Let $\triangle ABC$ be a scalene triangle and $X$, $Y$ and $Z$ be points on the lines $BC$, $AC$ and $AB$, respectively, such that $\measuredangle AXB = \measuredangle BYC = \measuredangle CZA$. The circumcircles of $BXZ$ and $CXY$ intersect at $P$. Prove that $P$ is on the circumference which diameter has ends in the ortocenter $H$ and in the baricenter $G$ of $\triangle ABC$.
Problem
Source: Problem 6, Brazilian MO 2015
Tags: geometry proposed, geometry, circumcircle
pi37
21.10.2015 05:57
Very similar to 2012 ISL G6. We'll rename $X,Y,Z$ to be $D,E,F$. Let $\Gamma_A=(BHC)$, $\omega_A=(AEF)$, and definte other circles similarly. Define $X_A$ to be the intersection of $AG$ and $\Gamma_A$. It is well-known (see 2015 TSTST #2) that $\angle HX_AG=90^{\circ}$, and that $X_A$ is the intersection of the circles through $A$ tangent to $BC$ at $B,C$ respectively. Let $Y_A=BE\cap CF$. Note that $Y_A$ is on $\omega_A$ and $\Gamma_A$; we claim that $X_A$ is the second intersection of the two circles. Indeed,
\[
\angle EY_AX_A=\angle BCX_a=\angle X_AAB=\angle X_AAE.
\]Of course by Miquel's theorem $P$ lies on $\omega_A,\omega_B,\omega_C$. By Miquel again on $\triangle BGC$ with respect to $DX_BX_C$, $P$ lies on $(GX_BX_C)$, which is the circle with diameter $GH$, as desired.
parmenides51
02.08.2019 09:00
also Serbia RMM TST 2019 P4
william122
22.01.2020 06:46
Solution with MathStudent2002 and Muriatic. Denote the HM points of the triangle as $A',B',C'$. We first claim that $(AYZ)$ always passes through $A'$. Note that $Y\to Z$ and $Y\to (AYA')\cap AB$ are projective, so we only need to check 3 cases. When $BY\perp AC$, the claim is obvious. When $\angle BYC=\angle B$, $(AYZ)$ is the circle passing through $A$ tangent to $BC$, so $A'$ lies on $(AYZ)$ by the definition of the HM point. As $\angle BYC=180-\angle C$ follows similarly, we have the desired result. Now, note that $X\to (BB'X)\cap (HG)$ is projective, as well as $X\to Y\to (CC'Y)\cap (HG)$. So, to show that $(BXZ)$ and $(CXY)$ intersect on $(HG)$, we only need to check three points. However, by choosing $X=(BB'C')\cap BC$, $X=(BB'A')\cap BC$, $X=(CC'B')\cap BC$, we can make $P$ the three HM points, which do indeed lie on $(HG)$. Thus, we are done.
NiltonCesar
31.12.2020 04:33
pi37 wrote:
Very similar to 2012 ISL G6. We'll rename $X,Y,Z$ to be $D,E,F$. Let $\Gamma_A=(BHC)$, $\omega_A=(AEF)$, and definte other circles similarly. Define $X_A$ to be the intersection of $AG$ and $\Gamma_A$. It is well-known (see 2015 TSTST #2) that $\angle HX_AG=90^{\circ}$, and that $X_A$ is the intersection of the circles through $A$ tangent to $BC$ at $B,C$ respectively. Let $Y_A=BE\cap CF$. Note that $Y_A$ is on $\omega_A$ and $\Gamma_A$; we claim that $X_A$ is the second intersection of the two circles. Indeed,
\[
\angle EY_AX_A=\angle BCX_a=\angle X_AAB=\angle X_AAE.
\]Of course by Miquel's theorem $P$ lies on $\omega_A,\omega_B,\omega_C$. By Miquel again on $\triangle BGC$ with respect to $DX_BX_C$, $P$ lies on $(GX_BX_C)$, which is the circle with diameter $GH$, as desired.
Sorry, but if you draw in Geogebra, you'll not have $\angle BCX_A=\angle X_AAB$.
Attachments:
VicKmath7
18.04.2023 17:06
Here is a sketch of a bit different solution.
Let $BY \cap CZ=A_1, AX \cap CZ=B_1, AX \cap BY=C_1$. The angle conditions imply $A_1 \in (AYZ)$ and similarly for the others. Thus $\triangle PB_1C_1 \sim \triangle PBC$ and similarly for the other points. Thus there is a spiral similarity at $P$ taking $\triangle A_1B_1C_1$ to $\triangle ABC$. Notice that $A_1 \in (HBC)$ so $HA_1=2R \sin \angle HCA_1$ since the radius of $(HBC)$ is $R$. But since $\angle HCA_1=\angle HCB_1=\angle HAB_1=\angle HAC_1$ we obtain $HA_1=HB_1=HC_1$, so $H$ is center of $(A_1B_1C_1)$. Add the circumcenter $O$ of $(ABC)$; hence the spiral similarity sends $H$ to $O$ and then finish with ratio lemma in $\triangle HPO$ using $\frac {HG} {GO}=2$.