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 $ABC$.
Problem
Source: 2022 China TST, Test 2, P2
Tags: geometry, Simson line, Nine Point Circle
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.