Olympiad problems

2014 Iranian Geometry Olympiad (junior), P4

In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$. by Morteza Saghafian

2024 Czech and Slovak Olympiad III A, 2

Tags: equal angles , geometry

Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$ Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.

2016 Sharygin Geometry Olympiad, P1

Tags: geometry , trapezoid , midpoint , equal angles

A trapezoid $ABCD$ with bases $AD$ and $BC$ is such that $AB = BD$. Let $M$ be the midpoint of $DC$. Prove that $\angle MBC$ = $\angle BCA$.

2017 Czech-Polish-Slovak Match, 2

Tags: concyclic , equal angles , geometry , circles

Let ${\omega}$ be the circumcircle of an acute-angled triangle ${ABC}$. Point ${D}$ lies on the arc ${BC}$ of ${\omega}$ not containing point ${A}$. Point ${E}$ lies in the interior of the triangle ${ABC}$, does not lie on the line ${AD}$, and satisfies ${\angle DBE =\angle ACB}$ and ${\angle DCE = \angle ABC}$. Let ${F}$ be a point on the line ${AD}$ such that lines ${EF}$ and ${BC}$ are parallel, and let ${G}$ be a point on ${\omega}$ different from ${A}$ such that ${AF = FG}$. Prove that points ${D,E, F,G}$ lie on one circle. (Slovakia)

1968 All Soviet Union Mathematical Olympiad, 094

Tags: geometry , equal angles , integer sides

Given an octagon with the equal angles. The lengths of all the sides are integers. Prove that the opposite sides are equal in pairs. [u]alternate wording[/u] Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.

2012 Denmark MO - Mohr Contest, 5

Tags: hexagon , equal angles , equilateral , equal segments , geometry

In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.

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 $\angle XNM = \angle BAN$.

2013 Oral Moscow Geometry Olympiad, 2

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

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 $AOD$.

2015 Caucasus Mathematical Olympiad, 3

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

Let $AL$ be the angle bisector of the acute-angled triangle $ABC$. and $\omega$ be the circle circumscribed about it. Denote by $P$ the intersection point of the extension of the altitude $BH$ of the triangle $ABC$ with the circle $\omega$ . Prove that if $\angle BLA= \angle BAC$, then $BP = CP$.

1950 Poland - Second Round, 4

Tags: geometry , trigonometry , equal angles

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

2002 Estonia Team Selection Test, 4

Tags: equal angles , cyclic , geometry , angle

Let $ABCD$ be a cyclic quadrilateral such that $\angle ACB = 2\angle CAD$ and $\angle ACD = 2\angle BAC$. Prove that $|CA| = |CB| + |CD|$.

2019 Saudi Arabia JBMO TST, 1

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

Let $E$ be a point lies inside the parallelogram $ABCD$ such that $\angle BCE = \angle BAE$. Prove that the circumcenters of triangles $ABE,BCE,CDE,DAE$ are concyclic.

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 $\angle BKC = \angle BDC$

Estonia Open Junior - geometry, 2012.2.3

Tags: geometry , circles , equal angles

Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A, B$, and $C$ lie on the circle $c$ and points $A', B'$, and $C$ lie on the circle $c'$ so that segment $AB\parallel A'B'$, $BC \parallel B'C'$, and $\angle ABC = \angle A'B'C'$. The lines $AA', BB$', and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\angle AOB = \angle A'O'B'$.

2016 Saudi Arabia IMO TST, 2

Tags: convex , hexagon , equal segments , equal angles , geometry , angle

Let $ABCDEF$ be a convex hexagon with $AB = CD = EF$, $BC =DE = FA$ and $\angle A+\angle B = \angle C +\angle D = \angle E +\angle F$. Prove that $\angle A=\angle C=\angle E$ and $\angle B=\angle D=\angle F$. Tran Quang Hung

Kyiv City MO Seniors 2003+ geometry, 2020.10.5.1

Tags: geometry , semicircle , equal angles , angle

Let $\Gamma$ be a semicircle with diameter $AB$. On this diameter is selected a point $C$, and on the semicircle are selected points $D$ and $E$ so that $E$ lies between $B$ and $D$. It turned out that $\angle ACD = \angle ECB$. The intersection point of the tangents to $\Gamma$ at points $D$ and $E$ is denoted by $F$. Prove that $\angle EFD=\angle ACD+ \angle ECB$.

Ukraine Correspondence MO - geometry, 2009.7

Tags: geometry , pentagon , equal angles

Let $ABCDE$ be a convex pentagon such that $AE\parallel BC$ and $\angle ADE = \angle BDC$. The diagonals $AC$ and $BE$ intersect at point $F$. Prove that $\angle CBD= \angle ADF$.

Novosibirsk Oral Geo Oly VIII, 2019.4

Tags: geometry , equal angles

Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.

2013 Regional Competition For Advanced Students, 4

Tags: equal segments , equal angles , pentagon , geometry , symmetry

We call a pentagon [i]distinguished [/i] if either all side lengths or all angles are equal. We call it [i]very distinguished[/i] if in addition two of the other parts are equal. i.e. $5$ sides and $2$ angles or $2$ sides and $5$ angles.Show that every very distinguished pentagon has an axis of symmetry.

2019 Saudi Arabia Pre-TST + Training Tests, 5.2

Tags: angle bisector , equal angles , geometry

Let the bisector of the outside angle of $A$ of triangle $ABC$ and the circumcircle of triangle $ABC$ meet at point $P$. The circle passing through points $A$ and $P$ intersects segments $BP$ and $CP$ at points $E$ and $F$ respectively. Let $AD$ is the angle bisector of triangle $ABC$. Prove that $\angle PED = \angle PFD$. [img]https://cdn.artofproblemsolving.com/attachments/0/3/0638429a220f07227703a682479ed150302aae.png[/img]

2007 Postal Coaching, 1

Tags: geometry , isosceles , equal angles , angle

Let $ABC$ be an isosceles triangle with $AC = BC$, and let $M$ be the midpoint of $AB$. Let $P$ be a point inside the triangle such that $\angle PAB =\angle PBC$. Prove that $\angle APM + \angle BPC = 180^o$.

1979 Chisinau City MO, 173

Tags: equal angles , pentagon , regular , geometry

The inner angles of the pentagon inscribed in the circle are equal to each other. Prove that this pentagon is regular.

2020 Novosibirsk Oral Olympiad in Geometry, 6

Tags: tangent , geometry , equal angles

In triangle $ABC$, point $M$ is the midpoint of $BC$, $P$ the point of intersection of the tangents at points $B$ and $C$ of the circumscribed circle of $ABC$, $N$ is the midpoint of the segment $MP$. The segment $AN$ meets the circumcircle $ABC$ at the point $Q$. Prove that $\angle PMQ = \angle MAQ$.

Ukrainian TYM Qualifying - geometry, 2014.8

Tags: equal angles , geometry

In the triangle $ABC$ on the ray $BA$ mark the point $K$ so that $\angle BCA= \angle KCA$ , and on the median $BM$ mark the point $T$ so that $\angle CTK=90^o$ . Prove that $\angle MTC=\angle MCB$ .

2021 Portugal MO, 2

Tags: geometry , equal segments , equal angles

Let $ABC$ be a triangle such that $AB = AC$. Let $D$ be a point in $[BC]$ and $E$ a point in $[AD]$ such that $\angle BE D = \angle BAC = 2 \angle DEC$. Shows that $DB = 2CD$. [img]https://cdn.artofproblemsolving.com/attachments/d/5/677e19d8e68a89134e17a4ab6051e41f283486.png[/img]