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 $H,K$ $\Longrightarrow$ $\angle THK=\angle TEK=\angle KMP$ $\Longrightarrow$ $H \in \odot(KMP).$


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