Found problems: 509
2011 Oral Moscow Geometry Olympiad, 2
In an isosceles triangle $ABC$ ($AB=AC$) on the side $BC$, point $M$ is marked so that the segment $CM$ is equal to the altitude of the triangle drawn on this side, and on the side $AB$, point $K$ is marked so that the angle $\angle KMC$ is right. Find the angle $\angle ACK$.
1995 Nordic, 1
Let $AB$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $OC$ and $AB$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $BC$ and let the lines $CP$ and $AB$ meet at $Q$. We choose $R$ on $AP$ so that $RQ$ and $AB$ are perpendicular to each other. Show that $BQ =QR$.
2016 Oral Moscow Geometry Olympiad, 1
Angles are equal in a hexagon, three main diagonals are equal and the other six diagonals are also equal. Is it true that it has equal sides?
1991 Austrian-Polish Competition, 3
Given two distinct points $A_1,A_2$ in the plane, determine all possible positions of a point $A_3$ with the following property: There exists an array of (not necessarily distinct) points $P_1,P_2,...,P_n$ for some $n \ge 3$ such that the segments $P_1P_2,P_2P_3,...,P_nP_1$ have equal lengths and their midpoints are $A_1, A_2, A_3, A_1, A_2, A_3, ...$ in this order.
2020 Kazakhstan National Olympiad, 3
A point $ N $ is marked on the median $ CM $ of the triangle $ ABC $ so that $ MN \cdot MC = AB ^ 2/4 $. Lines $ AN $ and $ BN $ intersect the circumcircle $ \triangle ABC $ for the second time at points $ P $ and $ Q $, respectively. $ R $ is the point of segment $ PQ $, nearest to $ Q $, such that $ \angle NRC = \angle BNC $. $ S $ is the point of the segment $ PQ $ closest to $ P $ such that $ \angle NSC = \angle ANC $. Prove that $ RN = SN $.
2017 QEDMO 15th, 10
Let $\ell$ be a straight line and $P \notin \ell$ be a point in the plane. On $\ell$ are, in this arrangement, points $A_1, A_2,...$ such that the radii of the incircles of all triangles $P A_iA_{i + 1}$ are equal. Let $k \in N$. Show that the radius of the incircle of the triangle $P A_j A_{j + k}$ does not depend on the choice of $j \in N$ .
1997 Romania National Olympiad, 1
Let $C_1,C_2,..., C_n$ , $(n\ge 3)$ be circles having a common point $M$. Three straight lines passing through $M$ intersect again the circles in $A_1, A_2,..., A_n$ ; $B_1,B_2,..., B_n$ and $X_1,X_2,..., X_n$ respectively. Prove that if
$$A_1A_2 =A_2A_3 =...=A_{n-1}A_n$$ and $$B_1B_2 =B_2B_3 =...=B_{n-1}B_n$$ then $$X_1X_2 =X_2X_3 =...=X_{n-1}X_n.$$
2009 Balkan MO Shortlist, G1
In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.
2021 Yasinsky Geometry Olympiad, 4
Given an acute triangle $ABC$, in which $\angle BAC = 60^o$. On the sides $AC$ and $AB$ take the points $T$ and $Q$, respectively, such that $CT = TQ = QB$. Prove that the center of the inscribed circle of triangle $ATQ$ lies on the side $BC$.
(Dmitry Shvetsov)
2013 Czech-Polish-Slovak Junior Match, 3
The $ABCDE$ pentagon is inscribed in a circle and $AB = BC = CD$. Segments $AC$ and $BE$ intersect at $K$, and Segments $AD$ and $CE$ intersect at point$ L$. Prove that $AK = KL$.
2015 Junior Balkan Team Selection Tests - Moldova, 7
In a right triangle $ABC$ with $\angle BAC =90^o $and $\angle ABC= 54^o$, point $M$ is the midpoint of the hypotenuse $[BC]$ , point $D$ is the foot of the angle bisector drawn from the vertex $C$ and $AM \cap CD = \{E\}$. Prove that $AB= CE$.
