geometry problems from Olympiad of Kharkiv Physics and Mathematics Lyceum No.27 in Ukraine, with aops links in the names
collected inside aops:
juniors (8-9) and seniors (10-11)
2014 - 2021
In parallelogram $ABCD$, diagonal $BD$ is equal to side $AD$. On the side $AD$, was chosen point $K$ such that $AB = BK$. Point $C_1$ is such that $K$ is the midpoint of the segment $CC_1$. Prove that $BC_1 = AC$.
The altitude $BD$ is drawn in an acute-angled triangle $ABC$. From vertices $A$ and $C$ are drawn perpendiculars $AN$ and $CM$ on sides $AB$ and$ BC$, respectively. It is known that $AN = CD$ and $CM = AD$. Prove that $BN = BM$.
The bisectors of angles $A$ and $C$ of triangle $ABC$ intersect the circumcircle of this triangle at points $A_0$ and $C_0$, respectively. A straight line passing through the center of the inscribed circle of triangle $ABC$ parallel to side $AC$, intersects line $A_0C_0$ at point $P$. Prove that line $PB$ is tangent to the circumcircle of triangle $ABC$.
Let $AL$ be the angle bisector of triangle $ABC$. Circle $S$ passing through points $A$ and $L$, touches the side $BC$ and intersects the segment $AB$ at the interior point $P$. Line $P C$ is intersects $S$ for second time at the point $Q$. Prove that the line $AQ$ divides the segment $LC$ into two equal parts.
In an isosceles right-angled triangle $ABC$ with right angle $C$, the perpendicular on the angle bisector $BK$ intersects leg $BC$ at point $D$. Prove that $BD = AK$.
In triangle $ABC$, the angle bisector $BK$ is equal to the side $AB$. Point $T$ is selected on the segment $BK$ so that $\angle ATK = \angle BCA$. Prove that $AT = CK$.
Point $P$ is chosen inside an acute-angled triangle $ABC$ such that $\angle PAC = \angle PBC$. Let $M$ and $N$ be the feet of the perpendiculars drawn from the point $P$ on the sides $BC$ and $AC$, respectively, and $D$ be the midpoint of $AB$. Prove that $DM = DN$.
Let $M$ be the midpoint of the hypotenuse $AB$ of a right-angled triangle $ABC$. Point $D$ is chosen on leg $AC$ so that $CM = CD$. Point $P$ is the second intersection point of the circles circumscribed around triangles $AMC$ and $BDA$. Prove that $AP$ is the bisector of the angle $CAB$.
In quadrilateral $ABCD$, sides $AD$ and $BC$ are parallel. Point $M$ is the midpoint of side $AD$. On the side $BC$ , there is a point $N$ such that NA is the bisector of the angle $BNM$. Prove that $ND$ is the bisector of the angle $CNM$.
Given a triangle $ABC$. Point $A_B$ is the foot of the perpendicular dropped from point $A$ on the bisector of the externa' angle $B$. Similarly are defined points $A_C$, $B_A$, $B_C$, $C_A$, $C_B$. Prove that $A_BA_C = B_AB_C = C_AC_B$.
Let $ABCDEF$ be a regular hexagon and let point $P$ lie on line $AF$ such that $\angle PCD = 45^o$. Find the angle $FPE$.
Sir Arthur commissioned an artist to draw a drawing for his quarter-circle shield and asked him to paint it in three colors: yellow - the color of generosity, red - courage and black - wisdom. When the artist brought a painted shield then it seemed to Arthur that there was more courage in the picture than wisdom. Prove Arthur was wrong.
Areas marked in the figure with the number 1 are painted in yellow, number 2 with red color , number 3 with black color.
The quadrilateral $ABCD$ is inscribed in a circle. Points $P$ and $Q$ are the feet of the perpendiculars, drawn from point $A$ on lines $BC$ and $CD$, respectively. Points $R$ and $T$ are the feet of the perpendiculars dropped drawn point $D$ on lines $AB$ and $BC$, respectively. Prove that the points $P, Q, R, T$ lie on the same circle.
In a right-angled triangle $ABC$ with a right angle $C$, the altitude $CH$ is drawn. The altitude $HK$ is drawn in the triangle $AHC$. It turned out that $KA = CH$. Prove that $BK$ is the bisector of angle $ABC$.
Given a triangle $ABC$ in which $\angle C = 90^o$. On the hypotenuse $AB$, points $D$ and $E$ are selected such that $AD = AC$ and $BE = BC$. The points $P$ and $Q$ lying on the segments AC and BC, respectively, are such that $AP = AE$ and $BQ = BD$. Point $M$ is the midpoint of the segment $PQ$. Prove that $M$ is the intersection point of the bisectors of triangle $ABC$.
Points $A$ and $B$ are located at a unit distance from each other. Circles $\omega_1$ and $\omega_2$ of unit radius with centers at $A$ and $B$, respectively, intersect at points $C$ and $D$. A circle with center $C$ passing through point $D$ intersects ray $CA$ at point $F$. Ray DF intersects circle $\omega_1$ at point $P$. Find angle $APB$.
