The altitudes of an acute triangle $ABC$ intersect at $H$. The tangent line at $H$ to the circumcircle of triangle $BHC$ intersects the lines $AB$ and $AC$ at points $Q$ and $P$ respectively. The circumcircles of triangles $ABC$ and $APQ$ intersect at point $K$ ($K\neq A$). The tangent lines at the points $A$ and $K$ to the circumcircle of triangle $APQ$ intersect at $T$. Prove that $TH$ passes through the midpoint of segment $BC$.
Problem
Source: Kazakhstan (Junior) National Olympiad 2023, 9th grade, P6
Tags: geometry
trinhquockhanh
24.03.2023 04:26
$\text{solution by Tran Gia Huy (here } \omega\; \text{is}\; (H,0))$
menpo
24.03.2023 14:32
Outline of Proof: 1. $S=AK\cap QP\cap BC,$ by radical axis on $(ABC), (APQ)$ and $(BCPQ)$. 2. $AF$ diameter of $\odot AQP,$ since $TA \parallel BC.$ 3. $AA’$ diameter of $\odot ABC,$ then $A’ = HM\cap KF.$ 4. $SH^2 = SB\cdot SC = SK\cdot SA,$ which implies $QP \parallel EF$ by angle chasing. Therefore, $AE$ passes through $A’.$ 5. Apply Pascal theorem on $KKF\text{ }AAE$ to conclude the problem.
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olympiad.geometer
24.03.2023 15:57
Adilet160205 wrote: Outline of Proof: 1. $S=AK\cap QP\cap BC,$ by radical axis on $(ABC), (APQ)$ and $(BCPQ)$. 2. $AF$ diameter of $\odot AQP,$ since $TA \parallel BC.$ 3. $AA’$ diameter of $\odot ABC,$ then $A’ = HM\cap KF.$ 4. $SH^2 = SB\cdot SC = SK\cdot SA,$ which implies $QP \parallel EF$ by angle chasing. Therefore, $AE$ passes through $A’.$ 5. Apply Pascal theorem on $KKF\text{ }AAE$ to conclude the problem. What program did you use to draw the diagram?
bin_sherlo
15.05.2024 20:48
Let $A'$ be the antipode of $A$ on $(ABC)$. $(AH)\cap (ABC)=\{A,N\},PQ\cap BC=S$ Let the altitudes from $A,B,C$ to $BC,CA,AB$ be $D,E,F$ respectively. $AD\cap (ABC)=\{A,G\},PQ\cap AN=L,AN\cap BC=R$ $\textbf{Claim:} \ SN$ is tangent to $(ABC)$. $A'N\perp LA$ and $LH\perp AA'$ since $\angle PAA'=90-\angle B=90-\angle QPA$ Hence $H$ is the orthocenter of $LAA'$. This gives that $LA'\perp AH\perp A'G$ thus $A',G,L$ are collinear. $HM=MA'$ and $MS\parallel LA'$ so $HS=SL$. Also $\angle HNL=90$ Thus $SG=SH=SL=SN$ which gives the desired result since $SN^2=SG^2=SB.SC$ $\textbf{Claim:} \ TA\parallel BC$ Circumcenter of $(APQ)$ is on $AH$ since $\angle PAH=90-\angle AQP,\ \angle HAQ=90-\angle QPA$ Also $TA$ is tangnet to $(APQ)$ which means that $AH\perp TA$. $\textbf{Claim:} \ T,N,H$ are collinear. Invert from $A$ with radius $\sqrt{AH.AG}$ We have $(ABC)\leftrightarrow PQ, \ (APQ)\leftrightarrow BC, \ TK\leftrightarrow (AST^*), \ TA\leftrightarrow TA, \ T\leftrightarrow T^*, \ N\leftrightarrow L, \ H\leftrightarrow G$ Since $TK$ and $(KAPQ)$ are tangent, we have that $(AT^*S)$ and $\overline{BCS}$ are tangent. Also $TA\parallel BC$. Thus $SA=ST^*$. $S$ is the circumcenter of $(LGHN)$ so $SL=SG$. $T^*A\parallel LG, \ ST^*=SA, \ SL=SG$ Hence $T^*A$ and $LG$ have common perpendicular bisector which gives that $T^*AGL$ is a rectangle. $(T^*AGL)\leftrightarrow T,H,N$ Thus $T,N,H$ are collinear as desired.$\blacksquare$