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

Triangle $ABC$ is given. Let $O$ be it's circumcenter. Let $I$ be the center of it's incircle.The external angle bisector of $A$ meet $BC$ at $D$. And $I_A$ is the $A$-excenter . The point $K$ is chosen on the line $AI$ such that $AK=2AI$ and $A$ is closer to $K$ than $I$. If the segment $DF$ is the diameter of the circumcircle of triangle $DKI_A$, then prove $OF=3OI$.
Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$.
Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO, AC,$ and $AB$ at $W, X,$ and $Y,$ respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \ne A$. Prove that if $W \ne D$ and $OW = OD,$ then $DZ$ is tangent to the circle $AXY.$
Let $ABCDE$ be a convex pentagon with $CD= DE$ and $\angle EDC \ne 2 \cdot \angle ADB$. Suppose that a point $P$ is located in the interior of the pentagon such that $AP =AE$ and $BP= BC$. Prove that $P$ lies on the diagonal $CE$ if and only if area $(BCD)$ + area $(ADE)$ = area $(ABD)$ + area $(ABP)$. (Hungary)
Consider the two triangles $ ABC$ and $ PQR$ shown below. In triangle $ ABC, \angle ADB \equal{} \angle BDC \equal{} \angle CDA \equal{} 120^\circ$. Prove that $ x\equal{}u\plus{}v\plus{}w$. [asy]unitsize(7mm); defaultpen(linewidth(.7pt)+fontsize(10pt)); pair C=(0,0), B=4*dir(5); pair A=intersectionpoints(Circle(C,5), Circle(B,6))[0]; pair Oc=scale(sqrt(3)/3)*rotate(30)*(B-A)+A; pair Ob=scale(sqrt(3)/3)*rotate(30)*(A-C)+C; pair D=intersectionpoints(Circle(Ob,length(Ob-C)), Circle(Oc,length(Oc-B)))[1]; real s=length(A-D)+length(B-D)+length(C-D); pair P=(6,0), Q=P+(s,0), R=rotate(60)*(s,0)+P; pair M=intersectionpoints(Circle(P,length(B-C)), Circle(Q,length(A-C)))[0]; draw(A--B--C--A--D--B); draw(D--C); label("$B$",B,SE); label("$C$",C,SW); label("$A$",A,N); label("$D$",D,NE); label("$a$",midpoint(B--C),S); label("$b$",midpoint(A--C),WNW); label("$c$",midpoint(A--B),NE); label("$u$",midpoint(A--D),E); label("$v$",midpoint(B--D),N); label("$w$",midpoint(C--D),NNW); draw(P--Q--R--P--M--Q); draw(M--R); label("$P$",P,SW); label("$Q$",Q,SE); label("$R$",R,N); label("$M$",M,NW); label("$x$",midpoint(P--R),NW); label("$x$",midpoint(P--Q),S); label("$x$",midpoint(Q--R),NE); label("$c$",midpoint(R--M),ESE); label("$a$",midpoint(P--M),NW); label("$b$",midpoint(Q--M),NE);[/asy]
Consider a cyclic quadrilateral $ABCD$ in which $BC = CD$ and $AB < AD$. Let $E$ be a point on the side $AD$ and $F$ a point on the line $BC$ such that $AE = AB = AF$. Prove that $EF \parallel BD$.
Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]
Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.
Given is a cyclic quadrilateral $ABCD$. The point $L_a$ lies in the interior of $BCD$ and is such that its distances to the sides of this triangle are proportional to the lengths of corresponding sides. The points $L_b, L_c$, and $L_d$ are defined analogously. Given that the quadrilateral $L_aL_bL_cL_d$ is cyclic, prove that the quadrilateral $ABCD$ has two parallel sides. (N. Beluhov)
In the quadrilateral $ABCD$ the angles $B{}$ and $D{}$ are straight. The lines $AB{}$ and $DC{}$ intersect at $E$ and the lines $AD$ and $BC$ intersect at $F{}.$ The line passing through $B{}$ parallel to $C{}$D intersects the circumscribed circle $\omega$ of $ABF{}$ at $K{}$ and the segment $KE{}$ intersects $\omega$ at $P{}.$ Prove that the line $AP$ divides the segment $CE$ in half.
Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and $AD$ intersect the circumcircle of $ABC$ for the second time at $E$. Let $P$ be the point symmetric to the point $E$ with respect to the point $D$ and $Q$ be the point of intersection of the lines $CP$ and $AB$. Prove that if $A,C,D,Q$ are concyclic, then the lines $BP$ and $AC$ are perpendicular.
