geometry problems from Kharkiv City Olympiads (from Ukraine) [grades $\ge 8$]
with aops links in the names
2013 Kharkiv City MO 8.3
In the isosceles triangle $ABC$ on the base $BC$, point $M$ is marked so that the segment $MS$ is equal to the height of the triangle $ABC$ drawn to this side. Point $K$ is marked on the side $AB$ so that the angle $KMC$ is right . Find the angle $ACK$. Justify the answer.
2013 Kharkiv City MO 9.4
On the side $AB$ of the triangle $ABC$, the point $D$ is chosen such that $AC = CD$. On the arc $BC$ of the circumscribed circle of the triangle $BCD$ (not containing the point $D$), the point $E$ is selected such that $\angle ACB = \angle ABE$. On the extension of the segment $BC$ beyond point $C$, a point $F$ is marked such that $CE = CF$. Prove that $AB = AF$
source: https://mathedu.kharkiv.ua/examples/sef5/
with aops links in the names
collected inside aops :
it didn't take place in 2020
2012 - 2021
In triangle $ABC, \angle A=120^o, \angle B= 40^o$. Let $AD$ be the angle bisector of $\angle A$, and point $E$ on the side of the $AC$ be such that $CE = BD$. Prove that $AD \perp BE$.
2012 Kharkiv City MO 9.4
In triangle $ABC$, points $P$ and $Q$ are the intersection points of a straight line parallel to $BC$ , passing through the vertex $A$, with bisectors of the external angles $B$ and $C$ of the triangle, respectively. The perpendicular on the line $BP$ at point $P$, and the perpendicular on the line $CQ$ at point $Q$, intersect at the point $R$. Let $I$ be the incenter of the triangle $ABC$. Prove that $AI=AR$.
In triangle $ABC$, points $P$ and $Q$ are the intersection points of a straight line parallel to $BC$ , passing through the vertex $A$, with bisectors of the external angles $B$ and $C$ of the triangle, respectively. The perpendicular on the line $BP$ at point $P$, and the perpendicular on the line $CQ$ at point $Q$, intersect at the point $R$. Let $I$ be the incenter of the triangle $ABC$. Prove that $AI=AR$.
2012 Kharkiv City MO 10.4
In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$, points $D$ and $E$ are chosen such that points $A, B, E$, and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$. Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$.
In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$, points $D$ and $E$ are chosen such that points $A, B, E$, and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$. Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$.
2012 Kharkiv City MO 11.4
The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $E$, respectively. Point $G$ lies on circle $\omega$ in such a way that $FG$ is a diameter. Lines $EG$ and $FD$ intersect at point $H$. Prove that $AB \parallel CH$.
The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $E$, respectively. Point $G$ lies on circle $\omega$ in such a way that $FG$ is a diameter. Lines $EG$ and $FD$ intersect at point $H$. Prove that $AB \parallel CH$.
2013 Kharkiv City MO 8.3
In the isosceles triangle $ABC$ on the base $BC$, point $M$ is marked so that the segment $MS$ is equal to the height of the triangle $ABC$ drawn to this side. Point $K$ is marked on the side $AB$ so that the angle $KMC$ is right . Find the angle $ACK$. Justify the answer.
2013 Kharkiv City MO 9.4
On the side $AB$ of the triangle $ABC$, the point $D$ is chosen such that $AC = CD$. On the arc $BC$ of the circumscribed circle of the triangle $BCD$ (not containing the point $D$), the point $E$ is selected such that $\angle ACB = \angle ABE$. On the extension of the segment $BC$ beyond point $C$, a point $F$ is marked such that $CE = CF$. Prove that $AB = AF$
2013 Kharkiv City MO 10.4
The pentagon $ABCDE$ is inscribed in the circle $\omega$. Let $T$ be the intersection point of the diagonals $BE$ and $AD$. A line is drawn through the point $T$ parallel to $CD$, which intersects $AB$ and $CE$ at points $X$ and $Y$, respectively. Prove that the circumscribed circle of the triangle $AXY$ is tangent to $\omega$.
