Olympiad problems

2016 Peru MO (ONEM), 1

Let $ABCD$ be a trapezoid of parallel bases $ BC$ and $AD$. If $\angle CAD = 2\angle CAB, BC = CD$ and $AC = AD$, determine all the possible values of the measure of the angle $\angle CAB$.

1985 IMO Longlists, 9

Tags: geometry , 3D geometry , polyhedron , angles , Euler s Theorem , IMO Shortlist

A polyhedron has $12$ faces and is such that: [b][i](i)[/i][/b] all faces are isosceles triangles, [b][i](ii)[/i][/b] all edges have length either $x$ or $y$, [b][i](iii)[/i][/b] at each vertex either $3$ or $6$ edges meet, and [b][i](iv)[/i][/b] all dihedral angles are equal. Find the ratio $x/y.$

2021 Novosibirsk Oral Olympiad in Geometry, 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]

2008 Hanoi Open Mathematics Competitions, 7

Tags: geometry , convex , pentagon , angles

The figure $ABCDE$ is a convex pentagon. Find the sum $\angle DAC + \angle EBD +\angle ACE +\angle BDA + \angle CEB$?

Kyiv City MO Juniors 2003+ geometry, 2005.89.5

Tags: geometry , angles , hexagon , Regular

Let $ABCDEF $ be a regular hexagon. On the line $AF $ mark the point $X$so that $ \angle DCX = 45^o$ . Find the value of the angle $FXE$. (Vyacheslav Yasinsky)

1975 IMO Shortlist, 8

Tags: geometry , trigonometry , Triangle , angles , IMO , IMO 1975

In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, $\angle ABR = \angle RAB = 15^{\circ}$. Prove that [b]a.)[/b] $\angle QRP = 90\,^{\circ},$ and [b]b.)[/b] $QR = RP.$

2024 Nepal TST, P5

Tags: geometry , angles

Let $ABC$ be an acute triangle so that $2BC = AB + AC,$ with incenter $I{}.$ Let $AI{}$ meet $BC{}$ at point $A'.{}$ The perpendicular bisector of $AA'{}$ meets $BI{}$ and $CI{}$ at points $B'{}$ and $C'{}$ respectively. Let $AB'{}$ intersect $(ABC)$ at $X{}$ and let $XI{}$ intersect $AC'{}$ at $X'{}.$ Prove that $2\angle XX'A'=\angle ABC.{}$ [i](Proposed by Kang Taeyoung, South Korea)[/i]

2021 Grand Duchy of Lithuania, 3

Tags: geometry , right triangle , angles

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 $AC$, $BD$, $CD$.

2006 Sharygin Geometry Olympiad, 25

Tags: geometry , 3D geometry , tetrahedron , ratio , angles

In the tetrahedron $ABCD$ , the dihedral angles at the $BC, CD$, and $DA$ edges are equal to $\alpha$, and for the remaining edges equal to $\beta$. Find the ratio $AB / CD$.

2022 Dutch BxMO TST, 2

Tags: geometry , equal angles , angles

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 $\angle BXD = \angle CYD$.

2015 Romania National Olympiad, 4

Tags: geometry , angles , Angle Chasing

Consider $\vartriangle ABC$ where $\angle ABC= 60 ^o$. Points $M$ and $D$ are on the sides $(AC)$, respectively $(AB)$, such that $\angle BCA = 2 \angle MBC$, and $BD = MC$. Determine $\angle DMB$.

1969 IMO Shortlist, 71

Tags: geometry , rhombus , angles , IMO Shortlist , IMO Longlist

$(YUG 3)$ Let four points $A_i (i = 1, 2, 3, 4)$ in the plane determine four triangles. In each of these triangles we choose the smallest angle. The sum of these angles is denoted by $S.$ What is the exact placement of the points $A_i$ if $S = 180^{\circ}$?

2014 India PRMO, 16

Tags: geometry , incenter , incircle , angles , Angle Chasing

In a triangle $ABC$, let $I$ denote the incenter. Let the lines $AI,BI$ and $CI$ intersect the incircle at $P,Q$ and $R$, respectively. If $\angle BAC = 40^o$, what is the value of $\angle QPR$ in degrees ?

Novosibirsk Oral Geo Oly IX, 2017.2

Tags: geometry , angles

You are given a convex quadrilateral $ABCD$. It is known that $\angle CAD = \angle DBA = 40^o$, $\angle CAB = 60^o$, $\angle CBD = 20^o$. Find the angle $\angle CDB $.

