Olympiad problems

2017 Novosibirsk Oral Olympiad in Geometry, 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 $.

1999 Junior Balkan Team Selection Tests - Moldova, 2

Tags: geometry , angles , right triangle , isosceles

Let $ABC$ be an isosceles right triangle with $\angle A=90^o$. Point $D$ is the midpoint of the side $[AC]$, and point $E \in [AC]$ is so that $EC = 2AE$. Calculate $\angle AEB + \angle ADB$ .

Kyiv City MO 1984-93 - geometry, 1991.7.5

Tags: geometry , rectangle , angles

Inside the rectangle $ABCD$ is taken a point $M$ such that $\angle BMC + \angle AMD = 180^o$. Determine the sum of the angles $BCM$ and $DAM$.

1978 Swedish Mathematical Competition, 3

Tags: geometry , angles , sphere , 3D geometry

Two satellites are orbiting the earth in the equatorial plane at an altitude $h$ above the surface. The distance between the satellites is always $d$, the diameter of the earth. For which $h$ is there always a point on the equator at which the two satellites subtend an angle of $90^\circ$?

Kyiv City MO Seniors 2003+ geometry, 2014.10.4.1

Tags: geometry , angles , angle bisector

In the triangle $ABC$ the side $AC = \tfrac {1} {2} (AB + BC) $, $BL$ is the bisector $\angle ABC$, $K, \, \, M $ - the midpoints of the sides $AB$ and $BC$, respectively. Find the value $\angle KLM$ if $\angle ABC = \beta$

2016 Romania National Olympiad, 4

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

Consider the isosceles right triangle $ABC$, with $\angle A = 90^o$ and the point $M \in (BC)$ such that $\angle AMB = 75^o$. On the inner bisector of the angle $MAC$ take a point $F$ such that $BF = AB$. Prove that: a) the lines $AM$ and $BF$ are perpendicular; b) the triangle $CFM$ is isosceles.

Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Tags: geometry , angles , right ange

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)

1983 IMO Longlists, 69

Tags: geometry , circles , angles , midpoint , IMO , IMO 1983

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 $\angle O_1AO_2=\angle M_1AM_2$.

1997 Tournament Of Towns, (541) 2

Tags: geometry , angle bisector , angles

$D$ and $E$ are points on the sides $BC$ and $AC$ of a triangle $ABC$ such that $AD$ and $BE$ are angle bisectors of the triangle $ABC$. If $DE$ bisects $\angle ADC$, find $\angle A$. (SI Tokarev)

Durer Math Competition CD Finals - geometry, 2011.C5

Tags: geometry , angles , Equilateral

Given a straight line with points $A, B, C$ and $D$. Construct using $AB$ and $CD$ regular triangles (in the same half-plane). Let $E,F$ be the third vertex of the two triangles (as in the figure) . The circumscribed circles of triangles $AEC$ and $BFD$ intersect in $G$ ($G$ is is in the half plane of triangles). Prove that the angle $AGD$ is $120^o$ [img]https://1.bp.blogspot.com/-66akc83KSs0/X9j2BBOwacI/AAAAAAAAM0M/4Op-hrlZ-VQRCrU8Z3Kc3UCO7iTjv5ZQACLcBGAsYHQ/s0/2011%2BDurer%2BC5.png[/img]

Estonia Open Junior - geometry, 2014.2.2

Tags: geometry , angles

In a scalene triangle one angle is exactly two times as big as another one and some angle in this triangle is $36^o$. Find all possibilities, how big the angles of this triangle can be.

1991 Tournament Of Towns, (313) 3

Tags: geometry , obtuse , angles , ratio , equal ratios , equal ratio

Point $D$ lies on side $AB$ of triangle $ABC$, and $$\frac{AD}{DC} = \frac{AB}{BC}.$$ Prove that angle $C$ is obtuse. (Sergey Berlov)

2019 Durer Math Competition Finals, 15

Tags: geometry , angles , isosceles

$ABC$ is an isosceles triangle such that $AB = AC$ and $\angle BAC = 96^o$. $D$ is the point for which $\angle ACD = 48^o$, $AD = BC$ and triangle $DAC$ is obtuse-angled. Find $\angle DAC$.

V Soros Olympiad 1998 - 99 (Russia), 10.4

Tags: geometry , incenter , angles

Let $M$ be the midpoint of side $BC$ of triangle $ABC$, $Q$ the point of intersection of its angle bisectors. It is known that $MQ=QA$. Find the smallest possible value of angle $\angle MQA$.

