Olympiad problems

Kyiv City MO Juniors 2003+ geometry, 2021.8.4

Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and $PM \perp BM$. Prove that $\angle ABM = \angle MBP$. (Mikhail Standenko)

Croatia MO (HMO) - geometry, 2017.3

Tags: arc midpoint , incenter , equal angles , geometry

In triangle $ABC$, $|AB| <|BC|$ holds. Point $I$ is the center of the circle inscribed in that triangle. Let $M$ be the midpoint of the side $AC$, and $N$ be the midpoint of the arc $AC$ of the circumcircle of that triangle containing point $B$. Prove that $\angle IMA = \angle INB$.

Kyiv City MO Seniors 2003+ geometry, 2018.11.4.1

Tags: geometry , equal angles , orthocenter

In the quadrilateral $ABCD$, the diagonal $AC$ is the bisector $\angle BAD$ and $\angle ADC = \angle ACB$. The points $X, \, \, Y$ are the feet of the perpendiculars drawn from the point $A$ on the lines $BC, \, \, CD$, respectively. Prove that the orthocenter $\Delta AXY$ lies on the line $BD$.

2019 Saudi Arabia JBMO TST, 3

Tags: equal angles , angle , geometry

Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$. Suppose $\angle MAC = \angle ABC$ and $\angle BAM = 105^o$. Find the measure of $\angle ABC$.

2007 Peru MO (ONEM), 4

Tags: geometry , rhombus , equilateral , equal angles

Let $ABCD$ be rhombus $ABCD$ where the triangles $ABD$ and $BCD$ are equilateral. Let $M$ and $N$ be points on the sides $BC$ and $CD$, respectively, such that $\angle MAN = 30^o$. Let $X$ be the intersection point of the diagonals $AC$ and $BD$. Prove that $\angle XMN = \angle\ DAM$ and $\angle XNM = \angle BAN$.

1999 Ukraine Team Selection Test, 5

Tags: pentagon , equal angles , geometry

A convex pentagon $ABCDE$ with $DC = DE$ and $\angle DCB = \angle DEA = 90^o$ is given. Let $F$ be a point on the segment $AB$ such that $AF : BF = AE : BC$. Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.

2005 Slovenia Team Selection Test, 1

Tags: diagonal , equal angles , geometry

The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The bisector of $\angle ACD$ intersects the ray $BA$ at $K$. Prove that if $MA\cdot MC + MA\cdot CD = MB \cdot MD $, then $\angle BKC = \angle BDC$

Swiss NMO - geometry, 2013.3

Tags: projection , geometry , equal angles

Let $ABCD$ be a cyclic quadrilateral with $\angle ADC = \angle DBA$. Furthermore, let $E$ be the projection of $A$ on $BD$. Show that $BC = DE - BE$ .

2012 Dutch IMO TST, 4

Tags: right angle , angle chasing , geometry , equal angles

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2014 Indonesia MO Shortlist, G5

Tags: midpoint , equal angles , geometry

Given a cyclic quadrilateral $ABCD$. Suppose $E, F, G, H$ are respectively the midpoint of the sides $AB, BC, CD, DA$. The line passing through $G$ and perpendicular on $AB$ intersects the line passing through $H$ and perpendicular on $BC$ at point $K$. Prove that $\angle EKF = \angle ABC$.

2014 Junior Balkan Team Selection Tests - Romania, 3

Tags: midpoint , diameter , equal angles , geometry

Let $ABC$ be an acute triangle and let $O$ be its circumcentre. Now, let the diameter $PQ$ of circle $ABC$ intersects sides $AB$ and $AC$ in their interior at$ D$ and $E$, respectively. Now, let $F$ and $G$ be the midpoints of $CD$ and $BE$. Prove that $\angle FOG=\angle BAC$

2000 Mexico National Olympiad, 6

Tags: equal angles , geometry

Let $ABC$ be a triangle with $\angle B > 90^o$ such that there is a point $H$ on side $AC$ with $AH = BH$ and BH perpendicular to $BC$. Let $D$ and $E$ be the midpoints of $AB$ and $BC$ respectively. A line through $H$ parallel to $AB$ cuts $DE$ at $F$. Prove that $\angle BCF = \angle ACD$.

