AoPS - Problem

Problem

Source: 2022 China TST, Test 2, P2

Tags: geometry, Simson line, Nine Point Circle

JustPostChinaTST

28.03.2022 12:23

Given a non-right triangle $ABC$ with $BC>AC>AB$ . Two points $P_1 \neq P_2$ on the plane satisfy that, for $i=1,2$ , if $AP_i, BP_i$ and $CP_i$ intersect the circumcircle of the triangle $ABC$ at $D_i, E_i$ , and $F_i$ , respectively, then $D_iE_i \perp D_iF_i$ and $D_iE_i = D_iF_i \neq 0$ . Let the line $P_1P_2$ intersects the circumcircle of $ABC$ at $Q_1$ and $Q_2$ . The Simson lines of $Q_1$ , $Q_2$ with respect to $ABC$ intersect at $W$ . Prove that $W$ lies on the nine-point circle of $ABC$ .

LoloChen

28.03.2022 17:16

This is not hard. Notice that O,P1and P2 are collinear and OP1*OP2=R^2. Then it is easy by orthopole.

Little_Dinosaur

28.03.2022 18:42

Interesting but easy problem. We will use the following steps to finish the problem.

Lemma 1. Let

$P$ be a point lie on

$(ABC)$ ,

$l$ be the Simson line of

$P$ . Let

$H$ be the orthocenter of

$\triangle ABC$ , then

$l$ bisects

$HP$ . Proof. Let

$BH$ intersect

$(ABC)$ at

$K$ ,

$l$ intersect

$BC,CA$ at

$D,E$ respectively.

$KP$ intersect

$AC,l$ at

$S,N$ respectively. Then

$\angle NPE = \angle BKP = \angle BCP = \angle NEP \Rightarrow NP=NE$ , which shows that

$N$ is the midpoint of

$PS$ . Then

$\angle KNE = \pi - 2\angle NSE = \pi - \angle KSH = \angle NSH \Rightarrow SH \parallel l$ , which is what we need.

Lemma 2. Let

$XY$ be a diameter of

$(ABC)$ , and

$W$ be the intersection of the Simson lines of

$X,Y$ . Then

$W$ lies on the nine-point circle of

$\triangle ABC$ . Proof. Let

$H$ be the orthocenter of

$\triangle ABC$ , and

$l_x,l_y$ intersect

$HX,HY$ at

$X',Y'$ respectively. Using Lemma 1, we have

$X',Y'$ are the midpoints of

$HX,HY$ respectively. Then

$X',Y'$ lie on the nine-point circle of

$\triangle ABC$ . Because that

$XY$ is the diameter of

$(ABC)$ ,

$X'Y'$ is the diameter of the nine-point circle of

$\triangle ABC$ . So we just need to prove that

$l_x \perp l_y$ . Which is easy by angle calculating.

According to Lemma 2, we just need to prove that

$Q_1Q_2$ is a diameter of

$(ABC)$ , or

$O,P_1,P_2$ are collinear. Let

$D_1E_1$ intersect

$D_2E_2$ at

$U$ , and

$D_1F_1$ intersect

$D_2F_2$ at

$V$ . It's easy to prove that

$D_1D_2 \parallel E_1E_2 \parallel F_1F_2$ , shows that

$U,V,O$ are collinear. (Let this line be

$k$ ) Let

$D_1E_2$ intersect

$D_2E_1$ at

$X$ , and

$D_1F_2$ intersect

$D_2F_1$ at

$Y$ . Then

$X,Y \in k$ Using Pascal's Theorem on

$BE_1D_2AD_1E_2$ , we have

$P_1,P_2,X$ are collinear. Using Pascal's Theorem on

$AD_1F_2CF_1D_2$ , we have

$P_1,P_2,Y$ are collinear. Then

$P_1, P_2 \in k$ , congratulation!

Attachments:

PROF65

29.03.2022 01:09

Little_Dinosaur wrote: Interesting but easy problem. We will use the following steps to finish the problem.

Lemma 1. Let

$P$ be a point lie on

$(ABC)$ ,

$l$ be the Simson line of

$P$ . Let

$H$ be the orthocenter of

$\triangle ABC$ , then

$l$ bisects

$HP$ . Proof. Let

$BH$ intersect

$(ABC)$ at

$K$ ,

$l$ intersect

$BC,CA$ at

$D,E$ respectively.

$KP$ intersect

$AC,l$ at

$S,N$ respectively. Then

$\angle NPE = \angle BKP = \angle BCP = \angle NEP \Rightarrow NP=NE$ , which shows that

$N$ is the midpoint of

$PS$ . Then

$\angle KNE = \pi - 2\angle NSE = \pi - \angle KSH = \angle NSH \Rightarrow SH \parallel l$ , which is what we need.