Let $ABCD$ be a convex cyclic quadrilateral with diagonals $AC$ and $BD$. Each of the four vertixes are reflected across the diagonal on which the do not lie. (a) Investigate when the four points thus obtained lie on a straight line and give as simple an equivalent condition as possible to the cyclic quadrilateral $ABCD$ for it. (b) Show that in all other cases the four points thus obtained lie on one circle. (Theresia Eisenkölbl)
Let $ABC$ be a triangle and $D$ be a point inside triangle $ABC$. $\Gamma$ is the circumcircle of triangle $ABC$, and $DB$, $DC$ meet $\Gamma$ again at $E$, $F$ , respectively. $\Gamma_1$, $\Gamma_2$ are the circumcircles of triangle $ADE$ and $ADF$ respectively. Assume $X$ is on $\Gamma_2$ such that $BX$ is tangent to $\Gamma_2$. Let $BX$ meets $\Gamma$ again at $Z$. Prove that the line $CZ$ is tangent to $\Gamma_1$ . [i]Proposed by HakureiReimu[/i].
$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$.
Let $ABCD$ be a convex quadrilateral with $A,B,C,D$ concyclic. Assume $\angle ADC$ is acute and $\frac{AB}{BC}=\frac{DA}{CD}$. Let $\Gamma$ be a circle through $A$ and $D$, tangent to $AB$, and let $E$ be a point on $\Gamma$ and inside $ABCD$. Prove that $AE\perp EC$ if and only if $\frac{AE}{AB}-\frac{ED}{AD}=1$.
A trapezoid has side lengths as indicated in the figure (the sides with length $11$ and $36$ are parallel). Calculate the area of the trapezoid.[img]https://1.bp.blogspot.com/-5PKrqDG37X4/XzcJtCyUv8I/AAAAAAAAMY0/tB0FObJUJdcTlAJc4n6YNEaVIDfQ91-eQCLcBGAsYHQ/s0/1995%2BMohr%2Bp1.png[/img]
9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$. ([i]A. Badzyan[/i])
Let $A$ be a point in the plane, and $\ell$ a line not passing through $A$. Evan does not have a straightedge, but instead has a special compass which has the ability to draw a circle through three distinct noncollinear points. (The center of the circle is [i]not[/i] marked in this process.) Additionally, Evan can mark the intersections between two objects drawn, and can mark an arbitrary point on a given object or on the plane. (i) Can Evan construct* the reflection of $A$ over $\ell$? (ii) Can Evan construct the foot of the altitude from $A$ to $\ell$? *To construct a point, Evan must have an algorithm which marks the point in finitely many steps. [i]Proposed by Zack Chroman[/i]
Find distinct points $A, B, C,$ and $D$ in the plane such that the length of the segment $AB$ is an even integer, and the lengths of the segments $AC, AD, BC, BD,$ and $CD$ are all odd integers. In addition to stating the coordinates of the points and distances between points, please include a brief explanation of how you found the configuration of points and computed the distances.
Let $E$ be a common point of circles $\omega _1$ and $\omega _2$. Let $AB$ be a common tangent to these circles, and $CD$ be a line parallel to $AB$, such that $A$ and $C$ lie on $\omega _1$, $B$ and $D$ lie on $\omega _2$. The circles $ABE$ and $CDE$ meet for the second time at point $F$. Prove that $F$ bisects one of arcs $CD$ of circle $CDE$.
In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]
Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$.
Let $w$ be the circumcircle of non-isosceles acute triangle $ABC$. Tangent lines to $w$ in $A$ and $B$ intersect at point $S$. Let M be the midpoint of $AB$, and $H$ be the orthocenter of triangle $ABC$. The line $HA$ intersects lines $CM$ and $CS$ at points $M_a$ and $S_a$, respectively. The points $M_b$ and $S_b$ are defined analogously. Prove that $M_aS_b$ and $M_bS_a$ are the altitudes of triangle $M_aM_bH$.
In a triangle $\triangle ABC$ with orthocenter $H$, let $BH$ and $CH$ intersect $AC$ and $AB$ at $E$ and $F$, respectively. If the tangent line to the circumcircle of $\triangle ABC$ passing through $A$ intersects $BC$ at $P$, $M$ is the midpoint of $AH$, and $EF$ intersects $BC$ at $G$, then prove that $PM$ is parallel to $GH$. [i]Proposed by Sreejato Bhattacharya[/i]
Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle. [i]Proposed by Dimitris Kontogiannis, Greece[/i]