$ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$ Proposed by Pouria Mahmoudkhan Shirazi
Problem
Source: Iran TST 2023 ; Exam 1 Problem 6
Tags: geometry, circumcircle
Shayan-TayefehIR
16.03.2023 14:08
Love this one ! Let $O$ be the center of circumcircle of triangle $\triangle ABC$ and suppose that $T$ is the reflection point of $O$ respect to $BC$ ( The center of circumcircle of triangle $\triangle BHC$ ). We define points $X,Y$ such that $X,C$ and $Y,B$ are located on same side of $AH$ and also we have : $$\triangle AHX \sim \triangle ATC , \triangle AHY \sim \triangle ATB$$Now note that because of the spiral similarity we have $\triangle AHT \sim \triangle AXC \sim \triangle AYB$ and one can see that : $$\angle CXH=\angle AHX+\angle ACX+\angle HAC=\angle CTA+\angle ATH+\angle CBH=\angle CTH+\angle CBH=3\angle CBH (I)$$So we'll show that the point $P'$ , reflection of $X$ respect to the line $CH$ , lies on the line $AT$. Let $E$ , $F$ be reflections of point $T$ respect to lines $CH$ and $BH$ and suppose that $XH$ and $CT$ intersect each other at point $Q$. Thus quadrilaterals $HECT$ and $HFBT$ are parallelogram and since $A$ is the Miquel point of $HXCT$ , quadrilateral $AHTQ$ is cyclic and we can get : $$\angle EHX=\angle HQT=\angle HAT , \frac{HX}{HE}=\frac{HX}{TC}=\frac{AH}{AT} \implies \triangle EHX \cong \triangle THP' \sim \triangle TAH$$$$ \implies \angle HTP'=\angle HTA$$Similary we can get $P'$ is the reflection point of $X , Y$ respect to $CH , BH$ which lies on $AT$ and by $(I)$ , we have $\angle HP'C=3\angle HBC$ and $\angle HP'B=3\angle HCB$ and as the result , $P \equiv P'$. Now Note that because of the spiral similarity , we have $\triangle SAH \sim \triangle OAT$ and while $\angle HAB=\angle OAC=90-\angle B$ , we get $\angle SAB=\angle TAC=\angle PAC$ and we're done.
squarc_rs3v2m
03.04.2023 14:57
In the following we use directed angles. Solved with NoctNight. Step 1: If $T = BX \cap CY$ then $T \in (ABC)$ (and so $A$ is the Miquel point of $BXYC$)
$\angle BTC = -(\angle CBT + \angle TCB) = -(2(\angle CBH + \angle HCB) - (\angle CBP + \angle PCB))$ by reflection; but since $\angle CBP + \angle PCB = -\angle BPC = 3 \angle CBH + 3\angle HCB$ we have that $\angle BTC = \angle A$ as required.
Step 2: $P$ is the image of $A$ in inversion in $(BHC)$.
By Step 1, $\frac{BP}{PC} = \frac{BX}{CY} = \frac{BA}{AC}$ so $P$ lies on $\Gamma$ the $A$-Apollonian circle. Let $O'$ be the circumcenter of $\triangle BHC$, let $P' = AO' \cap \Gamma$ and let $R \in BC$ with $AR$ tangent to $(ABC)$, so that $R$ is the center of $\Gamma$. Then $O'P' \cdot O'A = \mathrm{Pow}(O', \Gamma) = O'R^2 - A'R^2$ (since $A' \in \Gamma$ is trivial, where $A'$ is the reflection of $A$ in $BC$); since $\angle O'A'R = \angle RAO = 90^\circ$, this equals $O'A'^2$, but as $A' \in (BHC)$ we have that $P'$ is the image of $A$ in inversion in $\angle BHC$. However, $\angle HP'B = \angle HP'O' + \angle O'P'B = \angle O'HA + \angle ABO'$ by inverting in $(BHC)$. We also have $\angle O'HA = \angle O'HA' = \angle HH_1A' + 90^\circ$ where $H_1$ is the antipode of $H$ on $(BHC)$; this then equals $\angle HBA' + 90^\circ = \angle ABH' + 90^\circ$ where $H'$ is the reflection of $H$ in $BC$, so $\angle HP'B = \angle OBC - \angle B + \angle ABH' + 90^\circ = 90^\circ + \angle OBC + \angle HBC - 2\angle B = 90^\circ - 3\angle B = \angle HPB$, and similarly $\angle CP'H = \angle CPH$, so $P' \neq H$ is an intersection of $(HPC), (HPB)$ which means $P' = P$.
We now have enough to finish the problem.
It's well-known that $AO'$ passes through $X_5(\triangle ABC)$; hence we want $AS$ to pass through $X_{54}(\triangle ABC)$, which means that we want $A-S-O_1$ where $O_1$ is the circumcenter of $\triangle BOC$.
