Olympiad problems

Ukraine Correspondence MO - geometry, 2012.10

The diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ intersect at a point O. It is known that $\angle BAD = 60^o$ and $AO = 3OC$. Prove that the sum of some two sides of a quadrilateral is equal to the sum of the other two sides.

Ukraine Correspondence MO - geometry, 2007.11

Tags: ratio , geometry , circumcircle , Ukraine Correspondence

Denote by $B_1$ and $C_1$, the midpoints of the sides $AB$ and $AC$ of the triangle $ABC$. Let the circles circumscribed around the triangles $ABC_1$ and $AB_1C$ intersect at points $A$ and $P$, and let the line $AP$ intersect the circle circumscribed around the triangle $ABC$ at points $A$ and $Q$. Find the ratio $\frac{AQ}{AP}$.

Ukraine Correspondence MO - geometry, 2012.7

Tags: geometry , bisects sements , Ukraine Correspondence

Let $O$ and $H$ be the center of the circumcircle and the point of intersection of the altitudes of the acute triangle $ABC$ respectively, $D$ be the foot of the altitude drawn to $BC$, and $E$ be the midpoint of $AO$. Prove that the circumcircle of the triangle $ADE$ passes through the midpoint of the segment $OH$.

Ukraine Correspondence MO - geometry, 2006.10

Tags: geometry , mixtilinear excircle , radiii , fixed , isosceles , Ukraine Correspondence

Let $ABC$ be an isosceles triangle ($AB=AC$). An arbitrary point $M$ is chosen on the extension of the $BC$ beyond point $B$. Prove that the sum of the radius of the circle inscribed in the triangle $AMB$ and the radius of the circle tangent to the side $AC$ and the extensions of the sides $AM, CM$ of the triangle $AMC$ does not depend on the choice of point $M$.

Ukraine Correspondence MO - geometry, 2004.8

Tags: geometry , ratio , trapezoid , areas , Ukraine Correspondence

The extensions of the sides $AB$ and $CD$ of the trapezoid $ABCD$ intersect at point $E$. Denote by $H$ and $G$ the midpoints of $BD$ and $AC$. Find the ratio of the area $AEGH$ to the area $ABCD$.

Ukraine Correspondence MO - geometry, 2014.10

Tags: geometry , incenter , circumcircle , Ukraine Correspondence

In the triangle $ABC$, it is known that $AC <AB$. Let $\ell$ be tangent to the circumcircle of triangle $ABC$ drawn at point $A$. A circle with center $A$ and radius $AC$ intersects segment $AB$ at point $D$, and line $\ell$ at points $E$ and $F$. Prove that one of the lines $DE$ and $DF$ passes through the center inscribed circle of triangle $ABC$.

Ukraine Correspondence MO - geometry, 2008.11

Tags: geometry , parallelogram , Concyclic , Ukraine Correspondence

Let $ABCD$ be a parallelogram. A circle with diameter $AC$ intersects line $BD$ at points $P$ and $Q$. The perpendicular on $AC$ passing through point $C$, intersects lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P, Q, X$ and $Y$ lie on the same circle.

Ukraine Correspondence MO - geometry, 2011.9

Tags: geometry , hexagon , collinear , equal segments , Ukraine Correspondence

On the diagonals $AC$ and $CE$ of a regular hexagon $ABCDEF$ with side $1$ we mark points$ M$ and $N$ such that $AM = CN = a$. Find $a$ if the points $B, M, N$ lie on the same line.

Ukraine Correspondence MO - geometry, 2009.3

Tags: geometry , compass , inradius , right triangle , Ukraine Correspondence

A right triangle is drawn on the plane. How to use only a compass to mark two points, such that the distance between them is equal to the diameter of the circle inscribed in this triangle?

Ukraine Correspondence MO - geometry, 2013.9

Tags: geometry , cyclic quadrilateral , equal circles , Ukraine Correspondence

Let $E$ be the point of intersection of the diagonals of the cyclic quadrilateral $ABCD$, and let $K, L, M$ and $N$ be the midpoints of the sides $AB, BC, CD$ and $DA$, respectively. Prove that the radii of the circles circumscribed around the triangles $KLE$ and $MNE$ are equal.

Ukraine Correspondence MO - geometry, 2018.9

Tags: geometry , Circumcenter , orthocenter , Ukraine Correspondence

Let $ABC$ be an acute-angled triangle in which $AB <AC$. On the side $BC$ mark a point $D$ such that $AD = AB$, and on the side $AB$ mark a point $E$ such that the segment $DE$ passes through the orthocenter of triangle $ABC$. Prove that the center of the circumcircle of triangle $ADE$ lies on the segment $AC$.

