Let $BB_1$ be the symmedian of a nonisosceles acute-angled triangle $ABC$. Ray $BB_1$ meets the circumcircle of $ABC$ for the second time at point $L$. Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$. Ray $BH_B$ meets the circumcircle of $ABC$ for the second time at point $T$. Prove that $H_A, H_C, T, L$ are concyclic.
Problem
Source: Sharygin Geometry Olympiad 2016 First Round P12 grades 9-10
Tags: geometry, Concyclic, symmedian
checkmatetang
25.07.2018 04:52
Let $H_AH_C$ meet $AC$ at $P$. Then, since $BALC$ is harmonic, projecting from $T$ gives that $(A, C; H_B, TL\cap AC)=-1=(A,C; H_B, P)$ so $T, L,$ and $P$ are collinear. Observe that $AH_CH_AC$ and $ATLC$ are cyclic, so by the Radical Lemma the desired result follows.
khanhnx
20.01.2019 18:11
Let $E$ be the second intersection of $H_AH_C$ and $AC$ We have: $\dfrac{EC}{EA}$ = $\dfrac{H_BC}{H_BA}$ = $\dfrac{BC}{AB}$ . $\dfrac{TC}{AT}$ = $\dfrac{LC}{AL}$ . $\dfrac{TC}{AT}$ Hence: $E$, $T$, $L$ are collinear So: $\overline{ET}$ . $\overline{EL}$ = $\overline{EA}$ . $\overline{EC}$ = $\overline{EH_A}$ . $\overline{EH_C}$ or $H_A$, $H_C$, $L$, $T$ lie on the circle
Om245
23.11.2023 17:52
Let $H_AH_C \cup \overline{AC} = P$ , $BT \cup H_AH_C = X$ and $BL \cup H_AH_C = M$. Take Inversion at $B$ with radius $\sqrt{BH.BH_B}$ which swaps $H_A$ with $A$, $H_C$ with $C$ , $H_B$ with $H$ therefore Circumcircle of $\triangle ABC$ with $\overline {H_AH_C}$ and so $M$ is image of $L$ and $X$ is image of $T$ under inversion. therefore $H_AH_CLT$ cyclic is equivalent to prove $AXMC$ is cyclic. Now $\overline{H_AH_C}$ is antiparallel to $AC$ so, the symmedian in $\triangle BAC$ ,$\overline{BM}$ is median in $\triangle BH_AH_C$ Now as we know $$(P,H_B;A,C)=-1=(P,X;H_A,H_C)$$ $$PA.PC=PH_A.PH_C = PX.PM$$(by Midpoint Length) hence we get $ACMX$ is cyclic which give us $H_AH_CTL$ Cyclic.$\blacksquare$.
Attachments:
