## Thứ Sáu, 9 tháng 1, 2015

### Solution for HSGS TST 2014 Round 1 Day 2

Problem. Let $ABCD$ be cyclic quarilateral. $M,N$ are midpoints of $CD,AB$. $P$ lies on side $CD$ such that $\frac{PD}{PC}=\frac{BD^2}{AC^2}$. $AC$ cuts $BD$ at $E$. $H$ is projection of $E$ on $PN$. Prove that circles $(HMP)$ and $(EDC)$ are tangent.

Solution 1 (By Telv Cohl). Let $O$ be the center of $\odot (ABCD)$ .
Let $X=AD \cap BC, Y=AB \cap CD, Z= \odot (ABE) \cap (CDE)$ .

Easy to see $E, Y, Z$ are collinear .

Since $\angle DZA=\angle DCA+\angle DBA=\angle DOA$ ,
so we get $A, D, Z, O$ are concyclic ,
hence $\angle OZY=\angle OZD+\angle DZY=\angle OAD+\angle DCA=90^{\circ}$ ,
so we get $O, M, N, Y, Z$ are lie on a circle with diameter $OY$ .
From Brokard theorem we get $O$ is the orthocenter of $\triangle EXY$ ,
so combine with $\angle OZY = 90^{\circ}$ we get $O, X, Z$ are collinear

Since $PC:PD=AC^2:BD^2=XC^2:XD^2$ ,
so we get $XP$ is $X-$ symmedian of $\triangle XCD$ ,
hence combine with $\triangle XAB \sim \triangle XCD$ we get $P \in XN$ .
Since $E, H, X, Z$ lie on a circle with diameter $XE$ ,
so we get $\angle XHZ=\angle XEZ=\angle YOZ=\angle YMZ$ .
ie. $Z \in \odot (HMP)$

Since $Z$ is the center of spiral similar of $AC \mapsto BD$ ,
so we get $PC:PD=AC^2:BD^2=ZC^2:ZD^2$ ,
hence $ZP$ is $Z-$ symmedian of $\triangle ZCD$ ,
so we get $\angle CZM=\angle PZD$ and $\odot (HMPZ)$ is tangent to $\odot (EDCZ)$ at $Z$ .

Solution 2 (By Tran Quang Hung). $AD$ cuts $BC$ at $F$. We have $\frac{PD}{PC}=\frac{BD^2}{AC^2}=\frac{FD^2}{FC^2}$ so $FP$ is symmedian of $FCD$ therefore $FP$ passes through $N$. $AB$ cuts $CD$ at $S$. By Brokard theorem $S,E,G$ are collinear and $SE$ is perpendicular to $OF$ at $G$. We have $\angle GMS=\angle GOS=\angle GEF=\angle FHG$ so $HGMP$ is cyclic. Tangent at $G$ of $(GCD)$ cuts $CD$ at $T$. Notice $\triangle GAC\sim\triangle GBD$ we have $\frac{TC}{TD}=\frac{GC^2}{GD^2}=\frac{AC^2}{BD^2}=\frac{PC}{PD}$ so $(PG,CD)=-1$ deduce $TG^2=TC.TD=TP.TM$ so $TG$ tangent to $(GPM)$. Thus,  circles $(HMP)$ and $(EDC)$ are tangent at $G$.

Solution 3 (By Luis González). Let $S \equiv AB \cap CD,$ $T \equiv AD \cap BC$ and $SE$ cuts $(O) \equiv \odot(ABCD)$ at $U,V.$ Since $SE,TE$ are the polars of $T,S$ WRT $(O),$ then $SE$ is perpendicular to $OT$ at $K$ and $(U,V,K,S)=-1$ $\Longrightarrow$ $SE \cdot SK=SU \cdot SV=SC \cdot SD$ $\Longrightarrow$ $K \in \odot(ECD)$ and likewise $K \in \odot(EAB)$ $\Longrightarrow$ $K$ is center of spiral similarity that swaps $AC$ and $BD$ $\Longrightarrow$ $KC^2:KD^2=AC^2:BD^2=PC:PD$ $\Longrightarrow$ $KP$ is K-symmedian of $\triangle KDC$ $\Longrightarrow$ $KM,KP$ are isogonals WRT $\angle CKD$ $\Longrightarrow$ $\odot(KMP)$ touches $\odot(KDC).$ Now, it's enough to show that $H \in \odot(KMP).$

If $L \equiv TE \cap CD,$ then $(D,C,L,S)=-1$ $\Longrightarrow$ $SL \cdot SM=SD \cdot SC=SE \cdot SK$ $\Longrightarrow$ $EKML$ is cyclic. But $TKHE$ is cyclic due to right angles at $H,K$ $\Longrightarrow$ $\angle THK=\angle TEK=\angle KMP$ $\Longrightarrow$ $H \in \odot(KMP).$

Reference.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=618908