Olympiad problems

2022 IFYM, Sozopol, 2

Tags: geometry , angle

Let $ABC$ be a triangle with $\angle BAC=40^\circ $, $O$ be the center of its circumscribed circle and $G$ is its centroid. Point $D$ of line $BC$ is such that $CD=AC$ and $C$ is between $B$ and $D$. If $AD\parallel OG$, find $\angle ACB$.

2006 Sharygin Geometry Olympiad, 9.4

Tags: cut , geometry , angle , hexagon

In a non-convex hexagon, each angle is either $90$ or $270$ degrees. Is it true that for some lengths of the sides it can be cut into two hexagons similar to it and unequal to each other?

2011 Saudi Arabia Pre-TST, 3.3

Tags: angle , geometry , isosceles

In the isosceles triangle $ABC$, with $AB = AC$, the angle bisector of $\angle B$ intersects side $AC$ at $B'$. Suppose that $ B B' + B'A = BC$. Find the angles of the triangle.

2011 Junior Balkan Team Selection Tests - Moldova, 3

Tags: equal segments , geometry , angle , circles

Let $ABC$ be a triangle with $ \angle ACB = 90^o + \frac12 \angle ABC$ . The point $M$ is the midpoint of the side $BC$ . A circle with center at vertex $A$ intersects the line $BC$ at points $M$ and $D$. Prove that $MD = AB$.

1950 Moscow Mathematical Olympiad, 178

Tags: trigonometry , angle

Let $A$ be an arbitrary angle,let $B$ and $C$ be acute angles. Is there an angle $x$ such that $$\sin x =\frac{\sin B \cdot \sin C}{1 - \cos B \cdot \cos C \cdot \cos A} ?$$

2003 District Olympiad, 4

Tags: geometry , angle , 3d geometry , tetrahedron , equal angles

a) Let $MNP$ be a triangle such that $\angle MNP> 60^o$. Show that the side $MP$ cannot be the smallest side of the triangle $MNP$. b) In a plane the equilateral triangle $ABC$ is considered. The point $V$ that does not belong to the plane $(ABC)$ is chosen so that $\angle VAB = \angle VBC = \angle VCA$. Show that if $VA = AB$, the tetrahedron $VABC$ is regular. Valentin Vornicu

2016 Argentina National Olympiad, 4

Tags: angle , geometry

Find the angles of a convex quadrilateral $ABCD$ such that $\angle ABD = 29^o$, $\angle ADB = 41^o$, $\angle ACB = 82^o$ and $\angle ACD = 58^o$

1990 Tournament Of Towns, (260) 4

Tags: trapezoid , angle , geometry

Let $ABCD$ be a trapezium with $AC = BC$. Let $H$ be the midpoint of the base $AB$ and let $\ell$ be a line passing through $H$. Let $\ell$ meet $AD$ at $P$ and $BD$ at $Q$. Prove that the angles $ACP$ and $QCB$ are either equal or have a sum of $180^o$. (I. Sharygin, Moscow)

Ukraine Correspondence MO - geometry, 2017.8

Tags: trapezoid , min , angle , geometry

On the midline of the isosceles trapezoid $ABCD$ ($BC \parallel AD$) find the point $K$, for which the sum of the angles $\angle DAK + \angle BCK$ will be the smallest.

1998 Israel National Olympiad, 1

Tags: geometry , angle , plane , combinatorial geometry

In space are given $n$ segments $A_iB_i$ and a point $O$ not lying on any segment, such that the sum of the angles $A_iOB_i$ is less than $180^o$ . Prove that there exists a plane passing through $O$ and not intersecting any of the segments.

2011 Tournament of Towns, 3

Tags: angle , geometry , convex , quadrilateral

In a convex quadrilateral $ABCD, AB = 10, BC = 14, CD = 11$ and $DA = 5$. Determine the angle between its diagonals.

2014 Gulf Math Olympiad, 3

Tags: equal segments , geometry , right triangle , semicircle , angle

(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$. The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$. Prove that $BU = BA$ if, and only if, $CP = CA$. (ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$. The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$. Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$. Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.

