Olympiad problems

2020 Yasinsky Geometry Olympiad, 2

On the midline $MN$ of the trapezoid $ABCD$ ($AD\parallel BC$) the points $F$ and $G$ are chosen so that $\angle ABF =\angle CBG$. Prove that then $\angle BAF = \angle DAG$. (Dmitry Prokopenko)

Novosibirsk Oral Geo Oly IX, 2016.1

Tags: geometry , equal angles

In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.

2006 MOP Homework, 2

Tags: geometry , equal angles , orthocenter , perpendicular , right angle , projection

Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.

2013 Portugal MO, 6

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

2020 Greece National Olympiad, 2

Tags: geometry , parallelogram , equal angles , concurrency

Given a line segment $AB$ and a point $C$ lies inside it such that $AB=3 \cdot AC$ . Construct a parallelogram $ACDE$ such that $AC=DE=CE>AR$. Let $Z$ be a point on $AC$ such that $\angle AEZ=\angle ACE =\omega$. Prove that the line passing through point $B$ and perpendicular on side $EC$, and the line passing through point $D$ and perpendicular on side $AB$, intersect on point , let it be $K$, lying on line $EZ$.

2016 Novosibirsk Oral Olympiad in Geometry, 1

Tags: geometry , equal angles

In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.

2003 All-Russian Olympiad Regional Round, 8.7

Tags: geometry , equal segments , equal angles

In triangle $ABC$, angle $C$ is a right angle. Found on the side $AC$ point $D$, and on the segment $BD$, point $K$ such that $\angle ABC = \angle KAD =\angle AKD$. Prove that $BK = 2DC$.

Estonia Open Junior - geometry, 2010.1.2

Tags: geometry , equal angles , equal segments , angle

Given a convex quadrangle $ABCD$ with $|AD| = |BD| = |CD|$ and $\angle ADB = \angle DCA$, $\angle CBD = \angle BAC$, find the sizes of the angles of the quadrangle.

Indonesia MO Shortlist - geometry, g7

Tags: equal angles , trapezoid , isosceles , geometry

Given an isosceles trapezoid $ABCD$ with base $AB$. The diagonals $AC$ and $BD$ intersect at point $S$. Let $M$ the midpoint of $BC$ and the bisector of the angle $BSC$ intersect $BC$ at $N$. Prove that $\angle AMD = \angle AND$.

2017 Yasinsky Geometry Olympiad, 2

Tags: 3-dimensional geometry , 3d geometry , tetrahedron , geometry , equal angles , perpendicular

In the tetrahedron $DABC, AB=BC, \angle DBC =\angle DBA$. Prove that $AC \perp DB$.

Novosibirsk Oral Geo Oly VII, 2019.5

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 $BD + BA = AC$.

2016 Portugal MO, 4

Tags: tangent , geometry , parallelogram , equal angles , angle

Let $[ABCD]$ be a parallelogram with $AB <BC$ and let $E, F$ be points on the circle that passes through $A, B$ and $C$ such that $DE$ and $DF$ are tangents to this circle. Knowing that $\angle ADE = \angle CDF$ , determine $\angle ABC$. [img]https://cdn.artofproblemsolving.com/attachments/5/e/4140b92730e9d382df49ac05ca4e8ba48332dc.png[/img]

2011 Sharygin Geometry Olympiad, 12

Tags: altitude , geometry , equal angles , construction

Let $AP$ and $BQ$ be the altitudes of acute-angled triangle $ABC$. Using a compass and a ruler, construct a point $M$ on side $AB$ such that $\angle AQM = \angle BPM$.

1986 All Soviet Union Mathematical Olympiad, 428

Tags: line , equal angles , geometry

A line is drawn through the $A$ vertex of triangle $ABC$ with $|AB|\ne|AC|$. Prove that the line can not contain more than one point $M$ such, that $M$ is not a triangle vertex, and $\angle ABM = \angle ACM$. What lines do not contain such a point $M$ at all?

