Olympiad problems

2015 Czech-Polish-Slovak Junior Match, 1

In the right triangle $ABC$ with shorter side $AC$ the hypotenuse $AB$ has length $12$. Denote $T$ its centroid and $D$ the feet of altitude from the vertex $C$. Determine the size of its inner angle at the vertex $B$ for which the triangle $DTC$ has the greatest possible area.

2020 Ukrainian Geometry Olympiad - December, 1

Tags: angle , equal sides , geometry

The three sides of the quadrilateral are equal, the angles between them are equal, respectively $90^o$ and $150^o$. Find the smallest angle of this quadrilateral in degrees.

2016 Novosibirsk Oral Olympiad in Geometry, 2

Tags: geometry , angle

Bisector of one angle of triangle $ABC$ is equal to the bisector of its external angle at the same vertex (see figure). Find the difference between the other two angles of the triangle. [img]https://cdn.artofproblemsolving.com/attachments/c/3/d2efeb65544c45a15acccab8db05c8314eb5f2.png[/img]

Kyiv City MO Juniors 2003+ geometry, 2020.9.41

Tags: equal segments , angle , geometry

The points $A, B, C, D$ are selected on the circle as followed so that $AB = BC = CD$. Bisectors of $\angle ABD$ and $\angle ACD$ intersect at point $E$. Find $\angle ABC$, if it is known that $AE \parallel CD$.

2009 Sharygin Geometry Olympiad, 6

Tags: angle , geometry

Given triangle $ABC$ such that $AB- BC = \frac{AC}{\sqrt2}$ . Let $M$ be the midpoint of $AC$, and $N$ be the foot of the angle bisector from $B$. Prove that $\angle BMC + \angle BNC = 90^o$. (A.Akopjan)

2010 Sharygin Geometry Olympiad, 6

Tags: midpoint , angle chasing , geometry , angle , square

Let $E, F$ be the midpoints of sides $BC, CD$ of square $ABCD$. Lines $AE$ and $BF$ meet at point $P$. Prove that $\angle PDA = \angle AED$.

2021 Grand Duchy of Lithuania, 3

Tags: angle , right triangle , geometry

Let $ABCD$ be a convex quadrilateral satisfying $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^o$, $AD = BC$. Prove that there exists a right-angled triangle with side lengths $AC$, $BD$, $CD$.

1989 IMO Longlists, 67

Tags: angle , tetrahedron , 3d geometry , geometric inequality , geometry

Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than $ 120^{\circ}.$

1990 IMO Longlists, 5

Tags: minimization , triangle , angle , geometry

Given the condition that there exist exactly $1990$ triangles $ABC$ with integral side-lengths satisfying the following conditions: (i) $\angle ABC =\frac 12 \angle BAC;$ (ii) $AC = b.$ Find the minimal value of $b.$

2015 Auckland Mathematical Olympiad, 4

Tags: geometry , angle bisector , angle

The bisector of angle $A$ in parallelogram $ABCD$ intersects side $BC$ at $M$ and the bisector of $\angle AMC$ passes through point $D$. Find angles of the parallelogram if it is known that $\angle MDC = 45^o$. [img]https://cdn.artofproblemsolving.com/attachments/e/7/7cfb22f0c26fe39aa3da3898e181ae013a0586.png[/img]

Novosibirsk Oral Geo Oly VIII, 2021.4

Tags: angle bisector , angle , geometry

Angle bisectors $AD$ and $BE$ are drawn in triangle $ABC$. It turned out that $DE$ is the bisector of triangle $ADC$. Find the angle $BAC$.

Geometry Mathley 2011-12, 11.3

Tags: equal angles , equal ratio , angle , geometry

Let $ABC$ be a triangle such that $AB = AC$ and let $M$ be a point interior to the triangle. If $BM$ meets $AC$ at $D$. show that $\frac{DM}{DA}=\frac{AM}{AB}$ if and only if $\angle AMB = 2\angle ABC$. Michel Bataille

2020 Ukrainian Geometry Olympiad - April, 4

Tags: angle , equal segments , equal angles , geometry , 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 $2\angle BCQ = \angle AQP$.

2022 Dutch BxMO TST, 2

Tags: angle , equal angles , geometry

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A, X, B, C, Y$ . Show that $\angle BXD = \angle CYD$.

1950 Moscow Mathematical Olympiad, 176

Tags: sidelengths , geometric inequality , angle , inequalities

Let $a, b, c$ be the lengths of the sides of a triangle and $A, B, C$, the opposite angles. Prove that $$Aa + Bb + Cc \ge \frac{Ab + Ac + Ba + Bc + Ca + Cb}{2}$$

2010 Swedish Mathematical Competition, 5

Tags: sidelengths , angle , geometry

Consider the number of triangles where the side lengths $a,b,c$ satisfy $(a + b + c) (a + b -c) = 2b^2$. Determine the angles in the triangle for which the angle opposite to the side with the length $a$ is as big as possible.

2012 Cuba MO, 9

Tags: angle , geometry

Let $O$ be a point interior to triangle $ABC$ such that $\angle BAO = 30^o$, $\angle CBO = 20^o$ and $\angle ABO = \angle ACO = 40^o$ . Knowing that triangle $ABC$ is not equilateral, find the measures of its interior angles.

2010 Silk Road, 1

Tags: convex quadrilateral , angle , right triangle , geometry

In a convex quadrilateral it is known $ABCD$ that $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^{\circ}$ and $AD = BC$. Prove that from the lengths $DB$, $CA$ and $DC$, you can make a right triangle.

2018 Flanders Math Olympiad, 1

Tags: geometric inequalities , angle , geometry

In the triangle $\vartriangle ABC$ we have $| AB |^3 = | AC |^3 + | BC |^3$. Prove that $\angle C> 60^o$ .

2011 Sharygin Geometry Olympiad, 7

Tags: circles , equal segments , common tangent , angle , geometry

Circles $\omega$ and $\Omega$ are inscribed into the same angle. Line $\ell$ meets the sides of angles, $\omega$ and $\Omega$ in points $A$ and $F, B$ and $C, D$ and $E$ respectively (the order of points on the line is $A,B,C,D,E, F$). It is known that$ BC = DE$. Prove that $AB = EF$.

Kyiv City MO Seniors 2003+ geometry, 2020.10.5.1

Tags: semicircle , angle , geometry , equal angles

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 $\angle EFD=\angle ACD+ \angle ECB$.