Let $ABC$ be an acute-angled triangle with $AB < AC$ and $\Gamma$ the circumference that passes through $A,\ B$ and $C$. Let $D$ be the point diametrically opposite $A$ on $\Gamma$ and $\ell$ the tangent through $D$ to $\Gamma$. Let $P, Q$ and $R$ be the intersection points of $B C$ with $\ell$, of $A P$ with $\Gamma$ such that $Q \neq A$ and of $Q D$ with the $A$-altitude of the triangle $ABC$, respectively. Define $S$ to be the intersection of $AB$ with $\ell$ and $T$ to be the intersection of $A C$ with $\ell$. Show that $S$ and $T$ lie on the circumference that passes through $A, Q$ and $R$.
Problem
Source: 2023 Centroamerican and Caribbean Math Olympiad, P5
Tags: geometry, circumcircle
Ianis
26.07.2023 22:07
First observe that\begin{align*}\angle CTS & =\angle CTD \\ & =90^\circ -\angle TDC \\ & =\angle CDA \\ & =\angle CBA \\ & =180^\circ -\angle SBC, \end{align*}so $BCTS$ is cyclic. Then $PT\cdot PS=PC\cdot PB=PQ\cdot PA$ by PoP, so $AQTS$ is cyclic. We also have$$\angle AQR=\angle AQD=90^\circ ,$$so it suffices to prove that the circumcentre of $ATS$ is on $AR$, i.e. that the tangent to $(ATS)$ through $A$ is parallel to $BC$. This is clear from the fact that $BCTS$ is cyclic. So we are done.
NO_SQUARES
26.07.2023 22:50
Nice problem! Here is my solution with some motivation We will show that $\angle ARQ = \angle ASQ$, then, of course, we will have that (by analogue) $\angle ARQ = \angle ATQ$ and $ASRTQ$ will be inscribed. We have that $\angle ARQ = 90^{\circ} - \angle RAP = \angle BPA$, so we need to prove that $\angle ASQ = \angle APB$ or $SBQP$ is inscribed, which is equivalent to $\angle ABQ = \angle SPA$. But $\angle ABQ = \angle ADQ = \angle 90^{\circ} - \angle DAQ = \angle APD$, because $\angle AQD = \angle ADP = 90^{\circ}$ since $A \text{ and } D$ are diametrically opposite on $\Gamma$
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musan1909
26.07.2023 23:24
Consider the inversion of center $A$ and radius $AD$ called $\Phi$. Let $R' = BC\cap AR$. Notice that $AD^2 = AB \cdot AS = AQ \cdot AP = AC\cdot AT$. Since $RR'QP$ is cyclic, $AR' \cdot AR = AQ \cdot AP = AD^2$. We can conclude $\Phi(S) = B$, $\Phi(R) = R'$, $\Phi(T) = C$, $\Phi(Q) = P$. Since all of these points are on the same line we are done.
SQTHUSH
27.07.2023 04:37
Consider that $\angle ATD=\angle ADC=\angle ABC$ Which means that $B,S,T,C$ are cyclic Since $\angle ATS=\angle ABC=\angle CQP$ Hence $Q,C,T,P$ are cyclic So $\angle TQP=\angle TCP=\angle AST$ Which means that $A,S,T,Q$ are cyclic Note that $\angle ASQ=\angle ATQ=\angle CPQ$ Since $AR\bot BP,RQ\bot AP$ Hence $\angle ARQ=\angle CPQ=\angle ASQ$ Which means that $A,S,R,P,Q$ are cyclic
VicKmath7
27.07.2023 12:06
Since $AB \cdot AS=AD^2=AC \cdot AT$, $SBCT$ is cyclic, so we can rephrase the problem as follows, as the altitude of $ABC$ contains the diameter of $(AST)$:
Rephrased Centro 2023/5 wrote:
Given is a triangle $AST$ with altitude $AD$ and $A$-antipode $R$. Let $RD \cap (AST)=Q$ and let the circle with diameter $AD$ meet $AS, AT$ at $B, C$. If $BC \cap ST=P$, then show that $A, Q, P$ are collinear and that $Q$ lies on $(AD)$.
Since $\angle AQR=\angle AQD=90^{o}$, the latter is obvious. For the former, as $Q=(ABC) \cap (AST)$ is the Miquel point of the cyclic quadrilateral $SBCT$, it's well-known that $Q \in AP$.
AlanLG
27.07.2023 21:38
Lemma 1 $BCTS$ is cyclic Proof: As $\angle BAR=\angle DAT$ and $\angle(AR,BC)=\angle TDA=90^\circ$ then $\angle ABC=\angle ATD$ $\hspace{18cm}\blacksquare$ Lemma 2 $AQTS$ is cyclic Proof: By PoP in $BCTS$ and $AQCB$ $$ PT\cdot PS=PC\cdot PB=PQ\cdot PA$$$\hspace{18cm}\blacksquare$ Lemma 3 $AQRS$ is cyclic Proof: $$\angle QRA=90^\circ-\angle RAC-\angle CAQ=90^\circ-(90^\circ-\angle C)-\angle CAQ=\angle C -\angle CAQ, \hspace{0.1cm}\textrm{and}$$$$\angle QSA=\angle TSB-\angle TSQ=\angle C-\angle TAQ=\angle C-\angle CAQ$$$\hspace{18cm}\blacksquare$
UI_MathZ_25
27.07.2023 21:48
By angle chasing $BCTS$ and $BQPS$ are cyclic. Since $ABCQ$ is cyclic, it turns out that $Q$ is the Miquel's point of quadrilateral $BCTS$, therefore $AQTS$ is cyclic and in addition $CTPQ$ is cyclic. Finally, let $K = AR \cap BC$ and clearly $KQPR$ is cyclic, then \[\angle ARQ = \angle QPC = \angle QTA\]thereby $S$, $T$ $\in \odot(\triangle AQR)$