This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 721

In triangle $ABC$, the angle at vertex $B$ is $120^o$. Let $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$ respectively such that $AA_1, BB_1, CC_1$ are bisectors of the angles of triangle $ABC$. Determine the angle $\angle A_1B_1C_1$.
Let $A$ be an arbitrary angle,let $B$ and $C$ be acute angles. Is there an angle $x$ such that $$\sin x =\frac{\sin B \cdot \sin C}{1 - \cos B \cdot \cos C \cdot \cos A} ?$$
Given a point $M$ inside a right tetrahedron. Prove that at least one tetrahedron edge is seen from the $M$ in an angle, that has a cosine not greater than $-1/3$. (e.g. if $A$ and $B$ are the vertices, corresponding to that edge, $cos(\widehat{AMB}) \le -1/3$)
Let $ABC$ be an isosceles triangle. Suppose that points $K$ and $L$ are chosen on lateral sides $AB$ and $AC$ respectively so that $AK = CL$ and $\angle ALK + \angle LKB = 60^o$. Prove that $KL = BC$.
Let $ABCD$ be a convex cyclic quadrilateral with $AD > BC$, A$B$ not being diameter and $C D$ belonging to the smallest arc $AB$ of the circumcircle. The rays $AD$ and $BC$ are cut at $K$, the diagonals $AC$ and $BD$ are cut at $P$ and the line $KP$ cuts the side $AB$ at point $L$. Prove that angle $\angle ALK$ is acute.
Given a rectangle $ABCD$ with $AB> BC$. On the side $CD$, take a point $L$ such that $BL$ and $AC$ are perpendicular. Let $K$ be the intersection point of segments $BL$ and $AC$. It is known that segments $AL$. and $DK$ are perpendicular. Find $\angle ACB.$
In the triangle $ABC$ the angle bisectors $AD$ and $BE$ are drawn. Prove that $\angle ACB = 60 {} ^ \circ$ if and only if $AE + BD = AB$. (Hilko Danilo)
Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $. (Serdyuk Nazar)
Let $ABC$ be a triangle with sides $a,b,c$ and corresponding angles $\alpha,\beta\gamma$ . Prove that if $\alpha = 3\beta$ then $(a^2 -b^2)(a-b) = bc^2$ . Is the converse true?
The angle $B$ of a triangle $ABC$ is $120^o$. Let $M$ be a point on the side $AC$ and $K$ a point on the extension of the side $AB$, such that $BM$ is the internal bisector of the angle $\angle ABC$ and $CK$ is the external bisector corresponding to the angle $\angle ACB$ . The segment $MK$ intersects $BC$ at point $P$. Prove that $\angle APM = 30^o$.
Let $A_1, A_2, ..., A_m$ and $B_2 , B_3,..., B_n$ be the points on a circle such that $A_1A_2... A_n$ is a regular $m$-gon and $A_1B_2...B_n$ is a regular $n$-gon whereby $n > m$ and the point $B_2$ lies between $A_1$ and $A_2$. Find $\angle B_2A_1A_2$.
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)
The bisector of the interior angle at the vertex $B$ of the triangle $ABC$ and the perpendicular line on side $BC$ passing through the vertex $C$ intersects at $D$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $BD$, respectively, with $N$ on the side $AC$. Find all possibilities of the angles of the triangles $ABC$, if it is known that $\frac{| AM |}{| BC |}=\frac{|CD|}{|BD|}$. .
Consider triangle $ABC$ with $AB = AC$ and $\angle A = 40 ^o$. The points $S$ and $T$ are on the sides $AB$ and $BC$, respectively, so that $\angle BAT = \angle BCS= 10 ^o$. The lines $AT$ and $CS$ intersect at point $P$. Prove that $BT = 2PT$.
Points $K$ and $M$ are taken on the sides $AB$ and $CD$ of square $ABCD$, respectively, and on the diagonal $AC$ - point $L$ such that $ML = KL$. Let $P$ be the intersection point of the segments $MK$ and $BD$. Find the angle $\angle KPL$.
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 $\angle ABC = \beta$
Two quarter-circles touch as shown. Find the angle $x$. [img]https://cdn.artofproblemsolving.com/attachments/b/4/e70d5d69e46d6d40368f143cb83cf10b7d6d98.png[/img]
In a convex quadrilateral $ABCD$, $\angle ABD=30^\circ$, $\angle BCA=75^\circ$, $\angle ACD=25^\circ$ and $CD=CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE=BD$. [i]Proposed by BJ Venkatachala[/i]
Consider the triangle $ABC$, with $\angle A= 90^o, \angle B = 30^o$, and $D$ is the foot of the altitude from $A$. Let the point $E \in (AD)$ such that $DE = 3AE$ and $F$ the foot of the perpendicular from $D$ to the line $BE$. a) Prove that $AF \perp FC$. b) Determine the measure of the angle $AFB$.
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 $90^\circ$?
Let $ABC$ be an equilateral triangle, and let $P$ stand for an arbitrary point inside the triangle. Is it true that $| \angle PAB - \angle PAC| \ge | \angle PBC - \angle PCB|$ ?
An isosceles trapezoid is divided by each diagonal into two isosceles triangles. Determine the angles of the trapezoid.
In the quadrilateral $ABCD$ point $E$ - the midpoint of the side $AB$, point $F$ - the midpoint of the side $BC$, point $G$ - the midpoint $AD$ . It turned out that the segment $GE$ is perpendicular to $AB$, and the segment $GF$ is perpendicular to the segment $BC$. Find the value of the angle $GCD$, if it is known that $\angle ADC = 70 {} ^ \circ$.
Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal.
In the convex pentagon $ ABCDE,$ the sides $ BC, CD, DE$ are equal. Moreover each diagonal of the pentagon is parallel to a side ($ AC$ is parallel to $ DE$, $ BD$ is parallel to $ AE$ etc.). Prove that $ ABCDE$ is a regular pentagon.