2001 Cuba MO, 2

Tags: geometry , equal angles

Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let$ K$ be the intersection point of the extension of side $AB$ (from $A$) with the bisector of the $\angle ACD$. If $MA \cdot MC + MA \cdot CD = MB\cdot MD$ , prove that $\angle BKC = \angle CDB$.

2009 Oral Moscow Geometry Olympiad, 3

Tags: geometry , parallel , altitude , equal angles

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)

2013 District Olympiad, 3

Tags: geometry , midpoint , equal angles , collinear

On the sides $(AB)$ and $(AC)$ of the triangle $ABC$ are considered the points $M$ and $N$ respectively so that $ \angle ABC =\angle ANM$. Point $D$ is symmetric of point $A$ with respect to $B$, and $P$ and $Q$ are the midpoints of the segments $[MN]$ and $[CD]$, respectively. Prove that the points $A, P$ and $Q$ are collinear if and only if $AC = AB \sqrt {2}$

2021 SAFEST Olympiad, 4

Tags: equal segments , geometry , equal angles

Let $ABC$ be a triangle with $AB > AC$. Let $D$ be a point on the side $AB$ such that $DB = DC$ and let $M$ be the midpoint of $AC$. The line parallel to $BC$ passing through $D$ intersects the line $BM$ in $K$. Show that $\angle KCD = \angle DAC.$

Kharkiv City MO Seniors - geometry, 2014.10.4

Tags: angle chasing , square , angle , equal angles , geometry

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 $\angle BCQ=\frac{1}{2}\angle AQP$.

2020 Yasinsky Geometry Olympiad, 4

Tags: equal segments , median , geometry , equal angles

The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.

1977 Chisinau City MO, 137

Tags: geometry , angle , equal angles

Determine the angles of a triangle in which the median, bisector and altitude, drawn from one vertex, divide this angle into four equal parts.

1903 Eotvos Mathematical Competition, 3

Tags: tangent , equal angles , geometry

Let $A,B,C,D$ be the vertices of a rhombus, let $k_1$ be the circle through $B,C$ and $D$, let $k_2$ be the circle through $A,C$ and $D$, let $k_3$ be the circle through $A,B$ and $D$, let $k_4$ be the circle through $A,B$ and $C$. Prove that the tangents to $k_1$ and $k_3$ at $B$ form the same angle as the tangents to $k_2$ and $k_4$ at $A$.

1974 Chisinau City MO, 82

Tags: angle , equal angles , geometry

Is there a moment in a day when three hands - hour, minute and second - of a clock running correctly form angles of $120^o$ in pairs?

2019 Yasinsky Geometry Olympiad, p4

Tags: incenter , geometry , perimeter , equal angles

In the triangle $ABC$, the side $BC$ is equal to $a$. Point $F$ is midpoint of $AB$, $I$ is the point of intersection of the bisectors of triangle $ABC$. It turned out that $\angle AIF = \angle ACB$. Find the perimeter of the triangle $ABC$. (Grigory Filippovsky)

2011 Saudi Arabia Pre-TST, 1.4

Tags: equal segments , equal angles , geometry

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 $BT = 2PT$.

2017 JBMO Shortlist, G3

Tags: geometry , angle , equal angles , circumcircle

Consider triangle $ABC$ such that $AB \le AC$. Point $D$ on the arc $BC$ of thecircumcirle of $ABC$ not containing point $A$ and point $E$ on side $BC$ are such that $\angle BAD = \angle CAE < \frac12 \angle BAC$ . Let $S$ be the midpoint of segment $AD$. If $\angle ADE = \angle ABC - \angle ACB$ prove that $\angle BSC = 2 \angle BAC$ .

2012 NZMOC Camp Selection Problems, 5

Tags: geometry , parallel , equal angles , angle bisector

Let $ABCD$ be a quadrilateral in which every angle is smaller than $180^o$. If the bisectors of angles $\angle DAB$ and $\angle DCB$ are parallel, prove that $\angle ADC = \angle ABC$