Since $\angle SAO = \angle XAB$ by spiral sim, we want to prove the second equality in $\angle(XY, BC) = \angle XAB = \angle O_1AO' = \angle HAP$; hence we want to show that the reflection of $XY$ in $AH$ is perpendicular to $AP$; let $X', Y'$ be the reflections of $X, Y$ in $AH$. Utilising some facts about the composition of two reflections (namely what the resulting rotation is), we have that $\angle PHX' = 2\angle BHA, \angle PHY' = 2\angle CHA$. Since $HX' = HY'$ from the reflections, we want to show that the angle bisector of $\angle X'HY'$ is parallel to $AP$, namely that $\angle HPA = \frac12(\angle PHX' + \angle PHY') = \angle BHA + \angle CHA$. However, $\angle HPA = \angle HPO' = \angle O'HA$ by inversion in $(BHC)$; hence we finish with $\angle O'HA + \angle AHC = \angle OHC = 90^\circ - \angle CBH = \angle BHA$.
k12byda5h
10.04.2023 13:38
Let $\beta = \angle HBC$ and $\gamma = \angle HCB$ Let $\phi$ be the $\sqrt{bc}$ inversion of triangle $BHC$ (with the center $H$.) Let $P' = \phi(P)$. Thus, $\angle CBP' = 2\beta$ and $\angle BCP' = 2 \gamma .$ Claim 1: $A$ is the Miquel point of quadrilateral $BXYC$.
\begin{align*}
\angle BQC &= 180^\circ - \angle XBC - \angle YCB \\
&= 180^\circ - (\beta + \gamma) - (\angle HP'C + \angle HP'B) \\
&= 180^\circ - (\beta + \gamma) - (180^\circ - \angle P'BC - \angle P'CB) \\
&= 180^\circ - (\beta + \gamma) - (180^\circ - 2\beta - 2\gamma) \\
&= \beta + \gamma \\
&= \angle BAC
\end{align*}
, $B,C,Q,A$ are concyclic.
\begin{align*}\frac{BX}{CY} &= \frac{BP}{ CP} = \frac{P'C \cdot AP}{HC} \Big / \frac{P'B \cdot AP}{BH} \\&= \frac{P'C}{P'B} \cdot \frac{BH}{CH} = \frac{\sin 2\beta}{\sin 2\gamma} \cdot \frac{\sin \gamma}{\sin \beta} \\&= \frac{\cos \beta}{\cos \gamma} = \frac{BA}{CA}\end{align*}.
. $\quad \square$ Claim 2: $XY \perp HP'$ Proof: Note that $HX = HP = HY$, so $XY$ is perpendicular to internal bisector of $\angle XHY$ which is $HP'$. Let $O$ be the circumcenter of $\triangle ABC$ and $A' = \phi(A)$. Not hard to see that $A',P'$ are symmetric wrt $BC$, so $A'H' \perp BC$. We need to prove that $\angle BAS = CAP$ which is equivalent to $\angle HAP = \angle OAS$. Since $\angle HAP = \angle HP'A'$ and, by spiral similarity with a Miquel point, $\angle(XY,BC) = \angle OAS$, it suffices to show that $\angle HP'A' = \angle(XY,BC)$ which is $XY \perp HP'$, the claim we already proved. [asy][asy] import geometry; import olympiad; import graph; size(300); pair A,B,C,H,P,PP,X,Y,S,Q; B = dir(145); C = dir(35); PP = dir(-110); pair I= incenter(B,C,PP); pair HH = foot(I,B,C); H = 2 * HH - I; P = (B-H) * (C-H) / (PP-H) + H; X = 2*foot(P,B,H) - P; Y = 2*foot(P,C,H) - P; A = orthocenter(B,C,H); S = circumcenter(A,X,Y); Q = extension(B,X,C,Y); fill(A--B--C--cycle, 0.1*cyan + 0.9*white); fill(B--H--PP--cycle,0.3*green + 0.7*white); fill(C--H--P--cycle,0.3*green + 0.7*white); fill(C--H--Y--cycle,0.3*green + 0.7*white); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(circumcircle(A,X,Y)); draw(B--Q--C); draw(X--H--Y); draw(B--H--C); draw(B--PP--C); draw(PP--S,purple); draw(P--B); draw(P--C); draw(P--H); dot("$B$",B,dir(210)); dot("$C$",C,dir(-30)); dot("$P'$",PP,dir(PP)); dot("$H$",H,2*dir(110)); dot("$P$",P,1.5*dir(-90)); dot("$A$",A,dir(A)); dot("$X$",X,dir(200)); dot("$Y$",Y,2.5*dir(10)); dot("$Q$",Q,dir(Q)); dot("$S$",S,dir(S)); dot("$A'$",2*foot(PP,B,C) - PP,dir(S)); [/asy][/asy]
v4913
10.04.2023 19:38
