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

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Found problems: 3585

The point $M$ is inside the convex quadrilateral $ABCD$, such that \[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \] Prove that $AB \cdot CM = BC \cdot MD$ and $BM \cdot AD = MA \cdot CD.$
In this figure the radius of the circle is equal to the altitude of the equilateral triangle $ABC$. The circle is made to roll along the side $AB$, remaining tangent to it at a variable point $T$ and intersecting lines $AC$ and $BC$ in variable points $M$ and $N$, respectively. Let $n$ be the number of degrees in arc $MTN$. Then $n$, for all permissible positions of the circle: $\textbf{(A) }\text{varies from }30^{\circ}\text{ to }90^{\circ}$ $\textbf{(B) }\text{varies from }30^{\circ}\text{ to }60^{\circ}$ $\textbf{(C) }\text{varies from }60^{\circ}\text{ to }90^{\circ}$ $\textbf{(D) }\text{remains constant at }30^{\circ}$ $\textbf{(E) }\text{remains constant at }60^{\circ}$ [asy] pair A = (0,0), B = (1,0), C = dir(60), T = (2/3,0); pair M = intersectionpoint(A--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)), N = intersectionpoint(B--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)); draw((0,0)--(1,0)--dir(60)--cycle); draw(Circle((2/3,sqrt(3)/2),sqrt(3)/2)); label("$A$",A,dir(210)); label("$B$",B,dir(-30)); label("$C$",C,dir(90)); label("$M$",M,dir(190)); label("$N$",N,dir(75)); label("$T$",T,dir(-90)); //Credit to bobthesmartypants for the diagram [/asy]
Let $ABCD$ be a trapezoid with bases $BC \parallel AD$, where $AD > BC$, and non-parallel legs $AB$ and $CD$. Let $M$ be the intersection of $AC$ and $BD$. Let $\Gamma_1$ be a circumference that passes through $M$ and is tangent to $AD$ at point $A$; let $\Gamma_2$ be a circumference that passes through $M$ and is tangent to $AD$ at point $D$. Let $S$ be the intersection of the lines $AB$ and $CD$, $X$ the intersection of $\Gamma_1$ with the line $AS$, $Y$ the intesection of $\Gamma_2$ with the line $DS$, and $O$ the circumcenter of triangle $ASD$. Show that $SO \perp XY$.
In triangle $ABC$, $\angle A=60^{\circ}$. The point $D$ changes on the segment $BC$. Let $O_1,O_2$ be the circumcenters of the triangles $\Delta ABD,\Delta ACD$, respectively. Let $M$ be the meet point of $BO_1,CO_2$ and let $N$ be the circumcenter of $\Delta DO_1O_2$. Prove that, by changing $D$ on $BC$, the line $MN$ passes through a constant point.
Let $B$ and $D$ be points on segments $[AE]$ and $[AF]$ respectively. Excircles of triangles $ABF$ and $ADE$ touching sides $BF$ and $DE$ is the same, and its center is $I$. $BF$ and $DE$ intersects at $C$. Let $P_1, P_2, P_3, P_4, Q_1, Q_2, Q_3, Q_4$ be the circumcenters of triangles $IAB, IBC, ICD, IDA, IAE, IEC, ICF, IFA$ respectively. [b]a) [/b] Show that points $P_1, P_2, P_3, P_4$ concylic and points $Q_1, Q_2, Q_3, Q_4$ concylic. [b]b) [/b] Denote centers of theese circles as $O_1$ and $O_2$. Prove that $O_1, O_2$ and $I$ are collinear.
Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.
Let $ ABC$ be an acute angled triangle and let $ H$ be its ortho centre. Let $ h_{max}$ denote the largest altitude of the triangle $ ABC$. Prove that: $AH \plus{} BH \plus{} CH\leq2h_{max}$
Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of segments $BC$ and $AD$, respectively. Prove that the following three lines are concurrent: line $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$.
Let triangle $ABC$ be an acute scalene triangle, and denote $D,E,F$ by the midpoints of $BC,CA,AB$, respectively. Let the circumcircle of $DEF$ be $O_1$, and its center be $N$. Let the circumcircle of $BCN$ be $O_2$. $O_1$ and $O_2$ meet at two points $P, Q$. $O_2$ meets $AB$ at point $K(\neq B)$ and meets $AC$ at point $L(\neq C)$. Show that the three lines $EF,PQ,KL$ are concurrent.
Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$. Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$, and $F$ be the foot of perpendicular from $Q$ to the line $AB$. Let $M$ be intersection of the lines $DE$ and $AB$. Prove that $MP \perp CF$
Let $ABCD$ be a cyclic quadrilateral with $AB$ and $CD$ not parallel. Let $M$ be the midpoint of $CD.$ Let $P$ be a point inside $ABCD$ such that $P A = P B = CM.$ Prove that $AB, CD$ and the perpendicular bisector of $MP$ are concurrent.
In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$.
Let $k$ be the circumcircle of $\triangle ABC$ and let $k_a$ be A-excircle .Let the two common tangents of $k,k_a$ cut $BC$ in $P,Q$.Prove that $\measuredangle PAB=\measuredangle CAQ$.
Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$. [i]Proposed by Davood Vakili[/i]
In a cyclic quadrilateral $ABCD$, the diagonals intersect at $E$. $F$ and $G$ are on chord $AC$ and chord $BD$ respectively such that $AF = BE$ and $DG = CE$. Prove that, $A, G, F, D$ lie on the same circle.
Let $ A_1A_2 \ldots A_n$ be a convex polygon, $ n \geq 4.$ Prove that $ A_1A_2 \ldots A_n$ is cyclic if and only if to each vertex $ A_j$ one can assign a pair $ (b_j, c_j)$ of real numbers, $ j = 1, 2, \ldots, n,$ so that $ A_iA_j = b_jc_i - b_ic_j$ for all $ i, j$ with $ 1 \leq i < j \leq n.$
In acute triangle $ABC$, let $\odot O$ be its circumcircle, $\odot I$ be its incircle. Tangents at $B,C$ to $\odot O$ meet at $L$, $\odot I$ touches $BC$ at $D$. $AY$ is perpendicular to $BC$ at $Y$, $AO$ meets $BC$ at $X$, and $OI$ meets $\odot O$ at $P,Q$. Prove that $P,Q,X,Y$ are concyclic if and only if $A,D,L$ are collinear.
Let $ABC$ be a triangle with $AB=20$ and $AC=18$. $E$ is on segment $AC$ and $F$ is on segment $AB$ such that $AE=AF=8$. Let $BE$ and $CF$ intersect at $G$. Given that $AEGF$ is cyclic, then $BC=m\sqrt{n}$ for positive integers $m$ and $n$ such that $n$ is not divisible by the square of any prime. Compute $100m+n$. [i] Proposed by James Lin[/i]
In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic.
A rectangle $ABCD$ is inscribed in a circle with centre $O$. The exterior bisectors of $\angle ABD$ and $\angle ADB$ intersect at $P$; those of $\angle DAB$ and $\angle DBA$ intersect at $Q$; those of $\angle ACD$ and $\angle ADC$ intersect at $R$; and those of $\angle DAC$ and $\angle DCA$ intersect at $S$. Prove that $P,Q,R$, and $S$ are concyclic.
A chord $ST$ of constant length slides around a semicircle with diameter $AB$. $M$ is the midpoint of $ST$ and $P$ is the foot of the perpendicular from $S$ to $AB$. Prove that $\angle SPM$ is constant for all positions of $ST$.
$AB$ is a chord of $O$ and $AB$ is not a diameter of $O$. The tangent lines to $O$ at $A$ and $B$ meet at $C$. Let $M$ and $N$ be the midpoint of the segments $AC$ and $BC$, respectively. A circle passing through $C$ and tangent to $O$ meets line $MN$ at $P$ and $Q$. Prove that $\angle PCQ = \angle CAB$.
There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that \[ \angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ} \] if and only if the diagonals $ AC$ and $ BD$ are perpendicular. [i]Proposed by Dusan Djukic, Serbia[/i]
The circles $\mathcal{K}_1,\mathcal{K}_2$ and $\mathcal{K}_3$ are pairwise externally tangent to each other; the point of tangency betwwen $\mathcal{K}_1$ and $\mathcal{K}_2$ is $T$. One of the external common tangents of $\mathcal{K}_1$ and $\mathcal{K}_2$ meets $\mathcal{K}_3$ at points $P$ and $Q$. Prove that the internal common tangent of $\mathcal{K}_1$ and $\mathcal{K}_2$ bisects the arc $PQ$ of $\mathcal{K}_3$ which is closer to $T$.
Let $ABCD$ be a cyclic quadrilateral with its diagonals intersecting at $E$. Let $M$ be the midpoint of $AB$. Suppose that $ME$ is perpendicular to $CD$. Show that either $AC$ is perpendicular to $BD$, or $AB$ is parallel to $CD$.