collinear wanted, 2 circles passing through circumcenter related

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collinear wanted, 2 circles passing through circumcenter related

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Source: Estonia IMO TST 2016 p5 / Slovenia IMO TST 2015

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#1 h Jul 22, 2020, 10:47 PM

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Let $O$ be the circumcentre of the acute triangle $ABC$ . Let $c_1$ and $c_2$ be the circumcircles of triangles $ABO$ and $ACO$ . Let $P$ and $Q$ be points on $c_1$ and $c_2$ respectively, such that OP is a diameter of $c_1$ and $OQ$ is a diameter of $c_2$ . Let $T$ be the intesection of the tangent to $c_1$ at $P$ and the tangent to $c_2$ at $Q$ . Let $D$ be the second intersection of the line $AC$ and the circle $c_1$ . Prove that the points $D, O$ and $T$ are collinear

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#2 h Jul 24, 2020, 8:06 AM

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$\rm \angle PAO+\angle OAQ=2\cdot 90^{\circ}=180^{\circ}$ so the points $\rm P,A,Q$ are collinear.
Also $\rm \angle OPT+\angle OQT=2\cdot 90^{\circ}=180^{\circ}$ so the points $\rm T,P,O,Q$ are concyclic .
$\rm \angle POD=\angle OAQ=90^{\circ}-\angle AQO=\angle TQP=\angle TOP$ q.e.d

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#3 h Jul 24, 2020, 8:54 AM

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We will prove this theorem for any non-degenerate triangle $ABC$ . All angles in this solution are directed.
$\measuredangle QCO=90^\circ\implies\text{QC is tangent to ABC circumcircle}$ and $\measuredangle PAO=90^\circ\implies\text{PA is tangent to ABC circumcircle.}$ By tangent-chord angle theorem
$\measuredangle POD= \measuredangle PAB+ \measuredangle BAD= \measuredangle ACB+ \measuredangle BAC= \measuredangle ABC,\ \measuredangle ACQ= \measuredangle ABC.$ Observe $\measuredangle TPO=90^\circ= \measuredangle TQO\implies\text{points O, P, Q, T are concyclic}\implies \measuredangle POT= \measuredangle PQT= \measuredangle AQT.$ Finally $QT$ being tangent to $c_2$ implies $\measuredangle AQT= \measuredangle AOQ= \measuredangle ACQ= \measuredangle ABC.$ Therefore
$\measuredangle POT= \measuredangle ABC= \measuredangle POD.$ QED

This post has been edited 2 times. Last edited by WolfusA, Jul 24, 2020, 8:54 AM

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#4 h Mar 20, 2025, 6:54 AM

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First it’s easy to see that $TPOQ$ are concyclic, now inverse at $A$ and rewrite the problem:

Quote:

Given triangle $ABC$ , let $O$ be the reflection of $A$ wrt $BC$ . Let $P$ and $Q$ be points on $BO$ and $CO$ respectively, such that $AP\perp AO,AQ\perp AO$ (which implies that $A,P,Q$ are collinear) . Let $T$ be the intesection of [the circle passing through $A,P$ and tangent to $OP$ ], [the circle passing through $A,Q$ and tangent to $OQ$ ] and $\odot(OPQ)$ . Let $D$ be the intersection of the $AC$ and $BO$ . Prove that the points $D, O, T$ and $A$ are concyclic

Notice that $C$ is the circumcenter of $QAO$ ( $\because CO=CA$ and $\angle AOQ=90^{\circ}$ ),so $CA=CQ\implies CA$ is also tangent to $\odot(TAQ)$
Therefore, $\measuredangle TOD=\measuredangle TOP=\measuredangle TQP=\measuredangle OQT\implies\odot(ADOT)$

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#5 h Apr 8, 2025, 6:26 AM

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Weird how nobody mentioned this, but there's a better-looking formulation:

Let $\triangle ABC$ have intouch triangle $DEF$ and incentre $I$ . Show that $AI, DE, (BDF)$ concur.

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