle=$PAB$ angle= $\frac{BMC}{2}$ angle, prove that $BP=SD$.

2011 Iran Team Selection Test, 7

Tags: geometry proposed , geometry

Find the locus of points $P$ in an equilateral triangle $ABC$ for which the square root of the distance of $P$ to one of the sides is equal to the sum of the square root of the distance of $P$ to the two other sides.

2011 Greece Team Selection Test, 4

Tags: geometry , cyclic quadrilateral , circumcircle , geometry solved , geometry proposed , Nine point center , Poncelet point

Let $ABCD$ be a cyclic quadrilateral and let $K,L,M,N,S,T$ the midpoints of $AB, BC, CD, AD, AC, BD$ respectively. Prove that the circumcenters of $KLS, LMT, MNS, NKT$ form a cyclic quadrilateral which is similar to $ABCD$.

2011 Brazil Team Selection Test, 3

Tags: geometry , geometric transformation , reflection , circumcircle , complex numbers , geometry proposed

Let $ABC$ be an acute triangle with $\angle BAC=30^{\circ}$. The internal and external angle bisectors of $\angle ABC$ meet the line $AC$ at $B_1$ and $B_2$, respectively, and the internal and external angle bisectors of $\angle ACB$ meet the line $AB$ at $C_1$ and $C_2$, respectively. Suppose that the circles with diameters $B_1B_2$ and $C_1C_2$ meet inside the triangle $ABC$ at point $P$. Prove that $\angle BPC=90^{\circ}$ .

2007 Ukraine Team Selection Test, 2

Tags: geometry , trapezoid , geometry proposed

$ ABCD$ is convex $ AD\parallel BC$, $ AC\perp BD$. $ M$ is interior point of $ ABCD$ which is not a intersection of diagonals $ AC$ and $ BD$ such that $ \angle AMB \equal{}\angle CMD \equal{}\frac{\pi}{2}$ .$ P$ is intersection of angel bisectors of $ \angle A$ and $ \angle C$. $ Q$ is intersection of angel bisectors of $ \angle B$ and $ \angle D$. Prove that $ \angle PMB \equal{}\angle QMC$.

1996 Baltic Way, 2

Tags: geometry , geometry proposed

In the figure below, you see three half-circles. The circle $C$ is tangent to two of the half-circles and to the line $PQ$ perpendicular to the diameter $AB$. The area of the shaded region is $39\pi$, and the area of the circle $C$ is $9\pi$. Find the length of the diameter $AB$.

2014 JBMO TST - Turkey, 3

Tags: geometry , incenter , angle bisector , geometry proposed

Let a line $\ell$ intersect the line $AB$ at $F$, the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$, respectively and the internal bisector of the angle $BAC$ at $P$. Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$, $CD = CE$ and $P$ is in the interior the triangle $ABC$. Prove that \[FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.\]

2011 Federal Competition For Advanced Students, Part 2, 3

Tags: geometry , trapezoid , incenter , circumcircle , geometry proposed

We are given a non-isosceles triangle $ABC$ with incenter $I$. Show that the circumcircle $k$ of the triangle $AIB$ does not touch the lines $CA$ and $CB$. Let $P$ be the second point of intersection of $k$ with $CA$ and let $Q$ be the second point of intersection of $k$ with $CB$. Show that the four points $A$, $B$, $P$ and $Q$ (not necessarily in this order) are the vertices of a trapezoid.

2012 IberoAmerican, 1

Tags: geometry , parallelogram , geometry proposed

Let $ABCD$ be a rectangle. Construct equilateral triangles $BCX$ and $DCY$, in such a way that both of these triangles share some of their interior points with some interior points of the rectangle. Line $AX$ intersects line $CD$ on $P$, and line $AY$ intersects line $BC$ on $Q$. Prove that triangle $APQ$ is equilateral.

2011 Iran MO (3rd Round), 2

Tags: geometry , circumcircle , trapezoid , cyclic quadrilateral , perpendicular bisector , geometry proposed , Iran

In triangle $ABC$, $\omega$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $\omega$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $\omega$ iff $OM=ON$. [i]proposed by Mr.Etesami[/i]

2012 Puerto Rico Team Selection Test, 2

Tags: geometry , 3D geometry , geometry proposed

A cone is constructed with a semicircular piece of paper, with radius 10. Find the height of the cone.

2006 Iran MO (3rd Round), 7

Tags: geometry proposed , geometry

We have finite number of distinct shapes in plane. A "[i]convex Kearting[/i]" of these shapes is covering plane with convex sets, that each set consists exactly one of the shapes, and sets intersect at most in border. [img]http://aycu30.webshots.com/image/4109/2003791140004582959_th.jpg[/img] In which case Convex kearting is possible? 1) Finite distinct points 2) Finite distinct segments 3) Finite distinct circles