The pentagon $ABCDE$ is inscribed in the circle $\omega$. Let $T$ be the intersection point of the diagonals $BE$ and $AD$. A line is drawn through the point $T$ parallel to $CD$, which intersects $AB$ and $CE$ at points $X$ and $Y$, respectively. Prove that the circumscribed circle of the triangle $AXY$ is tangent to $\omega$.
2013 Kharkiv City MO 11.4
In the triangle $ABC$, the heights $AA_1$ and $BB_1$ are drawn. On the side $AB$, points $M$ and $K$ are chosen so that $B_1K\parallel BC$ and $A_1 M\parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.
In the triangle $ABC$, the heights $AA_1$ and $BB_1$ are drawn. On the side $AB$, points $M$ and $K$ are chosen so that $B_1K\parallel BC$ and $A_1 M\parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.
2014 Kharkiv City MO 8.5
In the right triangle $ABC$ with hypotenuse $AB$, points $D$ and $E$ are selected on side $AC$ and points $F$ and $G$ are selected on side $AC$ such that $\angle ABD = \angle DBE= \angle EBC$ and $\angle BAF = \angle FAG = \angle GAC$. Segments $AF$ and $BD$ intersect at point $K$. Prove that the triangle $EGK$ is equilateral.
In the right triangle $ABC$ with hypotenuse $AB$, points $D$ and $E$ are selected on side $AC$ and points $F$ and $G$ are selected on side $AC$ such that $\angle ABD = \angle DBE= \angle EBC$ and $\angle BAF = \angle FAG = \angle GAC$. Segments $AF$ and $BD$ intersect at point $K$. Prove that the triangle $EGK$ is equilateral.
2014 Kharkiv City MO 9.5
In the triangle $ABC$ with $AB <AC$ is satisfied, let point $I$ be the center of the inscribed circle of this triangle. Point $E$ is selected on the side $AC$ such that $AE = AB$. Point $G$ lies on the straight line $EI$ such that $\angle IBG = \angle CBA$, and points $E$ and $C$ lie on opposite sides of $I$. Prove that the line $AI$, the perpendicular on $AE$ at point $E$, and the bisector of angle $\angle BGI$ intersect at one point.
In the triangle $ABC$ with $AB <AC$ is satisfied, let point $I$ be the center of the inscribed circle of this triangle. Point $E$ is selected on the side $AC$ such that $AE = AB$. Point $G$ lies on the straight line $EI$ such that $\angle IBG = \angle CBA$, and points $E$ and $C$ lie on opposite sides of $I$. Prove that the line $AI$, the perpendicular on $AE$ at point $E$, and the bisector of angle $\angle BGI$ intersect at one point.
2014 Kharkiv City MO 10.4
Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.
Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.
In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.
2015 Kharkiv City MO 8.5
The diagonals $AC$ and $BD$ of a convex quadrilateral intersect at a point $O$. Points $M$ and .$N$ are chosen on the segments $AC$ and $BD$ respectively such that $AM = OC$ and$ DN = BO$. Prove that if $AB \nparallel CD$, then it is possible to form a triangle from segments $AB, CD$, and $MN$.
The diagonals $AC$ and $BD$ of a convex quadrilateral intersect at a point $O$. Points $M$ and .$N$ are chosen on the segments $AC$ and $BD$ respectively such that $AM = OC$ and$ DN = BO$. Prove that if $AB \nparallel CD$, then it is possible to form a triangle from segments $AB, CD$, and $MN$.
2015 Kharkiv City MO 9.5
The circles \omega_1 and \omega_2 intersect at points P and Q, are inside the circle \omega and tangent to it at points A_2 and A_2, respectively. The line PQ intersects the circle \omega at points B and D. Let E_1 and F_1 be the second intersection points of the lines A_1B and A_1D with \omega_1; Let E_1 and E_2 be the second intersection points of A_2B and A_2D with \omega_2. Prove that the points E_1,E_2,F_1 and F_2 lie on the same circle.
The circles \omega_1 and \omega_2 intersect at points P and Q, are inside the circle \omega and tangent to it at points A_2 and A_2, respectively. The line PQ intersects the circle \omega at points B and D. Let E_1 and F_1 be the second intersection points of the lines A_1B and A_1D with \omega_1; Let E_1 and E_2 be the second intersection points of A_2B and A_2D with \omega_2. Prove that the points E_1,E_2,F_1 and F_2 lie on the same circle.
