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 an isosceles, obtuse-angled triangle, the lengths of two internal angle bisectors are in a $2:1$ ratio. Find the obtuse angle of the triangle.
Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\angle NHP$ and $\angle AHQ$ are equal. Miguel Ochoa Sanchez and Leonard Giugiuc
Angle $A$ of triangle $ABC$ is $33^o$. A straight line passing through $A$ perpendicular to $AC$ intersects straight line $BC$ at point $D$ so that $CD = 2AB$. What is angle $C$ of triangle $ABC$? (Please list all options.)
In the interior of an equilateral triangle $ ABC $ a point $ P $ is chosen such that $ PA ^2 = PB ^2 + PC ^2 $. Find the measure of $ \angle BPC $.
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 $\angle AQP = \angle D_1RS$.
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] $QR = RP.$
The Poincare plane is a half-plane bounded by a line $R$. The lines are taken to be (1) the half-lines perpendicular to $R$, and (2) the semicircles with center on $R$. Show that given any line $L$ and any point $P$ not on $L$, there are infinitely many lines through $P$ which do not intersect $L$. Show that if $ ABC$ is a triangle, then the sum of its angles lies in the interval $(0, \pi)$.
Circle inscribed in square $ABCD$ , is tangent to sides $AB$ and $CD$ at points $M$ and $K$ respectively. Line $BK$ intersects this circle at the point $L, X$ is the midpoint of $KL$. Find the angle $\angle MXK $.
Let $X$ be a point lying in the interior of the acute triangle $ABC$ such that \begin{align*} \sphericalangle BAX = 2\sphericalangle XBA \ \ \ \ \hbox{and} \ \ \ \ \sphericalangle XAC = 2\sphericalangle ACX. \end{align*} Denote by $M$ the midpoint of the arc $BC$ of the circumcircle $(ABC)$ containing $A$. Prove that $XM=XA$.
Given two intersecting planes $\alpha$ and $\beta$ and a point $A$ on the line of their intersection. Prove that of all lines belonging to $\alpha$ and passing through $A$ the line which is perpendicular to the intersection line of $\alpha$ and $\beta$ forms the greatest angle with $\beta$.
Let $a, b, c$ be the lengths of the sides of a triangle and $A, B, C$, the opposite angles. Prove that $$Aa + Bb + Cc \ge \frac{Ab + Ac + Ba + Bc + Ca + Cb}{2}$$
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$. Proposed by Iman Maghsoudi
In the quadrilateral $ABCD$ it is known that $ABC + DBC = 180^o$ and $ADC + BDC = 180^o$. Prove that the center of the circle circumscribed around the triangle $BCD$ lies on the diagonal $AC$.
In triangle $ABC$ we have $AB = BC, \angle B = 20^o$. Point $M$ on $AC$ is such that $AM : MC = 1 : 2$, point $H$ is the projection of $C$ to $BM$. Find angle $AHB$. (M. Yevdokimov)
The side of the square $ABCD$ has a length equal to $1$. On the sides $(BC)$ ¸and $(CD)$ take respectively the arbitrary points $M$ and $N$ so that the perimeter of the triangle $MCN$ is equal to $2$. a) Determine the measure of the angle $\angle MAN$. b) If the point $P$ is the foot of the perpendicular taken from point $A$ to the line $MN$, determine the locus of the points $P$.
An equilateral triangle $BDE$ is constructed on the diagonal $BD$ of the square $ABCD$, and the point $C$ is located inside the triangle $BDE$. Let $M$ be the midpoint of $BE$. Find the angle between the lines $MC$ and $DE$. (Dmitry Shvetsov)
Find two angles which add to $180^{\circ}$ which difference is $1^{'}$
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle A = 20^o$. Let $M$ be the foot of the altitude from $C$ and let $N$ be a point on the side $AC$ such that $CN =\frac12 BC$. Determine the measure of the angle $AMN$.
Prove that if $\frac{AC}{BC}=\frac{AB + BC}{AC}$ in a triangle $ABC$ , then $\angle B = 2 \angle A$ .
Let $ABCD$ be a square and let point $F$ be any point on side $BC$. Let the line perpendicular to $DF$, that passes through $B$, intersect line $DC$ at $Q$. What is value of $\angle FQC$?
Let $ABC$ be an isosceles triangle with $CA = CB$. Let $D$ be the foot of the alttiude from $C$, and $\ell$ be the external angle bisector at $C$. Take a point $N$ on $\ell$ so that $AN> AC$ , on the same side as $A$ wrt $CD$. The bisector of the angle $NAC$ cuts $\ell$'at $F$. Show that $\angle NCD + \angle BAF> 180^o.$