Olympiad problems

2007 Junior Tuymaada Olympiad, 4

Tags: geometry , circumcircle , orthocenter , incenter , Circumcenter , angles

An acute-angle non-isosceles triangle $ ABC $ is given. The point $ H $ is its orthocenter, the points $ O $ and $ I $ are the centers of its circumscribed and inscribed circles, respectively. The circumcircle of the triangle $ OIH $ passes through the vertex $ A $. Prove that one of the angles of the triangle is $ 60^\circ $.

1973 Dutch Mathematical Olympiad, 3

Tags: angles , geometry , isosceles

The angles $A$ and $B$ of base of the isosceles triangle $ABC$ are equal to $40^o$. Inside $\vartriangle ABC$, $P$ is such that $\angle PAB = 30^o$ and $\angle PBA = 20^o$. Calculate, without table, $\angle PCA$.

Kyiv City MO Juniors Round2 2010+ geometry, 2012.8.5

Tags: geometry , incenter , Equilateral , angles

In the triangle $ABC$ on the sides $AB$ and $AC$ outward constructed equilateral triangles $ABD$ and $ACE$. The segments $CD$ and $BE$ intersect at point $F$. It turns out that point $A$ is the center of the circle inscribed in triangle $ DEF$. Find the angle $BAC$. (Rozhkova Maria)

2017 Singapore Junior Math Olympiad, 3

Tags: isosceles , angles , equal segments , equal angles

Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$, and $E$ a point on $AD$ such that $\angle BED=\angle BAC=2\angle CED$. Prove that $BD=2CD$.

2019 Poland - Second Round, 6

Tags: geometry , lengths , angles

Let $X$ be a point lying in the interior of the acute triangle $ABC$ such that \begin{align*} \sphericalangle BAX = 2\sphericalangle XBA \ \ \ \ \hbox{and} \ \ \ \ \sphericalangle XAC = 2\sphericalangle ACX. \end{align*} Denote by $M$ the midpoint of the arc $BC$ of the circumcircle $(ABC)$ containing $A$. Prove that $XM=XA$.

1989 Tournament Of Towns, (218) 2

Tags: geometry , incenter , angles , Circumcenter

The point $M$ , inside $\vartriangle ABC$, satisfies the conditions that $\angle BMC = 90^o +\frac12 \angle BAC$ and that the line $AM$ contains the centre of the circumscribed circle of $\vartriangle BMC$. Prove that $M$ is the centre of the inscribed circle of $\vartriangle ABC$.

2020 Ukrainian Geometry Olympiad - April, 4

Tags: geometry , equal segments , equal angles , angles , square

On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected respectively such that $NC=NP$. The point $Q$ is chosen on the segment $AN$ so that $\angle QPN = \angle NCB$. Prove that $2\angle BCQ = \angle AQP$.

2001 All-Russian Olympiad Regional Round, 8.3

Tags: geometry , angles , pentagon , geometric inequality

All sides of a convex pentagon are equal, and all angles are different. Prove that the maximum and minimum angles are adjacent to the same side of the pentagon.

2019 Costa Rica - Final Round, G2

Tags: geometry , Circumcenter , orthocenter , angles

Let $H$ be the orthocenter and $O$ the circumcenter of the acute triangle $\vartriangle ABC$. The circle with center $H$ and radius $HA$ intersects the lines $AC$ and $AB$ at points $P$ and $Q$, respectively. Let point $O$ be the orthocenter of triangle $\vartriangle APQ$, determine the measure of $\angle BAC$.

Novosibirsk Oral Geo Oly VII, 2023.6

Tags: geometry , equal segments , angles , isosceles

An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.

Novosibirsk Oral Geo Oly VIII, 2021.3

Tags: geometry , angles

Find the angle $BCA$ in the quadrilateral of the figure. [img]https://cdn.artofproblemsolving.com/attachments/0/2/974e23be54125cde8610a78254b59685833b5b.png[/img]

2020 Yasinsky Geometry Olympiad, 1

Tags: geometry , angles , Angle Chasing , right triangle , circumcircle

Given a right triangle $ABC$, the point $M$ is the midpoint of the hypotenuse $AB$. A circle is circumscribed around the triangle $BCM$, which intersects the segment $AC$ at a point $Q$ other than $C$. It turned out that the segment $QA$ is twice as large as the side $BC$. Find the acute angles of triangle $ABC$. (Mykola Moroz)

2022 Brazil EGMO TST, 2

Tags: geometry , angles

Let $\vartriangle ABC$ be a triangle in which $\angle ACB = 40^o$ and $\angle BAC = 60^o$ . Let $D$ be a point inside the segment $BC$ such that $CD =\frac{AB}{2}$ and let $M$ be the midpoint of the segment $AC$. How much is the angle $\angle CMD$ in degrees?

2018 Hanoi Open Mathematics Competitions, 8

Tags: angles , geometry , square

Let $P$ be a point inside the square $ABCD$ such that $\angle PAC = \angle PCD = 17^o$ (see Figure 1). Calculate $\angle APB$? [img]https://cdn.artofproblemsolving.com/attachments/d/0/0b20ebee1fe28e9c5450d04685ac8537acda07.png[/img]

Kharkiv City MO Seniors - geometry, 2017.10.4

Tags: geometry , angles , midpoint , Kharkiv

In the quadrangle $ABCD$, the angle at the vertex $A$ is right. Point $M$ is the midpoint of the side $BC$. It turned out that $\angle ADC = \angle BAM$. Prove that $\angle ADB = \angle CAM$.

2017 Bosnia And Herzegovina - Regional Olympiad, 2

Tags: geometry , isosceles , angles , cosine

Let $ABC$ be an isosceles triangle such that $AB=AC$. Find angles of triangle $ABC$ if $\frac{AB}{BC}=1+2\cos{\frac{2\pi}{7}}$

Ukraine Correspondence MO - geometry, 2010.11

Tags: geometry , angles , orthocenter , incenter , Ukraine Correspondence

Let $ABC$ be an acute-angled triangle in which $\angle BAC = 60^o$ and $AB> AC$. Let $H$ and $I$ denote the points of intersection of the altitudes and angle bisectors of this triangle, respectively. Find the ratio $\angle ABC: \angle AHI$.

1997 Argentina National Olympiad, 2

Tags: geometry , right triangle , isosceles , angles , midpoint

Let $ABC$ be a triangle and $M$ be the midpoint of $AB$. If it is known that $\angle CAM + \angle MCB = 90^o$, show that triangle $ABC$ is isosceles or right.

1967 IMO Shortlist, 6

Tags: geometry , 3D geometry , sphere , angles , IMO Shortlist , IMO Longlist

Three disks of diameter $d$ are touching a sphere in their centers. Besides, every disk touches the other two disks. How to choose the radius $R$ of the sphere in order that axis of the whole figure has an angle of $60^\circ$ with the line connecting the center of the sphere with the point of the disks which is at the largest distance from the axis ? (The axis of the figure is the line having the property that rotation of the figure of $120^\circ$ around that line brings the figure in the initial position. Disks are all on one side of the plane, passing through the center of the sphere and orthogonal to the axis).

2011 AMC 10, 18

Tags: geometry , rectangle , trigonometry , trig identities , Law of Sines , AMC , angles

Rectangle $ABCD$ has $AB=6$ and $BC=3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$? $ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 75 $