Olympiad problems

2019 May Olympiad, 3

On the sides $AB, BC$ and $CA$ of a triangle $ABC$ are located the points $P, Q$ and $R$ respectively, such that $BQ = 2QC, CR = 2RA$ and $\angle PRQ = 90^o$. Show that $\angle APR =\angle RPQ$.

2007 Sharygin Geometry Olympiad, 6

Tags: geometry , congruent triangles , equal angles

Two non-congruent triangles are called [i]analogous [/i] if they can be denoted as $ABC$ and $A'B'C'$ such that $AB = A'B', AC = A'C'$ and $\angle B = \angle B'$ . Do there exist three mutually [i]analogous[/i] triangles?

2010 Saudi Arabia IMO TST, 2

Tags: geometry , ratio , equal angles

Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\angle MAB = \angle NAC$ and $\angle MBA = \angle NBC$. Prove that $$\frac{AM \cdot AN}{AB \cdot AC}+ \frac{BM\cdot BN}{BA \cdot BC}+ \frac{CM \cdot CN }{CA \cdot CB}=1$$

1995 Singapore Team Selection Test, 2

Tags: geometry , angle , equal angles

$ABC$ is a triangle with $\angle A > 90^o$ . On the side $BC$, two distinct points $P$ and $Q$ are chosen such that $\angle BAP = \angle PAQ$ and $BP \cdot CQ = BC \cdot PQ$. Calculate the size of $\angle PAC$.

Cono Sur Shortlist - geometry, 2018.G3

Tags: pentagon , equal segments , equal angles , geometry

Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.

Swiss NMO - geometry, 2019.7

Tags: equal segments , geometry , angle , equal angles

Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.

2008 China Northern MO, 1A

Tags: geometry , trapezoid , tangential , equal angles

As shown in figure , $\odot O$ is the inscribed circle of trapezoid $ABCD$, and the tangent points are $E, F, G, H$, $AB \parallel CD$. The line passing through$ B$, parallel to $AD$ intersects extension of $DC$ at point $P$. The extension of $AO$ intersects $CP$ at point $Q$. If $AE=BE$ , prove that $\angle CBQ = \angle PBQ$. [img]https://cdn.artofproblemsolving.com/attachments/d/2/7c3a04bb1c59bc6d448204fd78f553ea53cb9e.png[/img]

2013 Portugal MO, 6

Tags: combinatorics , geometry , regular , equal angles , combinatorial geometry

In each side of a regular polygon with $n$ sides, we choose a point different from the vertices and we obtain a new polygon of $n$ sides. For which values of $n$ can we obtain a polygon such that the internal angles are all equal but the polygon isn't regular?

1996 North Macedonia National Olympiad, 5

Tags: combinatorics , combinatorial geometry , equal angles , line , concurrency

Find the greatest $n$ for which there exist $n$ lines in space, passing through a single point, such that any two of them form the same angle.

Kyiv City MO Juniors 2003+ geometry, 2019.8.3

Tags: equal segments , geometry , equal angles , angle

In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$. (Hilko Danilo)

2010 Junior Balkan Team Selection Tests - Romania, 2

Tags: geometry , equal angles , angle , midpoint

Let $ABC$ be a triangle and $D, E, F$ the midpoints of the sides $BC, CA, AB$ respectively. Show that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$

1950 Poland - Second Round, 4

Tags: trigonometry , geometry , equal angles

Inside the triangle $ABC$ there is a point $P$ such that $$\angle PAB=\angle PBC =\angle PCA = \phi.$$ Prove that $$\frac{1}{\sin^2 \phi}=\frac{1}{\sin^2 A} +\frac{1}{\sin^2 B} +\frac{1}{\sin^2 C}$$

2013 Junior Balkan Team Selection Tests - Romania, 4

Tags: geometry , equal angles , concurrency , orthocenter , perpendicular

Consider acute triangles $ABC$ and $BCD$, with $\angle BAC = \angle BDC$, such that $A$ and $D$ are on opposite sides of line $BC$. Denote by $E$ the foot of the perpendicular line to $AC$ through $B$ and by $F$ the foot of the perpendicular line to $BD$ through $C$. Let $H_1$ be the orthocenter of triangle $ABC$ and $H_2$ be the orthocenter of $BCD$. Show that lines $AD, EF$ and $H_1H_2$ are concurrent.

