Olympiad problems

Indonesia Regional MO OSP SMA - geometry, 2013.5

Given an acute triangle $ABC$. The longest line of altitude is the one from vertex $A$ perpendicular to $BC$, and it's length is equal to the length of the median of vertex $B$. Prove that $\angle ABC \le 60^o$

1964 All Russian Mathematical Olympiad, 041

Tags: angle , geometry

The two heights in the triangle are not less than the respective sides. Find the angles.

2011 Junior Balkan Team Selection Tests - Romania, 4

Tags: angle , geometry , equal segments , orthocenter , incenter

The measure of the angle $\angle A$ of the acute triangle $ABC$ is $60^o$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle $\angle B$.

2013 India PRMO, 9

Tags: incenter , geometry , angle , circumcircle

In a triangle $ABC$, let $H, I$ and $O$ be the orthocentre, incentre and circumcentre, respectively. If the points $B, H, I, C$ lie on a circle, what is the magnitude of $\angle BOC$ in degrees?

2021 Czech-Polish-Slovak Junior Match, 5

Tags: angle , hexagon , geometry

A regular heptagon $ABCDEFG$ is given. The lines $AB$ and $CE$ intersect at $ P$. Find the measure of the angle $\angle PDG$.

2007 District Olympiad, 1

Tags: equal segments , circumcircle , angle , angle chasing , circumcenter , geometry

Point $O$ is the intersection of the perpendicular bisectors of the sides of the triangle $\vartriangle ABC$ . Let $D$ be the intersection of the line $AO$ with the segment $[BC]$. Knowing that $OD = BD = \frac 13 BC$, find the measures of the angles of the triangle $\vartriangle ABC$.

2014 Contests, 3

Tags: equal segments , right triangle , geometry , 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 $\angle VAU$.

1990 IMO Longlists, 25

Tags: geometry , incenter , angle , midpoint , trigonometry

The incenter of the triangle $ ABC$ is $ K.$ The midpoint of $ AB$ is $ C_1$ and that of $ AC$ is $ B_1.$ The lines $ C_1K$ and $ AC$ meet at $ B_2,$ the lines $ B_1K$ and $ AB$ at $ C_2.$ If the areas of the triangles $ AB_2C_2$ and $ ABC$ are equal, what is the measure of angle $ \angle CAB?$

1988 Tournament Of Towns, (189) 2

Tags: equal angles , geometry , square , angle

A point $M$ is chosen inside the square $ABCD$ in such a way that $\angle MAC = \angle MCD = x$ . Find $\angle ABM$.

2019 Yasinsky Geometry Olympiad, p3

Tags: angle chasing , angle , cyclic quadrilateral , geometry

In the quadrilateral $ABCD$, the angles $B$ and $D$ are right . The diagonal $AC$ forms with the side $AB$ the angle of $40^o$, as well with side $AD$ an angle of $30^o$. Find the acute angle between the diagonals $AC$ and $BD$.

2018 District Olympiad, 4

Tags: angle bisector , geometry , angle

Let $ABC$ be a triangle with $\angle A = 80^o$ and $\angle C = 30^o$. Consider the point $M$ inside the triangle $ABC$ so that $\angle MAC= 60^o$ and $\angle MCA = 20^o$. If $N$ is the intersection of the lines $BM$ and $AC$ to show that a $MN$ is the bisector of the angle $\angle AMC$.

2018 EGMO, 5

Tags: power of a point , triangle , angle , geometry

Let $\Gamma $ be the circumcircle of triangle $ABC$. A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\Gamma$ at a point lying on the same side of the line $AB$ as $C$. The angle bisector of $\angle BCA$ intersects $\Omega$ at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC$.

1983 IMO, 2

Tags: midpoint , circles , angle , geometry

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.

Novosibirsk Oral Geo Oly VII, 2021.2

Tags: geometry , angle

The extensions of two opposite sides of the convex quadrilateral intersect and form an angle of $20^o$ , the extensions of the other two sides also intersect and form an angle of $20^o$. It is known that exactly one angle of the quadrilateral is $80^o$. Find all of its other angles.

