Two circles $O_1$ and $O_2$ intersect at $A,B$. Diameter $AC$ of $\odot O_1$ intersects $\odot O_2$ at $E$, Diameter $AD$ of $\odot O_2$ intersects $\odot O_1$ at $F$. $CF$ intersects $O_2$ at $H$, $DE$ intersects $O_1$ at $G,H$. $GH\cap O_1=P$. Prove that $PH=PK$.
Problem
Source: 2019 China North MO, Problem 2
Tags: geometry
math_pi_rate
21.02.2020 22:39
Note that $\triangle BFE$ is simply the orthic triangle of $\triangle ACD$. Let $CF \cap DE=T$, i.e. $T$ is the orthocenter of $\triangle ACD$, and suppose $CF$ meets $\odot (O_2)$ at $L$ apart from $H$. Then $$TK \cdot TG=TC \cdot TF=TE \cdot TD=TL \cdot TH \Rightarrow GHKL \text{ is cyclic}$$And, since $A$ lies on the perpendicular bisectors of $GK$ and $HL$, so we get that $A$ is the center of $\odot (GHKL)$. Thus, we have $$\angle APG=\angle AKG=90^{\circ}-\angle KAE=90^{\circ}-\angle KHP \Rightarrow AP \perp KH$$But, since $AK=AH$, so we get that $P$ also lies on perpendicular bisector of $KH$, or equivalently, $PK=PH$.
Rat09-kar_1729
27.10.2020 18:43
math_pi_rate wrote: Note that $\triangle BFE$ is simply the orthic triangle of $\triangle ACD$. Let $CF \cap DE=T$, i.e. $T$ is the orthocenter of $\triangle ACD$, and suppose $CF$ meets $\odot (O_2)$ at $L$ apart from $H$. Then $$TK \cdot TG=TC \cdot TF=TE \cdot TD=TL \cdot TH \Rightarrow GHKL \text{ is cyclic}$$And, since $A$ lies on the perpendicular bisectors of $GK$ and $HL$, so we get that $A$ is the center of $\odot (GHKL)$. Thus, we have $$\angle APG=\angle AKG=90^{\circ}-\angle KAE=90^{\circ}-\angle KHP \Rightarrow AP \perp KH$$But, since $AK=AH$, so we get that $P$ also lies on perpendicular bisector of $KH$, or equivalently, $PK=PH$. Can you please explain why $90-\angle KAE= 90-\angle KHP$ ?
rafaello
19.01.2021 14:21
Here is my approach. Let $J=CH\cap (O_2)$. Let $X=DE\cap CF$. Let $L$ the midpoint of $KH$ and let $P'=AL\cap GH$. We claim that $P=P'$. Let $B'=AX\cap CD$, since $CF\perp AD$, $DE\perp AC$, then $X$ is the orthocentre of $ACD$ and then $AX\perp CD$, thus $B'=B$. By PoP, $$XK\cdot XG=XA\cdot XB=XJ\cdot XH\implies GJKH \text{ is cyclic}$$Note that centre of $(GJKH)$ is $A$ since $A$ lies on perpendicular bisectors of $JH$ and $KG$. Thus, now by homothety (or just angle chase) with centre $K$ taking $GH$ to $EL$, we have $AELK$ being cyclic. Since $EL\parallel GH$, we have $$\measuredangle GP'K=\measuredangle HP'K=2\measuredangle HP'L=2\measuredangle ELA=2\measuredangle EKA=2\measuredangle GKA=\measuredangle GAK.$$Thus $P'$ lies on $(GAKBC)$, therefore $P=P'$, we are done.