Ukraine Correspondence MO - geometry, 2006.3

Tags: geometry , Locus , orthocenter , Ukraine Correspondence

Find the locus of the points of intersection of the altitudes of the triangles inscribed in a given circle.

Ukraine Correspondence MO - geometry, 2017.8

Tags: geometry , trapezoid , min , angles , Ukraine Correspondence

On the midline of the isosceles trapezoid $ABCD$ ($BC \parallel AD$) find the point $K$, for which the sum of the angles $\angle DAK + \angle BCK$ will be the smallest.

Ukraine Correspondence MO - geometry, 2019.11

Tags: geometry , perpendicular , arc midpoint , Ukraine Correspondence

Let $O$ be the center of the circle circumscribed around the acute triangle $ABC$, and let $N$ be the midpoint of the arc $ABC$ of this circle. On the sides $AB$ and $BC$ mark points $D$ and $E$ respectively, such that the point $O$ lies on the segment $DE$. The lines $DN$ and $BC$ intersect at the point $P$, and the lines $EN$ and $AB$ intersect at the point $Q$. Prove that $PQ \perp AC$.

Ukraine Correspondence MO - geometry, 2018.6

Tags: geometry , equal segments , equal angles , Ukraine Correspondence

Let $AD$ and $AE$ be the altitude and median of triangle $ABC$, in with $\angle B = 2\angle C$. Prove that $AB = 2DE$.

Ukraine Correspondence MO - geometry, 2010.11

Tags: geometry , angles , orthocenter , incenter , Ukraine Correspondence

Let $ABC$ be an acute-angled triangle in which $\angle BAC = 60^o$ and $AB> AC$. Let $H$ and $I$ denote the points of intersection of the altitudes and angle bisectors of this triangle, respectively. Find the ratio $\angle ABC: \angle AHI$.

Ukraine Correspondence MO - geometry, 2004.10.

Tags: geometry , isosceles , Ukraine Correspondence

In an isosceles triangle $ABC$ ($AB = AC$), the bisector of the angle $B$ intersects $AC$ at point $D$ such that $BC = BD + AD$. Find $\angle A$.

Ukraine Correspondence MO - geometry, 2007.7

Tags: geometry , isosceles , midpoints , Ukraine Correspondence

Let $ABC$ be an isosceles triangle ($AB = AC$), $D$ be the midpoint of $BC$, and $M$ be the midpoint of $AD$. On the segment $BM$ take a point $N$ such that $\angle BND = 90^o$. Find the angle $ANC$.

Ukraine Correspondence MO - geometry, 2013.11

Tags: geometry , midpoint , Ukraine Correspondence

Given a triangle $ABC$. The circle $\omega_1$ passes through the vertex $B$ and touches the side $AC$ at the point $A$, and the circle $\omega_2$ passes through the vertex $C$ and touches the side $AB$ at the point $A$. The circles $\omega_1$ and $\omega_2$ intersect a second time at the point $D$. The line $AD$ intersects the circumcircle of the triangle $ABC$ at point $E$. Prove that $D$ is the midpoint of $AE$..

Ukraine Correspondence MO - geometry, 2009.7

Tags: geometry , pentagon , equal angles , Ukraine Correspondence

Let $ABCDE$ be a convex pentagon such that $AE\parallel BC$ and $\angle ADE = \angle BDC$. The diagonals $AC$ and $BE$ intersect at point $F$. Prove that $\angle CBD= \angle ADF$.

Ukraine Correspondence MO - geometry, 2020.8

Tags: geometry , perpendicular , Ukraine Correspondence

Let $ABC$ be an acute triangle, $D$ be the midpoint of $BC$. Bisectors of angles $ADB$ and $ADC$ intersect the circles circumscribed around the triangles $ADB$ and $ADC$ at points $E$ and $F$, respectively. Prove that $EF\perp AD$.

Ukraine Correspondence MO - geometry, 2011.7

Tags: geometry , trapezoid , perpendicular , Ukraine Correspondence

Let $ABCD$ be a trapezoid in which $AB \parallel CD$ and $AB = 2CD$. A line $\ell$ perpendicular to $CD$ was drawn through point $C$. A circle with center at point $D$ and radius $DA$ intersects line $\ell$ at points $P$ and $Q$. Prove that $AP \perp BQ$.