Found problems: 429
Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ are collinear.
In the figure of [url]http://www.artofproblemsolving.com/Forum/download/file.php?id=50643&mode=view[/url]
$\odot O_1$ and $\odot O_2$ intersect at two points $A$, $B$.
The extension of $O_1A$ meets $\odot O_2$ at $C$, and the extension of $O_2A$ meets $\odot O_1$ at $D$,
and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$.
If $DE \parallel O_1A$, prove that $DC \perp CO_2$.
Let $ABCD$ be a quadrilateral inscribed in circle $\omega$. Define $E = AA \cap CD$, $F = AA \cap BC$, $G = BE \cap \omega$, $H = BE \cap AD$, $I = DF \cap \omega$, and $J = DF \cap AB$. Prove that $GI$, $HJ$, and the $B$-symmedian are concurrent.
[i]Proposed by Robin Park[/i]
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Let $M$ be any point on the circumcircle of triangle $ABC$. Suppose the tangents from $M$ to the incircle meet $BC$ at two points $X_1$ and $X_2$. Prove that the circumcircle of triangle $MX_1X_2$ intersects the circumcircle of $ABC$ again at the tangency point of the $A$-mixtilinear incircle.
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.
(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.
(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$.
[i]James Lin[/i]
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other
Given an acute $\vartriangle ABC$ with incenter $I$. Let $I'$ be the symmetric point $I$ with respect to the midpoint of $B,C$ and $D$ is the foot of $A$. If $DI$ and the circumcircle of vartriangle $BI'C$ intersect at $T$ and $TI' $ intersects the circumcircle of $\vartriangle ATI$ at $X$. Furthermore, $E,F$ are tangent points of the incircle and $AB,AC, P$ is the another intersection of the circumcircles of $\vartriangle ABC, \vartriangle AEF$. Show that $AX \parallel PI$.
[img]https://3.bp.blogspot.com/-tj9A8HIR6Vw/XndLEPMRvnI/AAAAAAAALfk/2vw_pZbhpnkTKIc1BcKf4K7SNZ11vu4TACK4BGAYYCw/s1600/2018%2Bimoc%2Bg4.png[/img]
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]
Tow circles $C,D$ are exterior tangent to each other at point $P$. Point $A$ is in the circle $C$. We draw $2$ tangents $AM,AN$ from $A$ to the circle $D$ ($M,N$ are the tangency points.). The second meet points of $AM,AN$ with $C$ are $E,F$, respectively. Prove that $\frac{PE}{PF}=\frac{ME}{NF}$.
Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$.
[i]Proposed by David Stoner[/i]
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 $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$. Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$. Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$ intersect at a point on line $BC$.
A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\geq 1$, the circles in $\textstyle\bigcup_{j=0}^{k-1} L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\textstyle\bigcup_{j=0}^6 L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S}\dfrac1{\sqrt{r(C)}}?\]
[asy]
import olympiad;
size(350);
defaultpen(linewidth(0.7));
// define a bunch of arrays and starting points
pair[] coord = new pair[65];
int[] trav = {32,16,8,4,2,1};
coord[0] = (0,73^2); coord[64] = (2*73*70,70^2);
// draw the big circles and the bottom line
path arc1 = arc(coord[0],coord[0].y,260,360);
path arc2 = arc(coord[64],coord[64].y,175,280);
fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.78));
fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.78));
draw(arc1^^arc2);
draw((-930,0)--(70^2+73^2+850,0));
// We now apply the findCenter function 63 times to get
// the location of the centers of all 63 constructed circles.
// The complicated array setup ensures that all the circles
// will be taken in the right order
for(int i = 0;i<=5;i=i+1)
{
int skip = trav[i];
for(int k=skip;k<=64 - skip; k = k + 2*skip)
{
pair cent1 = coord[k-skip], cent2 = coord[k+skip];
real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);
real shiftx = cent1.x + sqrt(4*r1*rn);
coord[k] = (shiftx,rn);
}
// Draw the remaining 63 circles
}
for(int i=1;i<=63;i=i+1)
{
filldraw(circle(coord[i],coord[i].y),gray(0.78));
}[/asy]
$\textbf{(A) }\dfrac{286}{35}\qquad\textbf{(B) }\dfrac{583}{70}\qquad\textbf{(C) }\dfrac{715}{73}\qquad\textbf{(D) }\dfrac{143}{14}\qquad\textbf{(E) }\dfrac{1573}{146}$
Let $A,B,C$ be three points in that order on a line $\ell$ in the plane, and suppose $AB>BC$. Draw semicircles $\Gamma_1$ and $\Gamma_2$ respectively with $AB$ and $BC$ as diameters, both on the same side of $\ell$. Let the common tangent to $\Gamma_1$ and $\Gamma_2$ touch them respectively at $P$ and $Q$, $P\ne Q$. Let $D$ and $E$ be points on the segment $PQ$ such that the semicircle $\Gamma_3$ with $DE$ as diameter touches $\Gamma_2$ in $S$ and $\Gamma_1$ in $T$.
[list=1][*]Prove that $A,C,S,T$ are concyclic.
[*]Prove that $A,C,D,E$ are concyclic.[/list]
Two circles $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.
An isosceles triangle $ABC$ is inscribed in a circle with $\angle ACB = 90^o$ and $EF$ is a chord of the circle such that neither E nor $F$ coincide with $C$. Lines $CE$ and $CF$ meet $AB$ at $D$ and $G$ respectively. Prove that $|CE|\cdot |DG| = |EF| \cdot |CG|$.
Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.
[i]Proposed by Evan Chen, Taiwan[/i]
Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that
$AC*BD*DE*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$
Let $ABC$ be a scalene triangle with incenter $I$ and incircle $\omega$. Let the tangency points of $\omega$ to $BC,AC\text{ and } AB$ be $D,E,F$ respectively. Let the line $EF$ intersect the circumcircle of $ABC$ at the points $G, H$. Assume that $E$ lies between the points $F$ and $G$. Let $\Gamma$ be a circle that passes through $G$ and $H$ and that is tangent to $\omega$ at the point $M$ which lies on different semi-planes with $D$ with respect to the line $EF$. Let $\Gamma$ intersect $BC$ at points $K$ and $L$ and let the second intersection point of the circumcircle of $ABC$ and the circumcircle of $AKL$ be $N$. Prove that the intersection point of $NM$ and $AI$ lies on the circumcircle of $ABC$ if and only if the intersection point of $HB$ and $GC$ lies on $\Gamma$.
In scalene triangle $ABC$, let the feet of the perpendiculars from $A$ to $BC$, $B$ to $CA$, $C$ to $AB$ be $A_1, B_1, C_1$, respectively. Denote by $A_2$ the intersection of lines $BC$ and $B_1C_1$. Define $B_2$ and $C_2$ analogously. Let $D, E, F$ be the respective midpoints of sides $BC, CA, AB$. Show that the perpendiculars from $D$ to $AA_2$, $E$ to $BB_2$ and $F$ to $CC_2$ are concurrent.
A cyclic quadrilateral $ABCD$ is given. An arbitrary circle passing through $C$ and $D$ meets $AC,BC$ at points $X,Y$ respectively. Find the locus of common points of circles $CAY$ and $CBX$.