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.

AND
OR
NO

Found problems: 1049

The bisector of angle $BAD$ in the parallelogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. It is known that $ABCD$ is not a rhombus. Prove that the centre of the circle passing through the points $C, K$ and $L$ lies on the circle passing through the points $B, C$ and $D$.
(a) Find the point $M$ in the plane of triangle $ABC$ for which the sum $MA + MB+ MC$ is minimal. (b) Given a parallelogram $ABCD$ whose angles do not exceed $120^o$, determine $min \{MA+ MB+NC+ND+ MN | M,N$ are in the plane $ABCD\}$ in terms of the sides and angles of the parallelogram.
Let $ABCD$ be a parallelogram with $AB=8$, $AD=11$, and $\angle BAD=60^\circ$. Let $X$ be on segment $CD$ with $CX/XD=1/3$ and $Y$ be on segment $AD$ with $AY/YD=1/2$. Let $Z$ be on segment $AB$ such that $AX$, $BY$, and $DZ$ are concurrent. Determine the area of triangle $XYZ$.
A fixed point $A$ inside a circle is given. Consider all chords $XY$ of the circle such that $\angle XAY$ is a right angle, and for all such chords construct the point $M$ symmetric to $A$ with respect to $XY$ . Find the locus of points $M$.
Let $ABCD$ be a parallelogram with center $O$ such that $\angle BAD <90^o$ and $\angle AOB> 90^o$. Consider points $A_1$ and $B_1$ on the rays $OA$ and $OB$ respectively, such that $A_1B_1$ is parallel to $AB$ and $\angle A_1B_1C = \frac12 \angle ABC$. Prove that $A_1D$ is perpendicular to $B_1C$.
Consider the parallelogram $ABCD$, with $\angle ABC = 60$ and sides $AB =\sqrt3$, $BC = 1$. Let $\omega$ be the circle of center $B$ and radius $BA$, and let $\tau$ be the circle of center $D$ and radius $DA$. Determine the area of the region between the circumferences $\omega$ and $\tau$, within the parallelogram $ABCD$ (the area of the shaded region). [img]https://cdn.artofproblemsolving.com/attachments/5/a/02b17ec644289d95b6fce78cb5f1ecb3d3ba5b.png[/img]
Given a parallelogram $ABCD$. A circle passing through $A$ meets the line segments $AB, AC$ and $AD$ at inner points $M,K,N$, respectively. Prove that \[|AB|\cdot |AM | + |AD|\cdot |AN|=|AK|\cdot |AC|\]
(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A'$, $ B'$; $ C_1$ is the meet of lines $ AA'$ and $ BB'$. Prove that $ \angle C_1CA \equal{} \angle C_0CB$.
$ABCD$ is parallelogram. Line $l$ is perpendicular to $BC$ at $B$. Two circles passes through $D,C$, such that $l$ is tangent in points $P$ and $Q$. $M$ - midpoint $AB$. Prove that $\angle DMP=\angle DMQ$
Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.
Prove that if all faces of a parallelepiped are equal parallelograms, they are rhombuses.
We are given a triangle $ABC$ such that $\angle BAC < 90^{\circ}$. The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^{\circ}$. Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^{\circ}$. The point $X$ is such that $ADXE$ is a parallelogram. Prove that $|BX| = |CX|$.
Circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$ intersect at points $A$ and $B$. Let point $C$ be such that $AO_2CO_1$ is a parallelogram. An arbitrary line is drawn through point $A$, which intersects the circles $\omega_1$ and $\omega_2$ at points $X$ and $Y$, respectively. Prove that $CX = CY$. [i]Proposed by Oleksii Masalitin[/i]
A frog makes $ 3$ jumps, each exactly $ 1$ meter long. The directions of the jumps are chosen independently and at random. What is the probability the frog's final position is no more than $ 1$ meter from its starting position? $ \textbf{(A)}\ \frac {1}{6} \qquad \textbf{(B)}\ \frac {1}{5} \qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$