1980 Tournament Of Towns, (006) 3

Tags: combinatorial geometry , combinatorics , angle , vector , geometry

We are given $30$ non-zero vectors in $3$ dimensional space. Prove that among these there are two such that the angle between them is less than $45^o$.

1998 Bundeswettbewerb Mathematik, 3

Tags: angle , geometry

A triangle $ABC$ satisfies $BC = AC +\frac12 AB$. Point $P$ on side $AB$ is taken so that $AP = 3PB$. Prove that $ \angle PAC = 2\angle CPA$.

IV Soros Olympiad 1997 - 98 (Russia), 9.3

Tags: orthocenter , angle , ratio , geometry

What is angle $B$ of triangle$ ABC$, if it is known that the altitudes drawn from $A$ and $C$ intersect inside the triangle and one of them is divided by of intersection point into equal parts, and the other one in the ratio of $2: 1$, counting from the vertex?

2001 Estonia National Olympiad, 3

Tags: sum , angle , geometry

There are three squares in the picture. Find the sum of angles $ADC$ and $BDC$. [img]https://cdn.artofproblemsolving.com/attachments/c/9/885a6c6253fca17e24528f8ba8a5d31a18c845.png[/img]

Kyiv City MO Juniors Round2 2010+ geometry, 2012.9.4

Tags: circumcircle , geometry , angle , parallel , orthocenter , equal segments

In an acute-angled triangle $ABC$, the point $O$ is the center of the circumcircle, and the point $H$ is the orthocenter. It is known that the lines $OH$ and $BC$ are parallel, and $BC = 4OH $. Find the value of the smallest angle of triangle $ ABC $. (Black Maxim)

2014 Oral Moscow Geometry Olympiad, 6

Tags: right triangle , angle chasing , isosceles , angle , geometry

Inside an isosceles right triangle $ABC$ with hypotenuse $AB$ a point $M$ is taken such that the angle $\angle MAB$ is $15 ^o$ larger than the angle $\angle MAC$ , and the angle $\angle MCB$ is $15^o$ larger than the angle $\angle MBC$. Find the angle $\angle BMC$ .

2020 Serbian Mathematical Olympiad, Problem 3

Tags: circumcircle , triangle , angle , geometry

We are given a triangle $ABC$. Points $D$ and $E$ on the line $AB$ are such that $AD=AC$ and $BE=BC$, with the arrangment of points $D - A - B - E$. The circumscribed circles of the triangles $DBC$ and $EAC$ meet again at the point $X\neq C$, and the circumscribed circles of the triangles $DEC$ and $ABC$ meet again at the point $Y\neq C$. Find the measure of $\angle ACB$ given the condition $DY+EY=2XY$.

2020 Novosibirsk Oral Olympiad in Geometry, 7

Tags: angle , angle bisector , geometry

You are given a quadrilateral $ABCD$. It is known that $\angle BAC = 30^o$, $\angle D = 150^o$ and, in addition, $AB = BD$. Prove that $AC$ is the bisector of angle $C$.

1992 IMO Longlists, 37

Tags: circles , angle , geometry , trigonometric inequality

Let the circles $C_1, C_2$, and $C_3$ be orthogonal to the circle $C$ and intersect each other inside $C$ forming acute angles of measures $A, B$, and $C$. Show that $A + B +C < \pi.$

2020 LIMIT Category 2, 17

Tags: calculus , limit , triangle , angle

Let $a_n$ denote the angle opposite to the side of length $4n^2$ units in an integer right angled triangle with lengths of sides of the triangle being $4n^2, 4n^4+1$ and $4n^4-1$ where $n \in N$. Then find the value of $\lim_{p \to \infty} \sum_{n=1}^p a_n$ (A) $\pi/2$ (B) $\pi/4$ (C) $\pi $ (D) $\pi/3$

2021 Sharygin Geometry Olympiad, 10-11.6

Tags: trapezoid , angle , equal angles , geometry

The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.