Lemma 2. Let

$XY$ be a diameter of

$(ABC)$ , and

$W$ be the intersection of the Simson lines of

$X,Y$ . Then

$W$ lies on the nine-point circle of

$\triangle ABC$ . Proof. Let

$H$ be the orthocenter of

$\triangle ABC$ , and

$l_x,l_y$ intersect

$HX,HY$ at

$X',Y'$ respectively. Using Lemma 1, we have

$X',Y'$ are the midpoints of

$HX,HY$ respectively. Then

$X',Y'$ lie on the nine-point circle of

$\triangle ABC$ . Because that

$XY$ is the diameter of

$(ABC)$ ,

$X'Y'$ is the diameter of the nine-point circle of

$\triangle ABC$ . So we just need to prove that

$l_x \perp l_y$ . Which is easy by angle calculating.

According to Lemma 2, we just need to prove that

$Q_1Q_2$ is a diameter of

$(ABC)$ , or

$O,P_1,P_2$ are collinear. Let

$D_1E_1$ intersect

$D_2E_2$ at

$U$ , and

$D_1F_1$ intersect

$D_2F_2$ at

$V$ . It's easy to prove that

$D_1D_2 \parallel E_1E_2 \parallel F_1F_2$ , shows that

$U,V,O$ are collinear. (Let this line be

$k$ ) Let

$D_1E_2$ intersect

$D_2E_1$ at

$X$ , and

$D_1F_2$ intersect

$D_2F_1$ at

$Y$ . Then

$X,Y \in k$ Using Pascal's Theorem on

$BE_1D_2AD_1E_2$ , we have

$P_1,P_2,X$ are collinear. Using Pascal's Theorem on

$AD_1F_2CF_1D_2$ , we have

$P_1,P_2,Y$ are collinear. Then

$P_1, P_2 \in k$ , congratulation!

I think the hard part is to prove that it exists exactly two points $P_1,P_2$ and that they satisfy $D_1D_2 \parallel E_1E_2$ which is not granted ! remark that we have two possibiities for $D_i$ but only one works ... My regards RH HAS

PROF65

31.03.2022 17:02

with the similar proof as this topic we can show that $\angle P_iBD_i=\angle P_iCD_i\iff P_i$ is on the $A$ -apollonius circle more it s easy to prove that $P_i$ is on the circle $(O')$ of the points $M$ s.t. $\angle BMC=\frac{\pi}{2}+\angle BAC$ but $\angle OBC=\frac{\pi}{2}+\angle BAC$ which means that $(O')$ is tangent to $OB$ hence it orthogonal to $(O)$ so $P_i$ is on the intersection of $(O')$ and $A$ -apollonius circle besides we know that the latest is also orthogonal to $(O)$ therefore there is exactly two points $P_1,P_2$ which are inverse of each other implies $O,P_1,P_2$ are collinear hence $Q_1,O,Q_2$ are collinear and the result follows . My regards RH HAS

JG666

03.04.2022 04:50

Complex bashed. (Simson line's equation is well known; the Lemma_2 by little_dinosaur above is also quite well known)

Little_Dinosaur

04.04.2022 08:15

JG666 wrote: Complex bashed. (Simson line's equation is well known; the Lemma_2 by little_dinosaur above is also quite well known) That is. That's why I thought this problem is quite easy

PHSH

06.08.2022 03:07

Great problem! My solution uses angle chasing to find interesting carectization of the points, and then the solution is quite natural! Key Lemma: Let $K$ be intersection of the tangents from $B$ and $C$ to $(ABC)$ and then let $\omega$ be the circle with center $K$ passing through $B.$ Further, let $\gamma$ be the apolloniam circle of $A$ with respect to the segment $BC.$ Then, points $P_1$ and $P_2$ are the intersections of $\omega$ and $\gamma.$

This is just angle chasing. First let

$X_1,Y_1,Z_1$ be the projections of

$P_1$ with respect to

$BC,AC,AB,$ respectively. It is well known that

$\triangle X_1Y_1Z_1 \sim \triangle D_1E_1F_1$ (I know it as Lemma 2.5 from Geometry of Conics, A. V. Akopyan, A. A. Zaslavsky; I hope this is well known; it is just angle chasing anyways ). [asy][asy] import graph; import geometry; import olympiad; size(200); pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L; A=dir(126.791); B=dir(200.87); C=dir(-20.87); O=(0,0); M=B*0.5+C*0.5; Ob=2*B-O; Oc=2*C-O; Oba=bisectorpoint(Ob,O); Oca=bisectorpoint(Oc,O); K=extension(B,Oba,C,Oca); Oa=2*A-O; Oaa=bisectorpoint(O,Oa); L=extension(A,Oaa,B,C); path u,p,q,r; u=Circle(O,1,400); p=Circle(K,distance(K,B),400); q=Circle(L,distance(L,A),400); r=circumcircle(B,O,C); pair Px=intersectionpoints(p,q)[0]; pair Py=intersectionpoints(p,q)[1]; path l =Px--Py; pair Qx=intersectionpoints(l,u)[0]; pair Qy=2*O-Qx; pair Da,Ea,Fa; Da=foot(Px,B,C); Ea=foot(Px,A,C); Fa=foot(Px,A,B); fill(A--Fa--Px--Ea--cycle, 0.9*white+ 0.1*cyan); fill(B--Da--Px--Fa--cycle, 0.9*white+0.1*green); fill(C--Da--Px--Ea--cycle,0.9*white+0.1*magenta); draw(A--B--C--cycle); draw(Ea--Da--Fa--cycle); draw(A--Px--C); draw(Px--B); draw(Ea--Px--Fa); draw(Px--Da); dot("

