Let $ABC$ be an acute-angled triangle with circumcenter $O$ and orthocenter $H$. The circle with center $X_a$ passes in the points $A$ and $H$ and is tangent to the circumcircle of $ABC$. Define $X_b, X_c$ analogously, let $O_a, O_b, O_c$ the symmetric of $O$ to the sides $BC, AC$ and $AB$, respectively. Prove that the lines $O_aX_a, O_bX_b, O_cX_c$ are concurrents.
Problem
Source: Brazil MO 2018 Grades 8 and 9
Tags: geometry, circumcircle
lxhoanghsgs
17.11.2018 05:10
My solution: Lemma: Let $\Delta ABC$ with circumcircle $(O)$,orthocenter $H$.The tangent line at $A$ of $(O)$ cuts $BC$ at $J_{a}$.$AH$ cuts $(O)$ again at $H_{a}$ and $J_{a}H_{a}$ cuts $(O)$ again at $T_{a}$.$T_{b}$ and $T_{c}$ are defined at the same with $T_{a}$. Prove that:$AT_{a}$,$BT_{b}$ and $CT_{c}$ are concurrent. Proof: We have a known result that $\frac {J_{a}B}{J_{a}C}=\frac{AB^{2}}{AC^{2}}=\frac{sin(J_{a}H_{a}B)}{sin(J_{a}H_{a}C)}=\frac{sin(BAT_{a})}{sin(CAT_{a})}$, so according to the Ceva Sine Theorem we have Q.E.D. Return to the main problem: Note that $O_{a}$ is the circumcenter of $\Delta BHC$ and let $(O_{a})$ cuts $(X_{a})$ again at $R_{a}$.It's easy to prove that $J_{a},H,R_{a}$ are collinear. We also can prove that $R_{a}$ is the reflection of $T_{a}$ through $BC$. So according to Hagge's circle, we have $H,R_{b},R_{a},R_{c}$ are cyclic. So $X_{a}O_{a},X_{b}O_{b},X_{c}O_{c}$ pass through the circumcenter of $HR_{a}R_{b}R_{c}$. Done! Sincerely, XH
Tsukuyomi
17.11.2018 16:29
Observe that $X_a\in O_bO_c$ as $O_b,O_c$ are the circumcenters of $\triangle{AHC},\triangle{AHB}$ and thus the points $X_,O_b,O_c$ lie on the perpendicular bisector of $AH$. We thus have, $$\displaystyle\prod_{\text{cyc}}\dfrac{O_bX_a}{X_aO_c}=\displaystyle\prod_{\text{cyc}}\dfrac{OO_b}{OO_c}\cdot\dfrac{\sin{\angle{O_bOX_a}}}{\sin{\angle{X_aOO_c}}}=\displaystyle\prod_{\text{cyc}}\dfrac{2R\cos{\angle{B}}\sin{\angle{B}}}{2R\cos{\angle{C}}\sin{\angle{C}}}=1,$$and so we have our desired result by Ceva's theorem.