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

Consider quadrilateral $ABCD $ inscribed in circle $\omega $. $P\equiv AC\cap BD$. $E$, $F$ lie on sides $AB$, $CD$ respectively such that $\hat {APE}=\hat {DPF} $. Circles $\omega_1$, $\omega_2$ are tangent to $\omega$ at $X $, $Y $ respectively and also both tangent to the circumcircle of $\triangle PEF $ at $P $. Prove that: $$\frac {EX}{EY}=\frac {FX}{FY} $$ [i]Proposed by Ali Zamani [/i]
Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$
Let $ABC$ be a triangle with circumcenter $O$. Let $P$ be a point inside $ABC$, so let the points $D, E, F$ be on $BC, AC, AB$ respectively so that the Miquel point of $DEF$ with respect to $ABC$ is $P$. Let the reflections of $D, E, F$ over the midpoints of the sides that they lie on be $R, S, T$. Let the Miquel point of $RST$ with respect to the triangle $ABC$ be $Q$. Show that $OP = OQ$. [i]Proposed by Yang Liu[/i]
On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear.
Let $\Omega$ and $O$ be the circumcircle and the circumcenter of an acute triangle $ABC$ $(\overline{AB} < \overline{AC})$. Define $D,E(\neq A)$ be the points such that ray $AO$ intersects $BC$ and $\Omega$. Let the line passing through $D$ and perpendicular to $AB$ intersects $AC$ at $P$ and define $Q$ similarly. Tangents to $\Omega$ on $A,E$ intersects $BC$ at $X,Y$. Prove that $X,Y,P,Q$ lie on a circle.
Let $\Omega$ and $O$ be the circumcircle of acute triangle $ABC$ and its center, respectively. $M\ne O$ is an arbitrary point in the interior of $ABC$ such that $AM$, $BM$, and $CM$ intersect $\Omega$ at $A_{1}$, $B_{1}$, and $C_{1}$, respectiuvely. Let $A_{2}$, $B_{2}$, and $C_{2}$ be the circumcenters of $MBC$, $MCA$, and $MAB$, respectively. It is to be proven that $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C{2}$ concur.
Let $A, B, C$ and $D$ be four different points lying on a common circle in this order. Assume that the line segment $AB$ is the (only) longest side of the inscribed quadrilateral $ABCD$. Prove that the inequality $AB + BD > AC + CD$ holds. [i](Proposed by Karl Czakler)[/i]
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ O(ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.
(a) Triangles $A_1B_1C_1$ and $A_2B_2C_2$ are inscribed into triangle $ABC$ so that $C_1A_1 \perp BC$, $A_1B_1 \perp CA$, $B_1C_1 \perp AB$, $B_2A_2 \perp BC$, $C_2B_2 \perp CA$, $A_2C_2 \perp AB$. Prove that these triangles are equal. (b) Points $A_1$, $B_1$, $C_1$, $A_2$, $B_2$, $C_2$ lie inside a triangle $ABC$ so that $A_1$ is on segment $AB_1$, $B_1$ is on segment $BC_1$, $C_1$ is on segment $CA_1$, $A_2$ is on segment $AC_2$, $B_2$ is on segment $BA_2$, $C_2$ is on segment $CB_2$, and the angles $BAA_1$, $CBB_2$, $ACC_1$, $CAA_2$, $ABB_2$, $BCC_2$ are equal. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are equal.
In triangle $\triangle ABC$ points $M,N$ lie on $BC$ st : $\angle BAM= \angle MAN= \angle NAC$ . Points $P,Q$ are on the angle bisector of $BAC$, on the same side of $BC$ as A , st : $$\frac{1}{3} \angle BAC = \frac{1}{2} \angle BPC = \angle BQC$$ Let $E = AM \cap CQ$ and $F = AN \cap BQ$ . Prove that the common tangents to $(EPF), (EQF)$ and the circumcircle of $\triangle ABC$ , are concurrent.
Prove that all acute-angled triangles with the equal altitudes $h_c$ and the equal angles $\gamma$ have orthic triangles with same perimeters.
The incircle of triangle $ABC$ touches $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. The perpendicular from the incenter $I$ to the median from vertex $C$ meets the line $A_1B_1$ in point $K$. Prove that $CK$ is parallel to $AB$.
Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]
$ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent. [i]Proposed by Yang Liu[/i]
In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.
Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]
Let $ABC$ be an acute triangle with orthocenter $H$, and let $P$ be a point on the nine-point circle of $ABC$. Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$, respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$, respectively. Show that as $P$ varies along the nine-point circle of $ABC$, the line $QR$ passes through a fixed point. [i]Proposed by Brandon Wang[/i]
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Given is a triangle $ABC$ with $AB>AC$. Its incircle touches $AB, AC$ at $D, E$, respectively. Let $CD$ meet the incircle at $K$ and $L$ is the foot of the perpendicular from $A$ to $CK$. If $M$ is the midpoint of $DE$ and $H$ is the orthocenter of $\triangle KML$, prove that $\angle AHK=90^{o}$. [i]Proposed by Dominik Burek[/i]
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]
Given a cyclic quadrilateral $ABCD$ such that the tangents at points $B$ and $D$ to its circumcircle $k$ intersect at the line $AC$. The points $E$ and $F$ lie on the circle $k$ so that the lines $AC, DE$ and $BF$ parallel. Let $M$ be the intersection of the lines $BE$ and $DF$. If $P, Q$ and $R$ are the feet of the altitides of the triangle $ABC$, prove that the points $P, Q, R$ and $M$ lie on the same circle
The tangent at $C$ to $\Omega$, the circumcircle of scalene triangle $ABC$ intersects $AB$ at $D$. Through point $D$, a line is drawn that intersects segments $AC$ and $BC$ at $K$ and $L$ respectively. On the segment $AB$ points $M$ and $N$ are marked such that $AC \parallel NL$ and $BC \parallel KM$. Lines $NL$ and $KM$ intersect at point $P$ lying inside the triangle $ABC$. Let $\omega$ be the circumcircle of $MNP$. Suppose $CP$ intersects $\omega$ again at $Q$. Show that $DQ$ is tangent to $\omega$.
In $\triangle ABC$, a point $D$ lies on line $BC$. The circumcircle of $ABD$ meets $AC$ at $F$ (other than $A$), and the circumcircle of $ADC$ meets $AB$ at $E$ (other than $A$). Prove that as $D$ varies, the circumcircle of $AEF$ always passes through a fixed point other than $A$, and that this point lies on the median from $A$ to $BC$. [i]Proposed by Allen Liu[/i]
Let $AD,BF,CE$ be altitudes of triangle $ABC$.$Q$ is a point on $EF$ such that $QF=DE$ and $F$ is between $E,Q$.$P$ is a point on $EF$ such that $EP=DF$ and $E$ is between $P,F$.Perpendicular bisector of $DQ$ intersect with $AB$ at $X$ and perpendicular bisector of $DP$ intersect with $AC$ at $Y$.Prove that midpoint of $BC$ lies on $XY$.