2015 Kharkiv City MO 10.3
On side $AB$ of triangle $ABC$, point $M$ is selected. A straight line passing through $M$ intersects the segment $AC$ at point $N$ and the ray $CB$ at point $K$. The circumscribed circle of the triangle $AMN$ intersects $\omega$, the circumscribed circle of the triangle $ABC$, at points $A$ and $S$. Straight lines $SM$ and $SK$ intersect with $\omega$ for the second time at points $P$ and $Q$, respectively. Prove that $AC = PQ$.
On side $AB$ of triangle $ABC$, point $M$ is selected. A straight line passing through $M$ intersects the segment $AC$ at point $N$ and the ray $CB$ at point $K$. The circumscribed circle of the triangle $AMN$ intersects $\omega$, the circumscribed circle of the triangle $ABC$, at points $A$ and $S$. Straight lines $SM$ and $SK$ intersect with $\omega$ for the second time at points $P$ and $Q$, respectively. Prove that $AC = PQ$.
2015 Kharkiv City MO 11.3
In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.
In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.
2016 Kharkiv City MO 8.3
On the side $BC$ of the triangle $ABC$, point $M$ is selected, on the side $AC$ point $N$ and on the side $AB$ point $K$, such that $BM = BK$ and $CM = CN$. The perpendicular dropped on the segment $MK$ from point $B$ intersects the perpendicular dropped on the segment $MN$ from point $C$, at point $I$. Prove that $\angle IKA= \angle INC$
On the side $BC$ of the triangle $ABC$, point $M$ is selected, on the side $AC$ point $N$ and on the side $AB$ point $K$, such that $BM = BK$ and $CM = CN$. The perpendicular dropped on the segment $MK$ from point $B$ intersects the perpendicular dropped on the segment $MN$ from point $C$, at point $I$. Prove that $\angle IKA= \angle INC$
2016 Kharkiv City MO 9.3
An isosceles triangle $ABC$ with the base $BC$ is given. The circle $\omega$ it tangent to the line $AC$ at point $C$ and intersects the ray $AB$ at points $X$ and $Y$. Prove that $\angle BCX = \angle BCY$.
An isosceles triangle $ABC$ with the base $BC$ is given. The circle $\omega$ it tangent to the line $AC$ at point $C$ and intersects the ray $AB$ at points $X$ and $Y$. Prove that $\angle BCX = \angle BCY$.
2016 Kharkiv City MO 10.3
Let $AD$ be the bisector of an acute-angled triangle $ABC$. The circle circumscribed around the triangle $ABD$ intersects the straight line perpendicular to $AD$ that passes through point $B$, at point $E$. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Prove that the points $A, O, E$ lie on the same line.
Let $AD$ be the bisector of an acute-angled triangle $ABC$. The circle circumscribed around the triangle $ABD$ intersects the straight line perpendicular to $AD$ that passes through point $B$, at point $E$. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Prove that the points $A, O, E$ lie on the same line.
The circle $\omega$ passes through the vertices $B$ and $C$ of triangle $ABC$ and intersects its sides $AC,AB$ at points $A,E$, respectively. On the ray $BD$, a point $K$ such that $BK = AC$ is chosen , and on the ray $CE$, a point $L$ such that $CL = AB$ is chosen. Prove that the center $O$ of the circumscribed circle of the triangle $AKL$ lies on the circle $\omega$.
2017 Kharkiv City MO 8.3
Let $ABCD$ be a convex quadrangle with $ \angle A = \angle B = 60^o$ and $\angle CAB = \angle CBD$. Prove that $AC = BD$.
2017 Kharkiv City MO 6.3 7.3
The square is divided into $5$ rectangles as shown. It is known that all rectangles of a partition have the same area and that the length of the segment $AB$ is $5$. Find the length of the segment $CD$. Justify the answer.
The square is divided into $5$ rectangles as shown. It is known that all rectangles of a partition have the same area and that the length of the segment $AB$ is $5$. Find the length of the segment $CD$. Justify the answer.
