Let $\Omega$ be the circumcircle of triangle $ABC$. Suppose that $X$ is a point on the segment $AB$ with $XB=XC$, and the angle bisector of $\angle BAC$ intersects $BC$ and $\Omega$ at $D$, $M$, respectively. If $P$ is a point on $BC$ such that $AP$ is tangent to $\Omega$ and $Q$ is a point on $DX$ such that $CQ$ is tangent to $\Omega$, show that $AB$, $CM$, $PQ$ are concurrent.
Problem
Source: IMOC 2021 G6
Tags: geometry, circumcircle, concurrent
rafaello
18.08.2021 17:55
Let $E=(ABC)\cap PM$. Let $R=(ABC)\cap DE$. Note that $PD^2=PA^2=PE\cdot PM$, hence $$\measuredangle BDE=\measuredangle DME=\measuredangle AME=\measuredangle ARD\implies AR\parallel BC.$$Note that $X$ lies on $RC$ as $ARBC$ is an isosceles trapezoid. Claim. $Q$ lies on $AE$. Proof. By Pascal's theorem on $CCBAER$, we get that $CC\cap AE,D$ and $X$ are collinear. Thus, $Q$ lies on $AE$. $\square$ Now, Pascal's theorem on $CCBAEM$ implies that $Q,P$ and $AB\cap MC$ are collinear. We are done. [asy][asy] import olympiad;import geometry; size(9cm);defaultpen(fontsize(10pt)); pair A,B,C,O,P,X,D,M,F,E,Q,R; O=(0,0);A=dir(110);B=dir(200);C=dir(340);path w=circumcircle(A,B,C); P=intersectionpoint(line(B,C),perpendicular(A,line(A,O)));M=intersectionpoints(w,100incenter(A,B,C)-99A--A)[1]; D=extension(A,M,B,C);X=extension(A,B,O,M);F=extension(A,B,M,C);E=intersectionpoints(w,P--M)[0];Q=extension(A,E,X,D); R=extension(D,E,X,C); draw(A--B--C--cycle,orange);draw(w,heavyblue);draw(C--Q,heavygreen);draw(A--P,heavyblue);draw(A--Q,heavygreen);draw(P--M,heavyblue); draw(Q--X,heavygreen);draw(A--M,heavyblue);draw(X--A,orange);draw(B--P,orange); draw(circumcircle(D,E,M),heavyblue);draw(X--C,orange);draw(R--E,heavyblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$M$",M,dir(M)); dot("$D$",D,dir(D)); dot("$X$",X,dir(X)); dot("$E$",E,dir(E)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); [/asy][/asy]
hsiangshen
25.08.2021 18:58
Bary finishes it. We use barycentric coordinates and take $ABC$ as the reference triangle. Here are some well-known points and eqations: $$P=(0:b^2:-c^2),M=(-a^2:b(b+c):c(b+c)),AA:b^2z+c^2y=0,CC:b^2x+a^2y=0,\text{Perpendicular bisector of BC}:(b^2-c^2)x+a^2y-a^2z=0$$$\implies AB\cap CM:=R=(-a^2:b(b+c):0)$ Let $PR\ \cap CC=(a^2:-b^2:t)$, easy to compute $t=\dfrac{-c^3}{b}$. $X\in(b^2-c^2)x+a^2y-a^2z=0\implies X=(a^2:c^2-b^2:0)$. Finally just expand $$\begin{vmatrix} a^2 &&-b^2&&\frac{-c^2}{b}\\0&&b&&c\\a^2&&c^2-b^2&&0\end{vmatrix}=0,\text{as desired.} \quad \blacksquare$$
MatBoy-123
12.10.2021 06:10
Just applications of $2$ theorems.. Let $AM \cap BB = L$ , also we have $\overline{A-D-M-L}$ are collinear, now as $\triangle BCQ$ and $\triangle AML$ are perspective, (their perspector is just line $AM$), so from desargues theorem $AB$ , $MC$ and $LQ$ are concurrent ( say at point $T$) . Now from Pascal theorem on the hexagon $BAAMCB$ , we get $\overline{T- P- L}$ are collinear, hence $P$ lies on $TQ$ , and we are done. $\blacksquare$