2008 Postal Coaching, 5

Tags: geometry , incircle , cyclic quadrilateral , angles , equal angles

A convex quadrilateral $ABCD$ is given. There rays $BA$ and $CD$ meet in $P$, and the rays $BC$ and $AD$ meet in $Q$. Let $H$ be the projection of $D$ on $PQ$. Prove that $ABCD$ is cyclic if and only if the angle between the rays beginning at $H$ and tangent to the incircle of triangle $ADP$ is equal to the angle between the rays beginning at $H$ and tangent to the incircle of triangle $CDQ$. Also find out whether $ABCD$ is inscribable or circumscribable and justify.

Kyiv City MO Juniors 2003+ geometry, 2019.8.3

Tags: geometry , equal angles , angles , equal segments

In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$. (Hilko Danilo)

Novosibirsk Oral Geo Oly IX, 2020.5

Tags: geometry , angle bisector , angles

Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.

1991 Chile National Olympiad, 6

Tags: angles , angle bisector , Angle Chasing , geometry

Given a triangle with $ \triangle ABC $, with: $ \angle C = 36^o$ and $ \angle A = \angle B $. Consider the points $ D $ on $ BC $, $ E $ on $ AD $, $ F $ on $ BE $, $ G $ on $ DF $ and $ H $ on $ EG $, so that the rays $ AD, BE, DF, EG, FH $ bisect the angles $ A, B, D, E, F $ respectively. It is known that $ FH = 1 $. Calculate $ AC$.

2021 Puerto Rico Team Selection Test, 2

Tags: geometry , incenter , angles , right triangle

Let $ABC$ be a right triangle with right angle at $ B$ and $\angle C=30^o$. If $M$ is midpoint of the hypotenuse and $I$ the incenter of the triangle, show that $ \angle IMB=15^o$.

Brazil L2 Finals (OBM) - geometry, 2007.1

Tags: geometry , circumcircle , angles , Angle Chasing , Circumcenter

Let $ABC$ be a triangle with circumcenter $O$. Let $P$ be the intersection of straight lines $BO$ and $AC$ and $\omega$ be the circumcircle of triangle $AOP$. Suppose that $BO = AP$ and that the measure of the arc $OP$ in $\omega$, that does not contain $A$, is $40^o$. Determine the measure of the angle $\angle OBC$. [img]https://3.bp.blogspot.com/-h3UVt-yrJ6A/XqBItXzT70I/AAAAAAAAL2Q/7LVv0gmQWbo1_3rn906fTn6wosY1-nIfwCK4BGAYYCw/s1600/2007%2Bomb%2Bl2.png[/img]

Russian TST 2018, P2

Tags: geometry , angles

Inside the acute-angled triangle $ABC$, the points $P{}$ and $Q{}$ are chosen so that $\angle ACP = \angle BCQ$ and $\angle CBP =\angle ABQ$. The point $Z{}$ is the projection of $P{}$ onto the line $BC$. The point $Q'$ is symmetric to $Q{}$ with respect to $Z{}$. The points $K{}$ and $L{}$ are chosen on the rays $AB$ and $AC$ respectively, so that $Q'K \parallel QC$ and $Q'L \parallel QB$. Prove that $\angle KPL=\angle BPC$.

Kharkiv City MO Seniors - geometry, 2014.10.4

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

Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.

1996 Dutch Mathematical Olympiad, 1

Tags: combinatorics , Integer , angles

How many different (non similar) triangles are there whose angles have an integer number of degrees?

1980 IMO Longlists, 15

Tags: geometry , 3D geometry , tetrahedron , sphere , triangle inequality , angles , IMO Shortlist

Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.

1996 Tournament Of Towns, (493) 6

Tags: geometry , angles

In an equilateral triangle $ABC$, let $D$ be a point on the side $AB$ such that $AD = AB /n$. Prove that the sum of $n - 1$ angles $\angle DP_lA$, $\angle DP_2A$, $...$, $\angle DP_nA$ where $P_1$, $P_2$, $...$ ,$P_{n-1}$ are the points dividing the side $BC$ into $n$ equal parts, is equal to $30$ degrees if (a) $n = 3$ (b) $n$ is an arbitrary integer, $n > 2$. (V Proizvolov)