Found problems: 361
1974 Chisinau City MO, 76
Altitude $AH$ and median $AM$ of the triangle $ABC$ satisfy the relation: $\angle ABM = \angle CBH$. Prove that triangle $ABC$ is isosceles or right-angled.
Ukraine Correspondence MO - geometry, 2007.9
In triangle $ABC$, the lengths of all sides are integers, $\angle B=2 \angle A$ and $\angle C> 90^o$. Find the smallest possible perimeter of this triangle.
2011 Saudi Arabia Pre-TST, 1.4
Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.
Ukraine Correspondence MO - geometry, 2014.12
Let $\omega$ be the circumscribed circle of triangle $ABC$, and let $\omega'$ 'be the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $AC$. The common tangents to the circles $\omega$ and $\omega'$ intersect the line $BC$ at points $D$ and $E$. Prove that $\angle BAD = \angle CAE$.
2017 Peru IMO TST, 11
Let $ABC$ be an acute and scalene of circumcircle $\Gamma$ and orthocenter $H$. Let $A_1,B_1,C_1$ be the second intersection points of the lines $AH, BH, CH$ with $\Gamma$, respectively. The lines that pass through $A_1,B_1,C_1$ and are parallel to $BC,CA, AB$ intersect again to $\Gamma$ at $A_2,B_2,C_2$, respectively. Let $M$ be the intersection point of $AC_2$ and $BC_1, N$ the intersection point of $BA_2$ and $CA_1$, and $P$ the intersection point of $CB_2$ and $AB_1$. Prove that $\angle MNB = \angle AMP$ .
2003 Oral Moscow Geometry Olympiad, 3
Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.
2017 Ukrainian Geometry Olympiad, 2
Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.
2018 NZMOC Camp Selection Problems, 4
Let $P$ be a point inside triangle $ABC$ such that $\angle CPA = 90^o$ and $\angle CBP = \angle CAP$. Prove that $\angle P XY = 90^o$, where $X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.
1953 Kurschak Competition, 3
$ABCDEF$ is a convex hexagon with all its sides equal. Also $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Show that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.
2010 NZMOC Camp Selection Problems, 5
The diagonals of quadrilateral $ABCD$ intersect in point $E$. Given that $|AB| =|CE|$, $|BE| = |AD|$, and $\angle AED = \angle BAD$, determine the ratio $|BC|:|AD|$.
2001 239 Open Mathematical Olympiad, 2
In a convex quadrangle $ ABCD $, the rays $ DA $ and $ CB $ intersect at point $ Q $, and the rays $ BA $ and $ CD $ at the point $ P $. It turned out that $ \angle AQB = \angle APD $. The bisectors of the angles $ \angle AQB $ and $ \angle APD $ intersect the sides quadrangle at points $ X $, $ Y $ and $ Z $, $ T $ respectively. Circumscribed circles of triangles $ ZQT $ and $ XPY $ intersect at $ K $ inside quadrangle. Prove that $ K $ lies on the diagonal $ AC $.
Ukraine Correspondence MO - geometry, 2015.8
On the sides $BC, AC$ and $AB$ of the equilateral triangle $ABC$ mark the points $D, E$ and $F$ so that $\angle AEF = \angle FDB$ and $\angle AFE = \angle EDC$. Prove that $DA$ is the bisector of the angle $EDF$.
2019 Estonia Team Selection Test, 7
An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.
Russian TST 2018, P1
Let $ABC$ be an isosceles triangle with $AB = AC$. Let P be a point in the interior of $ABC$ such that $PB > PC$ and $\angle PBA = \angle PCB$. Let $M$ be the midpoint of the side $BC$. Let $O$ be the circumcenter of the triangle $APM$. Prove that $\angle OAC=2 \angle BPM$ .
2016 Hanoi Open Mathematics Competitions, 12
In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.
2020 SAFEST Olympiad, 4
Let $O$ be the circumcenter and $H$ the orthocenter of an acute-triangle $ABC$. The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$. Let $L$ be the midpoint of $OH$. Prove that $\angle OAH = \angle LSA$.
Swiss NMO - geometry, 2013.7
Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.
1975 Spain Mathematical Olympiad, 5
In the plane we have a line $r$ and two points $A$ and $B$ outside the line and in the same half plane. Determine a point $M$ on the line such that the angle of $r$ with $AM$ is double that of $r$ with $BM$. (Consider the smaller angle of two lines of the angles they form).
2002 Estonia National Olympiad, 1
Points $K$ and $L$ are taken on the sides $BC$ and $CD$ of a square $ABCD$ so that $\angle AKB = \angle AKL$. Find $\angle KAL$.
1978 All Soviet Union Mathematical Olympiad, 253
Given a quadrangle $ABCD$ and a point $M$ inside it such that $ABMD$ is a parallelogram. $ \angle CBM = \angle CDM$. Prove that the $ \angle ACD = \angle BCM$.
2011 Silk Road, 2
Given an isosceles triangle $ABC$ with base $AB$. Point $K$ is taken on the extension of the side $AC$ (beyond the point $C$ ) so that $\angle KBC = \angle ABC$. Denote $S$ the intersection point of angle - bisectors of $\angle BKC$ and $\angle ACB$. Lines $AB$ and $KS$ intersect at point $L$, lines $BS$ and $CL$ intersect at point $M$ . Prove that line $KM$ passes through the midpoint of the segment $BC$.
2022 Dutch BxMO TST, 2
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A, X, B, C, Y$ . Show that $\angle BXD = \angle CYD$.
2009 Oral Moscow Geometry Olympiad, 3
In the triangle $ABC$, $AA_1$ and $BB_1$ are altitudes. On the side $AB$ , points $M$ and $K$ are selected so that $B_1K \parallel BC$ and $A_1M \parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.
(D. Prokopenko)
2019 Regional Olympiad of Mexico Center Zone, 3
Let $ABC$ be an acute triangle and $D$ a point on the side $BC$ such that $\angle BAD = \angle DAC$. The circumcircles of the triangles $ABD$ and $ACD$ intersect the segments $AC$ and $AB$ at $E$ and $F$, respectively. The internal bisectors of $\angle BDF$ and $\angle CDE$ intersect the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ are chosen on the side $BC$ such that $PX$ is parallel to $AC$ and $QY$ is parallel to $AB$. Finally, let $Z$ be the point of intersection of $BE$ and $CF$. Prove that $ZX = ZY$.
2021 Sharygin Geometry Olympiad, 10-11.6
The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.