wow this is such a beautiful problem! Let Ha be the intersection of AH and (ABC), and J the point such that Ha is the incenter of triangle JBC. Then notice that if O is the circumcenter of ABC, J, Ha, O are collinear and <BHaO = <ABC, <CHaO = <ACB. Furthermore, if J’ is the point such that HRC ~ PBJ’, it is also true that HRB ~ HCJ’, and <HBJ = <HPC, <HCJ = <HPB so J = J’, thus P and J are inverses through force overlaid inversion on triangle HBC. Furthermore, by the ratio lemma JB/JC = (BH/CH)(sin B/sin C) so JB/JC = (BH/CH)(AC/AB) so BX/CY = AB/AC which implies that AXB ~ AYC, and AXY ~ ABC. Thus, S is the reflection of H over XY since H is the circumcenter of PXY, and <XPY = <XAY. Finally, the isogonality condition is equivalent to <SAO = <PAH; if O’ is the reflection of O over BC then ASH ~ AOO’ so it suffices to show that A, P, O’ are collinear. By force overlaid inversion on triangle HBC, Ha and O’ swap so it suffices to show that H, Ha, J, and the intersection of O’H with (O’BC) are concyclic. However, this is clearly an isosceles trapezoid since by angle chasing BOCJ is cyclic, so (BOCJ), (O’BC) are reflections over BC, as are H and Ha, O and O. Isosceles trapezoids are cyclic, so we are done.
lzhang3
10.04.2023 20:34
Trivial by the following BEAUTIFUL article. https://web.evanchen.cc/handouts/bary/bary-full.pdf Why did they make a problem that is trivialized by a previous publication?
gnoka
12.04.2023 11:05
[asy][asy] import geometry; import olympiad; import graph; size(300); pair A,B,C,H,P,PP,X,Y,S,Q; B = dir(145); C = dir(35); PP = dir(-110); pair I= incenter(B,C,PP); pair HH = foot(I,B,C); H = 2 * HH - I; P = (B-H) * (C-H) / (PP-H) + H; X = 2*foot(P,B,H) - P; Y = 2*foot(P,C,H) - P; A = orthocenter(B,C,H); S = circumcenter(A,X,Y); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(circumcircle(A,X,Y)); draw(X--H--Y); draw(B--H--C); draw(P--B); draw(P--C); draw(P--H); dot("$B$",B,dir(210)); dot("$C$",C,dir(-30)); dot("$P'$",PP,dir(PP)); dot("$H$",H,2*dir(110)); dot("$P$",P,1.5*dir(-90)); dot("$A$",A,dir(A)); dot("$X$",X,dir(200)); dot("$Y$",Y,2.5*dir(10)); dot("$S$",S,dir(S)); [/asy][/asy]
ThisNameIsNotAvailable
16.04.2023 18:42
AHZOLFAGHARI wrote: $ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$ Proposed by Pouria Mahmoudkhan Shirazi Let $O'$ and $A'$ be the reflection point of the circumcenter $O$ of $\triangle ABC$ and $A$ about $BC$ and $BH$, respectively. Then it's clear that $B,H,C,A'$ are concyclic and $O'$ is the circumcenter of $\triangle BHC$. Redefine $P$ to be the intersection of $AO'$ and $A$-Apollonius circle $\omega$ of $\triangle ABC$. We first prove the angle condition of the problem. Since $O$ and $O'$ are reflect about $BC$, $O'H^2=OA^2=\mathbb{P}_{O/\omega}=\mathbb{P}_{O'/\omega}=O'P\cdot O'A$. Moreover, $AA'\cdot AC=\mathbb{P}_{A/\odot (BHC)}=AO'^2-AO^2=AO'^2-O'H^2=AP\cdot AO'$, so $P,O',C,A'$ are concylic. By an easy angle chasing, we get the given condition: $\angle HPC=\angle HPA+\angle APC=\angle OAH+\angle AA'O'=\cdots =3\angle HBC$. A simple angle chasing yields $\angle ABX=\angle PBA'=\cdots =\angle AO'H$. Combining with $\frac{BX}{BA}=\frac{BP}{BA}=\frac{O'P}{O'B}=\frac{O'H}{O'A}$, we can conclude $\triangle ABX\sim\triangle AO'H\sim\triangle ACY$, so $BX$ and $CY$ intersect at $T$, which is the intersection of $\odot (ABC)$ and $\odot (AXY)$. Hence by the spiral similarity, $AS,AO'$ are isogonal in $\angle HAO$ and the conclusion follows.