Olympiad problems

2006 Cuba MO, 9

In the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at $P$. Let $O$ be the center of the circumcircle $ABCD$, and $E$ a point of the extension of $OC$ beyond $C$. A parallel line to $CD$ is drawn through $E$ that cuts the extension of $OD$ beyonf $D$ at $F$. Let $Q$ be a point interior to $ABCD$, such that $\angle AFQ = \angle BEQ$ and $\angle FAQ = \angle EBQ$. Prove that $PQ \perp CD$.

2018 JBMO TST-Turkey, 6

Tags: geometry , parallelogram , circumcircle , concyclic , equal angles

A point $E$ is located inside a parallelogram $ABCD$ such that $\angle BAE = \angle BCE$. The centers of the circumcircles of the triangles $ABE,ECB, CDE$ and $DAE$ are concyclic.

2015 Dutch IMO TST, 1

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

In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.

2022 Regional Olympiad of Mexico West, 3

Tags: equal angles , isosceles , geometry

In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.

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?

2012 Switzerland - Final Round, 3

Tags: equal angles , geometry , circles , tangent

The circles $k_1$ and $k_2$ intersect at points $D$ and $P$. The common tangent of the two circles on the side of $D$ touches $k_1$ at $A$ and $k_2$ at $B$. The straight line $AD$ intersects $k_2$ for a second time at $C$. Let $M$ be the center of the segment $BC$. Show that $ \angle DPM = \angle BDC$ .

1967 Spain Mathematical Olympiad, 5

Tags: geometry , equal angles

Let $\gamma$ be a semicircle with diameter $AB$ . A creek is built with origin in $A$ , which has its vertices alternately in the diameter $AB$ and in the semicircle $\gamma$ , so that its sides make equal angles $\alpha$ with the diameter (but alternately in either direction). It is requested: a) Values of the angle $\alpha$ for the ravine to pass through the other end $B$ of the diameter. b) The total length of the ravine, in the case that it ends in $B$ , as a function of the length $d$ of the diameter and of the angle $\alpha$ . [img]https://cdn.artofproblemsolving.com/attachments/3/c/54c71cdf1bf8fbb3bdfa38f4b14a1dc961c5fe.png[/img]

2019 District Olympiad, 2

Tags: geometry , equal angles , equal segments , angle

Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$

2013 Switzerland - Final Round, 3

Tags: geometry , projection , 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 $BC = DE - BE$ .

2019 Yasinsky Geometry Olympiad, p4

Tags: geometry , perimeter , equal angles , incenter

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)

2013 Vietnam Team Selection Test, 5

Tags: geometry , equal angles , circumcircle , concyclic

Let $ABC$ be a triangle with $\angle BAC= 45^o$ . Altitudes $AD, BE, CF$ meet at $H$. $EF$ cuts $BC$ at $P$. $I$ is the midpoint of $BC$, $IF$ cuts $PH$ in $Q$. a) Prove that $\angle IQH = \angle AIE$. b) Let $(K)$ be the circumcircle of triangle $ABC$, $(J)$ be the circumcircle of triangle $KPD$. $CK$ cuts circle $(J)$ at $G$, $IG$ cuts $(J)$ at $M$, $JC$ cuts circle of diameter $BC$ at $N$. Prove that $G, N, M, C$ lie on the same circle.

2019 Saudi Arabia JBMO TST, 2

Tags: geometry , equal angles , equal segments , midpoint , altitude

In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$

2007 Bulgarian Autumn Math Competition, Problem 11.3

Tags: geometry , angle bisector , equal segments , equal angles

In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.

2011 Ukraine Team Selection Test, 9

Tags: geometry , circumcircle , equal angles , cyclic quadrilateral , tangent , tangent circle

Inside the inscribed quadrilateral $ ABCD $, a point $ P $ is marked such that $ \angle PBC = \angle PDA $, $ \angle PCB = \angle PAD $. Prove that there exists a circle that touches the straight lines $ AB $ and $ CD $, as well as the circles circumscribed by the triangles $ ABP $ and $ CDP $.