The angle bisector $BL$ is drawn in an isosceles triangle $ABC$. On the bsse $BC$ point $D$ is marked , and on the lateral side $AB$ point $E$ is mraked such that $AE = \frac12AL = CD$. Prove that $LE = LD$.
Let $AD$ be the angle bisector of triangle $ABC$ in which $\angle BAC = 2\angle ACB$, Points $M$ and $N$ are the midpoints of segments $AC$ and $BD$, respectively. It turned out that $\angle BAC = 2\angle MNC$. Find the angles of triangle $ABC$.
On the lateral sides $AB$ and $CD$ of trapezoid $ABCD$, points $P$ and $Q$ are selected, respectively. The segments $CP$ and $BQ$ intersect at point $T$. It turned out that a circle can be circumscribed around the pentagon $APTQD$. Prove that $AT = DT$.
Point $M$ is the midpoint of side $AB$ of square $ABCD$. Point $P$ is the foot of the perpendicular drawn from $B$ to line $CM$ and point $N$ is the midpoint of $CP$. The bisector of angle $DAN$ intersects line $DP$ at point $Q$. Prove that quadrilateral $BMQN$ is a parallelogram.
Point $D$ is marked on the lateral side $BC$ of isosceles triangle $ABC$, and point $E$ is marked on the extension of the base $AC$ beyond point $C$, and $AD = DE$. Prove that the area of triangle $ABD$ is equal to the area of triangle $BCE$.
In triangle $ABC$, angle $C$ is right. On the side $AC$ there is a point $D$ and on the segment $BD$ there is a point $K$ such such that $\angle ABC = \angle KAD = \angle AKD$. Prove that $BK = 2DC$.
Given a triangle $ABC$ Point $I$ is the center of its inscribed circle. Prove that the circle passing through $A$ and tangent to line $BI$ at point $I$, and the circle passing through $B$ and tangent to line $AI$ at point $I$, intersect on the circumcircle of triangle $ABC$.
The angle bisectors $AA_1$, $BB_1$, $CC_1$ of triangle $ABC$ meet at point $I$. The perpendicular bisector of the segment $BB_1$ intersects lines $AA_1$ and $CC_1$ at points $A_0$ and $C_0$, respectively. Prove that the circumcircles of triangles $A_0IC_0$ and $ABC$ are tangent.
Points $A_0$ and $B_0$ lie inside the parallelogram $ABCD$, and points $C_0$ and $D_0$ lie outside it. It is known that all sides of the octagon $AA_0BB_0CC_0DD_0$ are equal. Prove that points $A_0$, $B_0$, $C_0$, $D_0$ lie on the same circle.
Carlson cut a rectangular cake using $2$ straight line cuts parallel to the sides of the cake into $4$ pieces. The mass of the largest piece, together with the the smallest piece was equal to the masses of the other two pieces of cake. Prove that one of the cuts passed through the center of the cake.
On the hypotenuse $AB$ of a right triangle $ABC$, mark points $D$ and $E$ so that $AD = AC$ and $BE = BC$. From point $D$ , perpendicular on $CE$ is drawn, intersecting it at $F$. Prove that $DF = CF$.
In the acute-angled triangle $ABC$, the altitude $AH$, the median $BM$, and the angle bisector $CD$ are drawn. The segments $BM$ and $CD$ intersect at the point $K$. It turned out that $KB = KC$. Prove that $KM = KH$.
On the lateral side $AB$ of the isosceles triangle $ABC$, we mark the point $D$. On the lateral side $AC$ , we mark the point $E$ and on the extension of the base $BC$ beyond $B$ , we mark the point $F$ such that $CD = DF$. On the line $DE$, the point $ P$ is chosen, and on the segment $BD$, the point $Q$ is chosen so that $PF \parallel AC$ and $PQ \parallel CD$. Prove that $DE = QF$.
A parallelogram $ACDE$ is constructed on the outside of the triangle $ABC$. Let $O$ be the point of intersection of its diagonals, $N$ and $K$ be the midpoints of the sides $BC$ and $BA$, respectively. Prove that the lines $DK$, $EN$ and $BO$ intersect at a point.
In the $ABC$ triangle, the angle $\angle C$ is right. Let $AM$ be the median of the triangle, and let $D$ be the midpoint of AM. It turned out that the angles $\angle CAD =\angle CBD$. What can be the values of these equal angles?
The triangle $ABC$ is equilateral. Inside the angle $\angle ABC$, the point $M$ is chosen so that $\angle BMC = 30^o$ and $\angle BMA= 17^o$. Find the values of the angles $\angle BAM$ and $\angle BCM$.
Given a triangle $ABC$, in which $AB> AC$. On the side $AB$ is marked point $D$ such that $BD = AC$. Let $\gamma$ be a circle passing through point $B$ and tangent to the line $AC$ at point $A$. The circumcircle of the triangle $ABC$ intersects $\gamma$ at points $A$ and $E$. Prove that point $E$ is the intersection point of of the perpendicular bisectors of the segments $BC$ and $AD$.
The perpendicular bisector of the angle bisector $BL$ of the triangle $ABC$ intersects the bisectors of its external angles $A$ and $C$ at the points $P$ and $Q$, respectively. Prove that the circumscribed circles of triangles $PBQ$ and $ABC$ are tangent.
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