2023 Yasinsky Geometry Olympiad, 4
Let $C$ be one of the two points of intersection of circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$, respectively. The line $O_1O_2$ intersects the circles at points $A$ and $B$ as shown in the figure. Let $K$ be the second point of intersection of line $AC$ with circle $\omega_2$, $L$ be the second point of intersection of line $BC$ with circle $\omega_1$. Lines $AL$ and $BK$ intersect at point $D$. Prove that $AD=BD$.
(Yurii Biletskyi)
[img]https://cdn.artofproblemsolving.com/attachments/6/4/2cdccb43743fcfcb155e846a0e05ec79ba90e4.png[/img]
2014 Contests, 2 juniors
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
2005 Bosnia and Herzegovina Junior BMO TST, 4
The sum of the angles on the bigger base of a trapezoid is $90^o$. Prove that the line segment whose ends are the midpoints of the bases, is equal to the line segment whose ends are the midpoints of the diagonals.
2020 Dutch BxMO TST, 4
Three different points $A,B$ and $C$ lie on a circle with center $M$ so that $| AB | = | BC |$. Point $D$ is inside the circle in such a way that $\vartriangle BCD$ is equilateral. Let $F$ be the second intersection of $AD$ with the circle . Prove that $| F D | = | FM |$.
Durer Math Competition CD Finals - geometry, 2010.C5
Let $D$ the touchpoint of the inscribed circle of triangle $ABC$ be with side $AB$ . From $A$ the perpendicular lines on the angle bisectors of vertices $B$ and $C$ intersect them at points $A_1$ and $A_2$ respectively . Prove that $A_1A_2 = AD$.
2011 Saudi Arabia Pre-TST, 4.4
In a triangle $ABC$, let $O$ be the circumcenter, $H$ the orthocenter, and $M$ the midpoint of the segment $AH$. The perpendicular at $M$ onto $OM$ intersects lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that $MP = MQ$.
Champions Tournament Seniors - geometry, 2019.2
The quadrilateral $ABCD$ is inscribed in the circle and the lengths of the sides $BC$ and $DC$ are equal, and the length of the side $AB$ is equal to the length of the diagonal $AC$. Let the point $P$ be the midpoint of the arc $CD$, which does not contain point $A$, and $Q$ is the point of intersection of diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are perpendicular.
Durer Math Competition CD Finals - geometry, 2017.D+5
The inscribed circle of the triangle $ABC$ touches the sides $BC, CA, AB$ at points $A_1, B_1, C_1$ respectively. The points $P_b, Q_b, R_b$ are the points of the segments $BC_1, C_1A_1, A_1B$, respectively, such that $BP_bQ_bR_b$ is parallelogram. In the same way, the points $P_c, Q_c, R_c$ are the points of the sections $CB_1, B_1A_1, A_1C$, respectively such that $CP_cQ_cR_c$ is a parallelogram. The intersection of the lines $P_bR_b$ and $P_cR_c$ is $T$. Show that $TQ_b = TQ_c$.
2022 Saudi Arabia BMO + EGMO TST, 2.2
Given is an acute triangle $ABC$ with $BC < CA < AB$. Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$. Perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at $M$. Prove that $CK + KM = BL + LM$.
2016 Denmark MO - Mohr Contest, 3
Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area.
[img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]
1996 Estonia National Olympiad, 2
Three sides of a trapezoid are equal, and a circle with the longer base as a diameter halves the two non-parallel sides. Find the angles of the trapezoid.
2021 Novosibirsk Oral Olympiad in Geometry, 5
In an acute-angled triangle $ABC$ on the side $AC$, point $P$ is chosen in such a way that $2AP = BC$. Points $X$ and $Y$ are symmetric to $P$ with respect to vertices $A$ and $C$, respectively. It turned out that $BX = BY$. Find $\angle BCA$.
2002 Junior Balkan Team Selection Tests - Romania, 3
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle A = 20^o$. Let $M$ be the foot of the altitude from $C$ and let $N$ be a point on the side $AC$ such that $CN =\frac12 BC$. Determine the measure of the angle $AMN$.
2014 Estonia Team Selection Test, 4
In an acute triangle the feet of altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again in point $P$ and the circumcircle of $CAB$ again in point $Q$. Prove that $|MP| = |MQ|$.