2015 Sharygin Geometry Olympiad, P2

Tags: angle , circumcenter , orthocenter , incenter , geometry

Let $O$ and $H$ be the circumcenter and the orthocenter of a triangle $ABC$. The line passing through the midpoint of $OH$ and parallel to $BC$ meets $AB$ and $AC$ at points $D$ and $E$. It is known that $O$ is the incenter of triangle $ADE$. Find the angles of $ABC$.

2009 IMAR Test, 3

Tags: angle , angle chasing , convex quadrilateral , equal angles , geometry

Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$. Paolo Leonetti

1963 IMO, 3

Tags: polygon , inequalities , geometric inequality , angle , geometry

In an $n$-gon $A_{1}A_{2}\ldots A_{n}$, all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation \[a_{1}\geq a_{2}\geq \dots \geq a_{n}. \] Prove that $a_{1}=a_{2}= \ldots= a_{n}$.

2019 Polish Junior MO First Round, 2

Tags: angle , geometry

A convex quadrilateral $ABCD$ is given in which $\angle DAB = \angle ABC = 45^o$ and $DA = 3$, $AB = 7\sqrt2$, $BC = 4$. Calculate the length of side $CD$. [img]https://cdn.artofproblemsolving.com/attachments/1/2/046e31a628b3df4d23d3162cb570e1b9cb71e2.png[/img]

Brazil L2 Finals (OBM) - geometry, 2008.5

Tags: angle , circumcenter , orthocenter , geometry

Let $ABC$ be an acutangle triangle and $O, H$ its circumcenter, orthocenter, respectively. If $\frac{AB}{\sqrt2}=BH=OB$, calculate the angles of the triangle $ABC$ .

1997 Tournament Of Towns, (534) 6

Tags: isosceles , angle , isosceles triangle , geometry

Let $P$ be a point inside the triangle $ABC$ such that $AB = BC$, $\angle ABC = 80^o$, $\angle PAC = 40^o$ and $\angle ACP = 30^o$. Find $\angle BPC$. (G Galperin)

II Soros Olympiad 1995 - 96 (Russia), 11.5

Tags: angle , 3d geometry , polyhedron , geometry

$6$ points are taken on the surface of the sphere, forming three pairs of diametrically opposite points on the sphere. Consider a convex polyhedron with vertices at these points. Prove that if this polyhedron has one right dihedral angle, then it has exactly $6$ right dihedral angles.

2001 Bosnia and Herzegovina Team Selection Test, 1

Tags: ratio , geometry , angle , arc , circles

On circle there are points $A$, $B$ and $C$ such that they divide circle in ratio $3:5:7$. Find angles of triangle $ABC$

1993 Tournament Of Towns, (390) 2

Tags: right triangle , angle , geometry

Points $M$ and $N$ are taken on the hypotenuse $AB$ of a right triangle $ABC$ so that $BC = BM$ and $AC = AN$. Prove that the angle $MCN$ is equal to $45$ degrees. (Folklore)

2011 Bundeswettbewerb Mathematik, 1

Tags: combinatorial geometry , angle , geometry , hexagon , tiling , tiles , square

Prove that you can't split a square into finitely many hexagons, whose inner angles are all less than $180^o$.

2019 Poland - Second Round, 6

Tags: angle , length , geometry

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 $XM=XA$.

2002 Estonia National Olympiad, 1

Tags: geometry , angle , square , equal angles

Points $K$ and $L$ are taken on the sides $BC$ and $CD$ of a square $ABCD$ so that $\angle AKB = \angle AKL$. Find $\angle KAL$.

Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Tags: geometry , angle , right angle

Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Inside $\vartriangle BMC$ was found a point $P$ such that $\angle BMP = 90^o$, $\angle ABC+ \angle APC =180^o$. Prove that $\angle PBM + \angle CBM = \angle PCA$. (Anton Trygub)