$A$ ",A,dir(A)); dot("

$B$ ",B,dir(B)); dot("

$C$ ",C,dir(C)); dot("

$P_1$ ",Px,dir(140)); dot("

$X_1$ ",Da,dir(Da)); dot("

$Y_1$ ",Ea,dir(Ea)); dot("

$Z_1$ ",Fa,dir(Fa)); [/asy][/asy] By problem condition we must have that

$\measuredangle ABP_1 + \measuredangle P_1CA = \frac{\pi}{2}$ and

$\measuredangle CBP_1 + \measuredangle CAP_1 = \measuredangle P_1CB + \measuredangle BAP_1$ We can easily manipulate this to get

$\measuredangle BP_1C= \measuredangle A + \frac{\pi}{2}$ so it follows easily that

$K$ is indeed the circumcenter of

$\triangle BP_1C.$ Finally, we have that by law of sines at

$(BX_1P_1Z_1)$ and

$(CX_1P_1Y_1),$

$\frac{P_1X_1}{X_1Y_1}=\frac{P_1X_1}{X_1Z_1} \iff \frac{\sin \angle P_1BC}{\sin \angle ABC} = \frac{\sin \angle P_1CB}{\sin \angle ACB}$ so it follows that

$P_1$ lies on

$\gamma$ .

$\square.$

[asy][asy] import graph; import geometry; import olympiad; size(300); pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L; A=dir(126.791); B=dir(200.87); C=dir(-20.87); O=(0,0); M=B*0.5+C*0.5; Ob=2*B-O; Oc=2*C-O; Oba=bisectorpoint(Ob,O); Oca=bisectorpoint(Oc,O); K=extension(B,Oba,C,Oca); Oa=2*A-O; Oaa=bisectorpoint(O,Oa); L=extension(A,Oaa,B,C); path u,p,q,r; u=Circle(O,1); p=Circle(K,distance(K,B)); q=Circle(L,distance(L,A)); r=circumcircle(B,O,C); pair Px=intersectionpoints(p,q)[0]; pair Py=intersectionpoints(p,q)[1]; path l =Px--Py; pair Qx=intersectionpoints(l,u)[0]; pair Qy=2*O-Qx; pair Ota=2*B-Oba; draw(u,red); draw(p,green); draw(q,blue); draw(A--B--C--cycle); draw(Px--Py--Qx--Qy--cycle,dotted+magenta); draw(K--Ota,gray); draw(K--Oca,gray); dot(" $A$ ",A,dir(90)); dot(" $B$ ",B,dir(150)); dot(" $C$ ",C,dir(30)); dot(" $K$ ",K,dir(K)); dot(" $O$ ",O,dir(90)); dot(" $P_1$ ",Px,dir(150)); dot(" $P_2$ ",Py,dir(Py)); dot(" $Q_1$ ",Qx,dir(-90)); dot(" $Q_2$ ",Qy,dir(Qy)); [/asy][/asy] By the Key Lemma, we must have that line $P_1P_2$ is the radical axes of $\omega$ and $\gamma.$ Hence, since $(ABC)$ is ortogonal to both $\omega$ and $\gamma,$ we must have that $O,$ the circumcenter of $\triangle ABC,$ must lie on $P_1P_2.$ For finish, let $H$ be the orthocenter of $\triangle ABC,$ $U$ be the intersection of the Simson lines of $Q_1$ and $Q_2,$ and $T$ be the reflection of $H$ across $U.$ By above, we must have that $Q_1,Q_2$ are diametrically oppositive and it follows easily from angle chasing that their Simson lines are perpendicular $(i)$ . It is well known that the Simson line of $Q_1$ (respectively $Q_2$ ) bissects segment $HQ_1$ (respectively $HQ_2$ ); so it follows that $TQ_1, TQ_2$ is parallel to the Simson line of $Q_1,Q_2,$ repsectively; then it follows from conclusion $(i)$ that $\angle Q_1TQ_2 = {\pi}{/2}.$ But since points $Q_1$ and $Q_2$ are diametrically oppositive, this implies that $T$ lies on $(ABC)$ and so $U$ must lie on the nine-point circle (beacause the homothety with ratio ${1}{/2}$ centered at $H$ sends $(ABC)$ to the nine-point circle). So done.