5. In tetrahedron $ABCD$ following equalities hold: $\angle BAC+\angle BDC=\angle ABD+\angle ACD$ $\angle BAD+\angle BCD=\angle ABC+\angle ADC$ Prove that center of sphere circumscribed about ABCD lies on a line through midpoints of $AB$ and $CD$.
We are given an isosceles triangle $ABC$ such that $BC=a$ and $AB=BC=b$. The variable points $M\in (AC)$ and $N\in (AB)$ satisfy $a^2\cdot AM \cdot AN = b^2 \cdot BN \cdot CM$. The straight lines $BM$ and $CN$ intersect in $P$. Find the locus of the variable point $P$. [i]Dan Branzei[/i]
a) Let $a,b\in R$, $a <b$. Prove that $x \in (a,b)$ if and only if there exists $\lambda \in (0,1)$ such that $x=\lambda a +(1-\lambda)b$. b) If the function $f: R \to R$ has the property: $$f (\lambda x+(1-\lambda) y) < \lambda f(x) + (1-\lambda)f(y), \forall x,y \in R, x\ne y, \forall \lambda \in (0,1), $$ prove that one cannot find four points on the function’s graph that are the vertices of a parallelogram
Let $ABCD$ be a parallelogram. Point $E$ lies on segment $CD$ such that \[2\angle AEB=\angle ADB+\angle ACB,\] and point $F$ lies on segment $BC$ such that \[2\angle DFA=\angle DCA+\angle DBA.\] Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$. [i]Merlijn Staps[/i]
Let $ABCD$ be a parallelogram. Let $\omega_1$ be the circle passing through $D$ tangent to $AB$ at $A$. Let $\omega_2$ be the circle passing through $A$ tangent to $CD$ at $D$. The tangents from $B$ to $\omega_1$ touch it at $A$ and $P$. The tangents from $C$ to $\omega_2$ touch it at $D$ and $Q$. Lines $AP$ and $DQ$ intersect at $X$. The perpendicular bisector of $BC$ intersects $AD$ at $R$. Show that the circumcircles of triangles $\triangle PQX$, $\triangle BCR$ are concentric.
Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF \equal{} EG \equal{} EC$. Prove that $ \ell$ is the bisector of angle $ DAB$. [i]Author: Charles Leytem, Luxembourg[/i]
Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.
In parallelogram $ABCD$, $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}$. ['''Note:''' ''Both pictures represent the same parallelogram.''] If $DC=12$, $EB=4$, and $DE=6$, then $DF=$ [asy] unitsize(12); pair A,B,C,D,P,Q,W,X,Y,Z; A = (0,0); B = (12,0); C = (20,6); D = (8,6); W = (18,0); X = (30,0); Y = (38,6); Z = (26,6); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); P = (8,0); Q = (758/25,6/25); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot(P); dot(Q); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); draw(D--P); draw(Z--Q); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",P,S); label("$A$",W,SW); label("$B$",X,S); label("$C$",Y,NE); label("$D$",Z,NW); label("$F$",Q,E); [/asy] $\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$
Parallelogram $JHMT$ satisfies $JH=11$ and $HM=6,$ and point $P$ lies on $\overline{MT}$ such that $JP$ is an altitude of $JHMT.$ The circumcircles of $\triangle{HMP}$ and $\triangle{JMT}$ intersect at the point $Q\neq M.$ Let $A$ be the point lying on $\overline{JH}$ and the circumcircle of $\triangle{JMT}.$ If $MQ=10,$ then the perimeter of $\triangle{JAM}$ can be expressed in the form $\sqrt{a}+\tfrac{b}{c},$ where $a, \ b,$ and $c$ are positive integers, $a$ is not divisible by the square of any prime, and $b$ and $c$ are relatively prime. Find $a+b+c.$
A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.
$M$ is midpoint of side $BC$ of triangle $ABC$, and $I$ is incenter of triangle $ABC$, and $T$ is midpoint of arc $BC$, that does not contain $A$. Prove that \[\cos B+\cos C=1\Longleftrightarrow MI=MT\]