Found problems: 87
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that
\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]
Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Let $ABC$ be a triangle with incenter $I$. The line through $I$, perpendicular to $AI$, intersects the circumcircle of $ABC$ at points $P$ and $Q$. It turns out there exists a point $T$ on the side $BC$ such that $AB + BT = AC + CT$ and $AT^2 = AB \cdot AC$. Determine all possible values of the ratio $IP/IQ$.
Determine the poles of the inversions that transform four collienar points $A,B, C, D$, aligned in this order, at four points $A' $, $B' $, $C'$ , $D'$ that are vertices of a rectangle, and such that $A'$ and $C'$ are opposite vertices.
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]
In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.
[i]Proposed by Kada Williams[/i]
The circle $\Omega$ with center $I_A$, is the $A$-excircle of triangle $ABC$. Which is tangent to $AB,AC$ at $F,E$ respectivly. Point $D$ is the reflection of $A$ through $I_AB$. Lines $DI_A$ and $EF$ meet at $K$. Prove that ,circumcenter of $DKE$ , midpoint of $BC$ and $I_A$ are collinear.
Two circles $\gamma $ and $\gamma'$ cross one another at points $A$ and $B$ . The tangent to $\gamma'$ at $A$ meets $\gamma$ again at $C$ , the tangent to $\gamma$ at $A$ meets $\gamma'$ again at $C'$ , and the line $CC'$ separates the points $A$ and $B$ . Let $\Gamma$ be the circle externally tangent to $\gamma$ , externally tangent to $\gamma'$ , tangent to the line $CC'$, and lying on the same side of $CC'$ as $B$ . Show that the circles $\gamma$ and $\gamma'$ intercept equal segments on one of the tangents to $\Gamma$ through $A$ .
Transform by inversion two concentric and coplanar circles into two equal.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that:
$\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$
Let $H $ be the orthocenter of $ABC $ ,$AB\neq AC $ ,and let $F $ be a point on circumcircle of $ABC $ such that $\angle AFH=90^{\circ} $.$K $ is the symmetric point of $H $ wrt $B $.Let $P $ be a point such that $\angle PHB=\angle PBC=90^{\circ} $,and $Q $ is the foot of $B $ to $CP $.Prove that $HQ $ is tangent to tge circumcircle of $FHK $.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
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.
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]
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]
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]
$\textbf{(A) }\dfrac{286}{35}\qquad\textbf{(B) }\dfrac{583}{70}\qquad\textbf{(C) }\dfrac{715}{73}\qquad\textbf{(D) }\dfrac{143}{14}\qquad\textbf{(E) }\dfrac{1573}{146}$
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 triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$.
Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.
[hide="comment"]
[i]Edited by Orl.[/i]
[/hide]
Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Two circles are internally tangent at a point $A$. Let $C$ be a point on the smaller circle other than $A$. The tangent line to the smaller circle at $C$ meets the bigger circle at $D$ and $E$; and the line $AC$ meets the bigger circle at $A$ and $P$. Show that the line $PE$ is tangent to the circle through $A$, $C$, and $E$.
Consider a triangle $ABC$ with altitudes $AD, BE$, and $CF$, and orthocenter $H$. Let the perpendicular line from $H$ to $EF$ intersects $EF, AB$ and $AC$ at $P, T$ and $L$, respectively. Point $K$ lies on the side $BC$ such that $BD=KC$. Let $\omega$ be a circle that passes through $H$ and $P$, that is tangent to $AH$. Prove that circumcircle of triangle $ATL$ and $\omega$ are tangent, and $KH$ passes through the tangency point.