2020 Durer Math Competition Finals, 5

Tags: equal angles , hexagon , geometry

The hexagon $ABCDEF$ has all angles equal . We know that four consecutive sides of the hexagon have lengths $7, 6, 3$ and $5$ in this order. What is the sum of the lengths of the two remaining sides?

2019 Saint Petersburg Mathematical Olympiad, 4

Tags: geometry , circumcircle , centroid , equal angles

Given a convex quadrilateral $ABCD$. The medians of the triangle $ABC$ intersect at point $M$, and the medians of the triangle $ACD$ at point$ N$. The circle, circumscibed around the triangle $ACM$, intersects the segment $BD$ at the point $K$ lying inside the triangle $AMB$ . It is known that $\angle MAN = \angle ANC = 90^o$. Prove that $\angle AKD = \angle MKC$.

2015 Romania Team Selection Tests, 1

Tags: equal angles , perimeter , geometry

Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectrices of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent.

2019 Junior Balkan Team Selection Tests - Romania, 3

Tags: geometry , incenter , excenter , circumcenter , equal angles

In the acute triangle $ABC$ point $I$ is the incenter, $O$ is the circumcenter, while $I_a$ is the excenter opposite the vertex $A$. Point $A'$ is the reflection of $A$ across the line $BC$. Prove that angles $\angle IOI_a$ and $\angle IA'I_a$ are equal.

2010 Sharygin Geometry Olympiad, 4

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

Circles $\omega_1$ and $\omega_2$ inscribed into equal angles $X_1OY$ and $Y OX_2$ touch lines $OX_1$ and $OX_2$ at points $A_1$ and $A_2$ respectively. Also they touch $OY$ at points $B_1$ and $B_2$. Let $C_1$ be the second common point of $A_1B_2$ and $\omega_1, C_2$ be the second common point of $A_2B_1$ and $\omega_2$. Prove that $C_1C_2$ is the common tangent of two circles.

2021 Polish Junior MO First Round, 4

Tags: geometry , equal angles

A convex quadrilateral $ABCD$ is given where $\angle DAB =\angle ABC = 120^o$ and $CD = 3$,$BC = 2$, $AB = 1$. Calculate the length of segment $AD$.

2021 Sharygin Geometry Olympiad, 9.6

Tags: geometry , trapezoid , equal angles

The diagonals of trapezoid $ABCD$ ($BC\parallel AD$) meet at point $O$. Points $M$ and $N$ lie on the segments $BC$ and $AD$ respectively. The tangent to the circle $AMC$ at $C$ meets the ray $NB$ at point $P$; the tangent to the circle $BND$ at $D$ meets the ray $MA$ at point $R$. Prove that $\angle BOP =\angle AOR$.

2013 District Olympiad, 3

Tags: geometry , collinear , equal angles , midpoint

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 $AC = AB \sqrt {2}$

Denmark (Mohr) - geometry, 2014.3

Tags: geometry , equal angles , equal segments

The points $C$ and $D$ lie on a halfline from the midpoint $M$ of a segment $AB$, so that $|AC| = |BD|$. Prove that the angles $u = \angle ACM$ and $v = \angle BDM$ are equal. [img]https://1.bp.blogspot.com/-tQEJ1VBCa8U/XzT7IhwlZHI/AAAAAAAAMVI/xpRdlj5Rl64VUt_tCRsQ1UxIsv_SGrMlACLcBGAsYHQ/s0/2014%2BMohr%2Bp3.png[/img]

Kyiv City MO Juniors 2003+ geometry, 2016.8.51

Tags: geometry , equal angles , equal segments

In the quadrilateral $ABCD$, shown in fig. , the equations are true: $\angle ABC = \angle BCD$ and $2AB = CD$. On the side $BC$, a point $X$ is selected such that $\angle BAX = \angle CDA$. Prove that $AX = AD$. [img]https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png[/img]

2016 Switzerland - Final Round, 5

Tags: equal angles , geometry , 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 $AB$.