Let $P$ be an arbitrary interior point of $\triangle ABC$, and $AP$, $BP$, $CP$ intersect $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Suppose that $M$ be the midpoint of $BC$, $\odot(AEF)$ and $\odot(ABC)$ intersect at $S$, $SD$ intersects $\odot(ABC)$ at $X$, and $XM$ intersects $\odot(ABC)$ at $Y$. Show that $AY$ is tangent to $\odot(AEF)$.
Problem
Source: IMOC 2021 G8
Tags: geometry, tangent
MP8148
12.08.2021 05:05
We want to show that the spiral sim at $S$ sending $\overline{FE}$ to $\overline{BC}$ also sends $A$ to $Y$. By ratio lemma and spiral sim we have $$\frac{YB}{YC} = \frac{XC}{XB} = \frac{DC}{DB} \cdot \frac{SB}{SC} = \frac{DC}{DB} \cdot \frac{FB}{EC} = \frac{AF}{AE}$$where the last step follows by ceva. Since $\angle FAE = \angle BAC = \angle BYC$, we may conclude that $\triangle AFE \sim \triangle YBC$.
hsiangshen
25.08.2021 17:46
Let $EF \cap BC=G$. Notice that $S$ is the miquel point of complete quadrilateral $AB,BC,CA,EF$ Via some angle chasing we can easily conclude that the intersection of the line passing $A$ which is parallel to $EF$ and the line $GS$ lies on $\odot(ABC)$. Then by cross ratio:$$(B,C;D,G)=-1=(B,C;X,GS\cap\odot(ABC))=(B,C;M,Y(GS\cap\odot(ABC))\cap BC)$$$$\implies Y(GS\cap\odot(ABC))\parallel BC$$The rest of the problem can be finished by angle chasing. $\quad\blacksquare$
Amiralizakeri2007
18.08.2024 18:24
Define $f(X)=\frac{XB}{XC}$. $$f(Y)=\frac{1}{f(X)}=\frac{f(D)}{f(S)}=\frac{AE}{AF}$$Thus $\triangle YBC \sim \triangle AEF$ and we are done.