2020 Yasinsky Geometry Olympiad, 4

Tags: equal angles , equal segments , median , geometry

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 $\angle CQM = \angle BAM$.

2014 Denmark MO - Mohr Contest, 3

Tags: equal angles , equal segments , geometry

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]

2019 Chile National Olympiad, 4

Tags: geometry , parallel , equal angles

In the convex quadrilateral $ABCD$ , $\angle ADC = \angle BCD > 90^o$ . Let $E$ be the intersection of the line $AC$ with the line parallel to $AD$ that passes through $B$. Let $F$ be the intersection of line $BD$ with the line parallel to $BC$ passing through $A$. Prove that $EF$ is parallel to $CD$.

Cono Sur Shortlist - geometry, 2003.G1

Tags: geometry , circumcircle , equal angles , angle chase

Let $O$ be the circumcenter of the isosceles triangle $ABC$ ($AB = AC$). Let $P$ be a point of the segment $AO$ and $Q$ the symmetric of $P$ with respect to the midpoint of $AB$. If $OQ$ cuts $AB$ at $K$ and the circle that passes through $A, K$ and $O$ cuts $AC$ in $L$, show that $\angle ALP = \angle CLO$.

2019 Singapore MO Open, 1

Tags: geometry , circumcircle , orthocenter , equal angles

In the acute-angled triangle $ABC$ with circumcircle $\omega$ and orthocenter $H$, points $D$ and $E$ are the feet of the perpendiculars from $A$ onto $BC$ and from $B$ onto $AC$ respecively. Let $P$ be a point on the minor arc $BC$ of $\omega$ . Points $M$ and $N$ are the feet of the perpendiculars from $P$ onto lines $BC$ and $AC$ respectively. Let $PH$ and $MN$ intersect at $R$. Prove that $\angle DMR=\angle MDR$.

Estonia Open Junior - geometry, 2000.2.4

Tags: geometry , concyclic , equal angles , 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, 1995.1.4

Tags: geometry , right triangle , equal angles

The midpoint of the hypotenuse $AB$ of the right triangle $ABC$ is $K$. The point $M$ on the side $BC$ is taken such that $BM = 2 \cdot MC$. Prove that $\angle BAM = \angle CKM$.

2014 China Northern MO, 5

Tags: geometry , equal angles , parallelogram

As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$. [img]https://cdn.artofproblemsolving.com/attachments/4/3/5fa16c208ef3940443854756ae7bdb9c4272ed.png[/img]

2009 Postal Coaching, 5

Tags: equal angles , geometry , sum , angle

Let $P$ be an interior point of a circle and $A_1,A_2...,A_{10}$ be points on the circle such that $\angle A_1PA_2 = \angle A_2PA_3 = ... = \angle A_{10}PA_1 = 36^o$. Prove that $PA_1 + PA_3 + PA_5 + PA_7 +PA_9 = PA_2 + PA_4 + PA_6 + PA_8 + PA_{10}$.

Cono Sur Shortlist - geometry, 2021.G5

Tags: geometry , equal angles

Let $\vartriangle ABC$ be a triangle with circumcenter $O$, orthocenter $H$, and circumcircle $\omega$. $AA'$, $BB'$ and $CC'$ are altitudes of $\vartriangle ABC$ with $A'$ in $BC$, $B'$ in $AC$ and $C'$ in $AB$. $P$ is a point on the segment $AA'$. The perpenicular line to $B'C'$ from $P$ intersects $BC$ at $K$. $AA'$ intersects $\omega$ at $M \ne A$. The lines $MK$ and $AO$ intersect at $Q$. Prove that $\angle CBQ = \angle PBA$.

2003 Junior Balkan Team Selection Tests - Romania, 4

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

Let $E$ be the midpoint of the side $CD$ of a square $ABCD$. Consider the point $M$ inside the square such that $\angle MAB = \angle MBC = \angle BME = x$. Find the angle $x$.