Let $ABCD$ be a convex quadrangle with $ \angle A = \angle B = 60^o$ and $\angle CAB = \angle CBD$. Prove that $AC = BD$.
2017 Kharkiv City MO 9.3
A circle intersects the sides $AB, BC, CA$ of the triangle $ABC$ at points $R$ and $S, M$ and $N, P$ and $Q$, respectively, so that point $S$ lies on the segment $RB$, point $N$ on the segment $MC$, point $Q$ on the segment $PA$. Prove that if $RS=MN=PQ$ and $QR=SM=NP$, then triangle $ABC$ is equilateral.
2017 Kharkiv City MO 10.4
In the quadrangle $ABCD$, the angle at the vertex $A$ is right. Point $M$ is the midpoint of the side $BC$. It turned out that $\angle ADC = \angle BAM$. Prove that $\angle ADB = \angle CAM$.
A circle intersects the sides $AB, BC, CA$ of the triangle $ABC$ at points $R$ and $S, M$ and $N, P$ and $Q$, respectively, so that point $S$ lies on the segment $RB$, point $N$ on the segment $MC$, point $Q$ on the segment $PA$. Prove that if $RS=MN=PQ$ and $QR=SM=NP$, then triangle $ABC$ is equilateral.
In the quadrangle $ABCD$, the angle at the vertex $A$ is right. Point $M$ is the midpoint of the side $BC$. It turned out that $\angle ADC = \angle BAM$. Prove that $\angle ADB = \angle CAM$.
The quadrilateral $ABCD$ is inscribed in the circle $\omega$. Lines $AD$ and $BC$ intersect at point $E$. Points $M$ and $N$ are selected on segments $AD$ and $BC$, respectively, so that $AM: MD = BN: NC$. The circumscribed circle of the triangle $EMN$ intersects the circle $\omega$ at points $X$ and $Y$. Prove that the lines $AB, CD$ and $XY$ intersect at the same point or are parallel.
2018 Kharkiv City MO 8.5
Let $ABCD$ be a convex quadrilateral with $\angle A = 45^o, \angle ADC = \angle ACD = 75^o$ and $AB = CD = 1$. Find the length of the segment $BC$. Justify the answer.
Let $ABCD$ be a convex quadrilateral with $\angle A = 45^o, \angle ADC = \angle ACD = 75^o$ and $AB = CD = 1$. Find the length of the segment $BC$. Justify the answer.
2018 Kharkiv City MO 9.2
On the base $AC$ of the isosceles triangle $ABC$, the point $D$ is selected so that $CD = 2AD$. On the extension of the segment $BD$ beyond point $D$, point $E$ is selected such that $BD = DE$. Prove that $AE =DE$.
On the base $AC$ of the isosceles triangle $ABC$, the point $D$ is selected so that $CD = 2AD$. On the extension of the segment $BD$ beyond point $D$, point $E$ is selected such that $BD = DE$. Prove that $AE =DE$.
On the sides $AB, AC ,BC$ of the triangle $ABC$, the points $M, N, K$ are selected, respectively, such that $AM = AN$ and $BM = BK$. The circle circumscribed around the triangle $MNK$ intersects the segments $AB$ and $BC$ for the second time at points $P$ and $Q$, respectively. Lines $MN$ and $PQ$ intersect at point $T$. Prove that the line $CT$ bisects the segment $MP$.
2018 Kharkiv City MO 11.4
The line $\ell$ parallel to the side $BC$ of the triangle $ABC$, intersects its sides $AB,AC$ at the points $D,E$, respectively. The circumscribed circle of triangle $ABC$ intersects line $\ell$ at points $F$ and $G$, such that points $F,D,E,G$ lie on line $\ell$ in this order. The circumscribed circles of the triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P$ and $Q$ are collinear.
The line $\ell$ parallel to the side $BC$ of the triangle $ABC$, intersects its sides $AB,AC$ at the points $D,E$, respectively. The circumscribed circle of triangle $ABC$ intersects line $\ell$ at points $F$ and $G$, such that points $F,D,E,G$ lie on line $\ell$ in this order. The circumscribed circles of the triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P$ and $Q$ are collinear.
