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: 351

Given an isosceles triangle $ABC$ ($AB = AC$), the inscribed circle $\omega$ touches its sides $AB$ and $AC$ at points $K$ and $L$, respectively. On the extension of the side of the base $BC$, towards $B$, an arbitrary point $M$. is chosen. Line $M$ intersects $\omega$ at the point $N$ for the second time, line $BN$ intersects the second point $\omega$ at the point $P$. On the line $PK$, there is a point $X$ such that $K$ lies between $P$ and $X$ and $KX = KM$. Determine the locus of the point $X$.
Construct an equilateral trapezoid given the height and the midline, if it is known that the midline is divided by diagonals into three equal parts. (Grigory Filippovsky)
Let $\Gamma$ be a circle and $P$ a point outside of $\Gamma$ . A tangent from $P$ to the circle intersects it in $A$. Another line through $P$ intersects $\Gamma$ at the points $B$ and $C$. The bisector of $\angle APB$ intersects $AB$ at $D$ and $AC$ at $E$. Prove that the triangle $ADE$ is isosceles.
$ABCD$ is parallelogram, $M$ is the midpoint of side $CD$ and $H$ is the foot of the perpendicular from $B$ to line $AM$. Prove that $BCH$ is an isosceles triangle. (M Volchkevich)
Let the isosceles triangle $ABC$ with $| AB | = | AC |$. The point $M$ is the midpoint of the base $[BC]$, the point $N$ is the orthogonal projection of the point $M$ on the line $AC$, and the point $P$ is located on the segment $(MC)$ such that $| MP | = | P C | \sin^2 C$. Prove that the lines $AP$ and $BN$ are perpendicular.
An isosceles triangle $ABC$ with base $AC$ is given. Point $H$ is the intersection of altitudes. On the sides $AB$ and $BC$, points $M$ and $K$ are selected, respectively, so that the angle $KMH$ is right. Prove that a right-angled triangle can be constructed from the segments $AK, CM$ and $MK$.
The angles $A$ and $B$ of base of the isosceles triangle $ABC$ are equal to $40^o$. Inside $\vartriangle ABC$, $P$ is such that $\angle PAB = 30^o$ and $\angle PBA = 20^o$. Calculate, without table, $\angle PCA$.
Let $AH_a$ and $BH_b$ be altitudes, $AL_a$ and $BL_b$ be angle bisectors of a triangle $ABC$. It is known that $H_aH_b // L_aL_b$. Is it necessarily true that $AC = BC$? (B. Frenkin)
Let $ABC$ be right isosceles triangle with right angle in $A$. Given a point $P$ inside the triangle, denote by $a, b$ and $c$ the lengths of $PA, PB$ and $PC$, respectively. Prove that there is a triangle whose sides have a length of $a\sqrt2 , b$ and $c$.
In the exterior of the triangle $ABC$ with $m(\angle B) > 45^o$, $m(\angle C) >45°^o$ one constructs the right isosceles triangles $ACM$ and $ABN$ such that $m(\angle CAM) = m(\angle BAN) = 90^o$ and, in the interior of $ABC$, the right isosceles triangle $BCP$, with $m(\angle P) = 90^o$. Show that the triangle $MNP$ is a right isosceles triangle.
Given the parallelogram $ABCD$. The trisection points of side $AB$ are: $H_1, H_2$, ($AH_1 = H_1H_2 =H_2B$). The trisection points of the side $DC$ are $G_1, G_2$, ($DG_1 = G_1G_2 = G_2C$), and $AD = 1, AC = 2$. Prove that triangle $AH_2G_1$ is isosceles.
In the triangle $ABC$ with $| AB | = c, | BC | = a, | CA | = b$ the relations hold simultaneously $$a \ge max \{ b, c, \sqrt{bc}\}, \sqrt{(a - b) (a + c)} + \sqrt{(a - c) (a + b) } \ge 2\sqrt{a^2-bc}$$ Prove that the triangle $ABC$ is isosceles.
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 $BT = 2PT$.
Let $ABC$ be a triangle with $AB = AC$ with centroid $G$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$ respectively and $O$ be the circumcenter of triangle $BCN$ . Prove that $MBOG$ is a cyclic quadrilateral .
Triangle $\vartriangle ABC$ is isosceles with $AB = AC$ and $\angle ABC = 2\angle BAC$. Compute $\frac{AB}{BC}$ .
On a straight line lie $100$ points and another point outside the line. Which is the biggest the number of isosceles triangles can be formed from the vertices of these $101$ points?
The center $O$ of the circle $\omega$ passing through the vertex $C$ of the isosceles triangle $ABC$ ($AB = AC$) is the interior point of the triangle $ABC$. This circle intersects segments $BC$ and $AC$ at points $D \ne C$ and $E \ne C$, respectively, and the circumscribed circle $\Omega$ of the triangle $AEO$ at the point $F \ne E$. Prove that the center of the circumcircle of the triangle $BDF$ lies on the circle $\Omega$.
The trapezoid is inscribed in a circle. Prove that the sum of distances from any point of the circle to the midpoints of the lateral sides are not less than the diagonal of the trapezoid.
A triangle is constructed with the lengths of the sides chosen from the set $\{2, 3, 5, 8, 13, 21, 34, 55, 89, 144\}$. Show that this triangle must be isosceles. (A triangle is isosceles if it has at least two sides the same length.)
Given an isosceles triangle $ABC$ such that $|AB|=|AC|$ . Let $M$ be the midpoint of the segment $BC$ and let $P$ be a point other than $A$ such that $PA\parallel BC$. The points $X$ and $Y$ are located respectively on rays $PB$ and $PC$, so that the point $B$ is between $P$ and $X$, the point $C$ is between $P$ and $Y$ and $\angle PXM=\angle PYM$. Prove that the points $A,P,X$ and $Y$ are concyclic.
Each diagonal of a convex quadrilateral divides it into two isosceles triangles. The two diagonals of the same quadrilateral divide it into four isosceles triangles. Must this quadrilateral be a square?
Let $ABC$ be an isosceles triangle with apex $A$ and altitude $AD$. On $AB$, choose a point $F$ distinct from $B$ such that $CF$ is tangent to the incircle of $ABD$. Suppose that $\vartriangle BCF$ is isosceles. Show that those conditions uniquely determine: a) which vertex of $BCF$ is its apex, b) the size of $\angle BAC$
In an isosceles triangle $ABC$ with $AB = AC$ and $\angle A = 30^o$, points $L$ and $M$ lie on the sides $AB$ and $AC$, respectively such that $AL = CM$. Point $K$ lies on $AB$ such that $\angle AMK = 45^o$. If $\angle LMC = 75^o$, prove that $KM +ML = BC$. [i]Proposed by Mahdi Etesamifard - Iran[/i]
In a triangle $ABC$ with $ \angle A = 36^o$ and $AB = AC$, the bisector of the angle at $C$ meets the oposite side at $D$. Compute the angles of $\triangle BCD$. Express the length of side $BC$ in terms of the length $b$ of side $AC$ without using trigonometric functions.
Let the triangle $ABC$ be acute. Let us take in the segment $BC$ two points $F$ and $G$ such that $BG > BF = GC$ and an interior point$ P$ to the triangle on the bisector of $\angle BAC$. Then are drawn through $P$, $PD\parallel AB$ and $PE \parallel AC$, $D \in AC$ and $E \in AB$, $\angle FEP = \angle PDG$. prove that $\vartriangle ABC$ is isosceles.