2020 Czech and Slovak Olympiad III A, 5

Tags: isosceles , equal angles , concurrency , geometry

Given an isosceles triangle $ABC$ with base $BC$. Inside the side $BC$ is given a point $D$. Let $E, F$ be respectively points on the sides $AB, AC$ that $|\angle BED | = |\angle DF C| > 90^o$ . Prove that the circles circumscribed around the triangles $ABF$ and $AEC$ intersect on the line $AD$ at a point different from point $A$. (Patrik Bak, Michal Rolínek)

Estonia Open Junior - geometry, 2000.2.4

Tags: geometry , equal angles , concyclic , angle

In the plane, there is an acute angle $\angle AOB$ . Inside the angle points $C$ and $D$ are chosen so that $\angle AOC = \angle DOB$. From point $D$ the perpendicular on $OA$ intersects the ray $OC$ at point $G$ and from point C the perpendicular on $OB$ intersects the ray $OD$ at point $H$. Prove that the points $C, D, G$ and $H$ are conlyclic.

Estonia Open Junior - geometry, 2008.2.2

Tags: equal segments , right triangle , geometry , equal angles

In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.

2003 IMO Shortlist, 6

Tags: hexagon , equal angles , geometry

Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to $\dfrac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.

2020 Ukrainian Geometry Olympiad - April, 3

Tags: equal angles , circles , tangent , geometry

The angle $POQ$ is given ($OP$ and $OQ$ are rays). Let $M$ and $N$ be points inside the angle $POQ$ such that $\angle POM = \angle QON$ and $\angle POM < \angle PON$. Consider two circles: one touches the rays $OP$ and $ON$, the other touches the rays $OM$ and $OQ$. Denote by $B$ and $C$ the points of their intersection. Prove that $\angle POC = \angle QOB$.

Champions Tournament Seniors - geometry, 2006.3

Tags: geometry , isosceles , equal angles

Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the base $BC$ such that $BD:DC = 2: 1$. Note on the segment $AD$ a point $P$ such that $\angle BAC= \angle BPD $. Prove that $\angle BPD = 2 \angle CPD$.

2013 Grand Duchy of Lithuania, 2

Tags: geometry , isosceles , tangent , equal angles

Let $ABC$ be an isosceles triangle with $AB = AC$. The points $D, E$ and $F$ are taken on the sides $BC, CA$ and $AB$, respectively, so that $\angle F DE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that the line $BC$ is tangent to the circumcircle of $\vartriangle DEF$ if and only if $D$ is the midpoint of the side $BC$.

Swiss NMO - geometry, 2016.5

Tags: geometry , equal angles , concurrency

Let $ABC$ be a right triangle with $\angle ACB = 90^o$ and M the center of $AB$. Let $G$ br any point on the line $MC$ and $P$ a point on the line $AG$, such that $\angle CPA = \angle BAC$ . Further let $Q$ be a point on the straight line $BG$, such that $\angle BQC = \angle CBA$ . Show that the circles of the triangles $AQG$ and $BPG$ intersect on the segment $AB$.

2013 Oral Moscow Geometry Olympiad, 2

Tags: angle bisector , geometry , equal angles , inscribed circle , common tangent

Inside the angle $AOD$, the rays $OB$ and $OC$ are drawn such that $\angle AOB = \angle COD.$ Two circles are inscribed inside the angles $\angle AOB$ and $\angle COD$ . Prove that the intersection point of the common internal tangents of these circles lies on the bisector of the angle $AOD$.

2018 India PRMO, 17

Tags: geometry , equal angles

Triangles $ABC$ and $DEF$ are such that $\angle A = \angle D, AB = DE = 17, BC = EF = 10$ and $AC - DF = 12$. What is $AC + DF$?

2020 Ukrainian Geometry Olympiad - April, 2

Tags: tangent , equal angles , circles , symmetry , geometry

Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.

2012 Romania National Olympiad, 2

Tags: geometry , equal angles , right triangle

Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$