2021 OMpD, 5

Tags: geometry , angles , circumcircle

Let $ABC$ be a triangle with $\angle BAC > 90^o$ and with $AB < AC$. Let $r$ be the internal bisector of $\angle ACB$ and let $s$ be the perpendicular, through $A$, on $r$. Denote by $F$ the intersection of $r$ and $ s$, and denote by $E$ the intersection of $s$ with the segment $BC$. Let also $D$ be the symmetric of $A$ with respect to the line $BF$. Assuming that the circumcircle of triangle $EAC$ is tangent to line $AB$ and $ D$ lies on $r$, determine the value of $\angle CDB$.

1998 Swedish Mathematical Competition, 4

Tags: geometry , angles , area

$ABCD$ is a quadrilateral with $\angle A = 90o$, $AD = a$, $BC = b$, $AB = h$, and area $\frac{(a+b)h}{2}$. What can we say about $\angle B$?

2017 JBMO Shortlist, G3

Tags: geometry , circumcircle , equal angles , angles

Consider triangle $ABC$ such that $AB \le AC$. Point $D$ on the arc $BC$ of thecircumcirle of $ABC$ not containing point $A$ and point $E$ on side $BC$ are such that $\angle BAD = \angle CAE < \frac12 \angle BAC$ . Let $S$ be the midpoint of segment $AD$. If $\angle ADE = \angle ABC - \angle ACB$ prove that $\angle BSC = 2 \angle BAC$ .

2022 Yasinsky Geometry Olympiad, 1

Tags: geometry , angles , median

In the triangle $ABC$, the median $AM$ is extended to the intersection with the circumscribed circle at point $D$. It is known that $AB = 2AM$ and $AD = 4AM$. Find the angles of the triangle $ABC$. (Gryhoriy Filippovskyi)

1996 North Macedonia National Olympiad, 3

Tags: trigonometry , inequalities , Geometric Inequalities , angles

Prove that if $\alpha, \beta, \gamma$ are angles of a triangle, then $\frac{1}{\sin \alpha}+ \frac{1}{\sin \beta} \ge \frac{8}{ 3+2 \ cos\gamma}$ .

2020 Yasinsky Geometry Olympiad, 1

Tags: geometry , rectangle , Equilateral , angles

In the rectangle $ABCD$, $AB = 2BC$. An equilateral triangle $ABE$ is constructed on the side $AB$ of the rectangle so that its sides $AE$ and $BE$ intersect the segment $CD$. Point $M$ is the midpoint of $BE$. Find the $\angle MCD$.

2003 All-Russian Olympiad Regional Round, 11.2

Tags: geometry , angles , equal angles

On the diagonal $AC$ of a convex quadrilateral $ABCD$ is chosen such a point $K$ such that $KD = DC$, $\angle BAC = \frac12 \angle KDC$, $\angle DAC = \frac12 \angle KBC$. Prove that $\angle KDA = \angle BCA$ or $\angle KDA = \angle KBA$.

2019 Romania National Olympiad, 3

Tags: geometry , 3D geometry , prism , equal angles , angles

In the regular hexagonal prism $ABCDEFA_1B_1C_1D_1E_1F_1$, We construct $, Q$, the projections of point $A$ on the lines $A_1B$ and $A_1C$ repsectilvely. We construct $R,S$, the projections of point $D_1$ on the lines $A_1D$ and $C_1D$ respectively. a) Determine the measure of the angle between the planes $(AQP)$ and $(D_1RS)$. b) Show that $\angle AQP = \angle D_1RS$.

2005 Sharygin Geometry Olympiad, 16

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

We took a non-equilateral acute-angled triangle and marked $4$ wonderful points in it: the centers of the inscribed and circumscribed circles, the center of gravity (the point of intersection of the medians) and the intersection point of altitudes. Then the triangle itself was erased. It turned out that it was impossible to establish which of the centers corresponds to each of the marked points. Find the angles of the triangle

1935 Moscow Mathematical Olympiad, 009

Tags: geometry , 3D geometry , cone , regular hexagon , hexagon , angles

The height of a truncated cone is equal to the radius of its base. The perimeter of a regular hexagon circumscribing its top is equal to the perimeter of an equilateral triangle inscribed in its base. Find the angle $\phi$ between the cone’s generating line and its base.

Kyiv City MO Juniors Round2 2010+ geometry, 2013.7.3

Tags: geometry , angles , square , Angle Chasing , equal angles

In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.