Given an acute triangle $ABC$ draw a triangle $PQR$ such that $AB=2PQ,BC=2QR,CA=2RP$, and the lines $PQ,QR,RP$ pass through $A,B,C$ respectively. Note $A,B,C,P,Q,R$ are distinct.
This is a problem from the Middle European Mathematical Olympiad 2013. I tried to locate the homothetic center mapping the triangles $ABC$ and $PQR$ into each other. I suspected it could be the incenter, but no conclusive evidence. It could be a variable point. But in general I cannot think anything of this problem. Can someone help? Thanks.
solution:
make high $CD \perp AB$,since it is acute triangle, $D$ will fall on $AB$. $EC \perp CB,$ circle C @$r=CB$, cross $EC$ at $E$, make $EF \perp EC$, extend $CA$, cross $ FE $ at $F$, circle B @$r=BA$, cross line $CB$ at $G$, connect $BF$, make $GH$//$BF$, cross $EF$ at $H$,connect $CH$, make it's midpoint $ M $, circle M@$r=MH$, cross $CG$ at $N$, connect $HN$, make it's midpoint $S$,,circle H @r=HS, cross circle $M$ at $T$, connect $CT$, circle C@$r=CB$, cross $CT$ at $B'$,circle C @$r=CA$, cross $CB$ at $H$, cross $CT$ at $U$,make $\angle B'CA$ bisector $CV$, cross circle $CA$ at $V$, circle V@$r=VH$, cross circle $CA$ at $A'$, connect $A'B'$, make $RB$//$A'B'$, cross $CT$ at $R$, connect $A'C$,make $AQ$//$A'C$, cross $CT$ at $Q$, cross $BR$ at $P, \triangle PQR $ is the solution.
to proof this method, put $ECB$ on $yOx$, then set $B(a,0),A(b\cos{C},b\sin{C})$
let $B'(a\cos{\alpha},b\sin{\alpha}),R(x_1,y_1),Q(x_2,y_2)$, note $x_1-x_2=\dfrac{a\cos{\alpha}}{2}, ab\sin{C}=ch_c,c^2=a^2+b^2-2ab\cos{C}$
more comments : this solution always work when high foot fall on the side. so it is mean for any triangle we can find the solution no matter if acute or not. the standard solution is conservative.