2019 Kharkiv City MO 8.4
Let $ABCD$ be parallelogram with $\angle ADC = 40^o$. Let K be a point such that the segment $AK$ intersects the side $BC$, $AK = BC$ and $\angle BAK = 80^o$. Let $L$ be a point such that the segment $CL$ intersects the side $AD$, $CL = AB$ and $\angle BCL = 80^o$. Find the angles of the triangle $BKL$. Justify your answer.
Let $ABCD$ be parallelogram with $\angle ADC = 40^o$. Let K be a point such that the segment $AK$ intersects the side $BC$, $AK = BC$ and $\angle BAK = 80^o$. Let $L$ be a point such that the segment $CL$ intersects the side $AD$, $CL = AB$ and $\angle BCL = 80^o$. Find the angles of the triangle $BKL$. Justify your answer.
2019 Kharkiv City MO 9.4
Two circles $\gamma$ and $\omega$ are such that the circle $\omega$ passes through the center of the circle $\gamma$. Let $A$ and $B$ be the intersection points of $\gamma$ and $\omega$ . On the circle $\omega$, a random point $P$, is chosen. The lines $PA$ and $PB$ intersect $\gamma$ for the second time at points $E$ and $F$, respectively. Prove that $AB = EF$.
Two circles $\gamma$ and $\omega$ are such that the circle $\omega$ passes through the center of the circle $\gamma$. Let $A$ and $B$ be the intersection points of $\gamma$ and $\omega$ . On the circle $\omega$, a random point $P$, is chosen. The lines $PA$ and $PB$ intersect $\gamma$ for the second time at points $E$ and $F$, respectively. Prove that $AB = EF$.
2019 Kharkiv City MO 10.5
In triangle $ABC$, point$ I$ is incenter , $I_a$ is the $A$-excenter. Let $K$ be the intersection point of the $BC$ with the external bisector of the angle $BAC$, and $E$ be the midpoint of the arc $BAC$ of the circumcircle of triangle $ABC$. Prove that $K$ is the orthocenter of triangle $II_aE$.
In triangle $ABC$, point$ I$ is incenter , $I_a$ is the $A$-excenter. Let $K$ be the intersection point of the $BC$ with the external bisector of the angle $BAC$, and $E$ be the midpoint of the arc $BAC$ of the circumcircle of triangle $ABC$. Prove that $K$ is the orthocenter of triangle $II_aE$.
2019 Kharkiv City MO 11.6
In the acute-angled triangle $ABC$, let $CD, BE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$, respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$, respectively. The lines $HF$ and $KG$ intersect at point $P$. Prove that line $BP$ bisects the segment $DE$.
In the acute-angled triangle $ABC$, let $CD, BE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$, respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$, respectively. The lines $HF$ and $KG$ intersect at point $P$. Prove that line $BP$ bisects the segment $DE$.
In the triangle $ABC$, on the median $BM$ lies point $K$ such that $CK = CM$. It is known that $\angle CBM = 2\angle ABM$. Prove that $BC = MK$
On the side $AB$ of the triangle $ABC$, the point $K$ is marked so that $AB = CK$. The points $N$ and $M$ are the midpoints of the segments $AK$ and $BC$, respectively. The segments $NM$ and $CK$ intersect at $R$. Prove that $KN=KR$.
The inscribed circle $\Omega$ of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $ L$, respectively. The line $BL$ intersects the circle $\Omega$ for the second time at the point $M$. The circle $\omega$ passes through the point $M$ and is tangent to the lines $AB$ and $BC$ at the points $P$ and $Q$, respectively. Let $N$ be the second intersection point of circles $\omega$ and $\Omega$, which is different from $M$. Prove that if $KM \parallel AC$ then the points $P, N$ and $L$ lie on one line.
In the triangle $ABC$, the segment $CL$ is the angle bisector. The $C$-exscribed circle with center at the point $ I_c$ touches the side of the $AB$ at the point $D$ and the extension of sides $CA$ and $CB$ at points $P$ and $Q$, respectively. It turned out that the length of the segment $CD$ is equal to the radius of this exscribed circle. Prove that the line $PQ$ bisects the segment $I_CL$.
source: https://mathedu.kharkiv.ua/examples/sef5/
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