Fixed point config on external similar isosceles triangles

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Fixed point config on external similar isosceles triangles

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Source: Bulgaria Spring Mathematical Competition 2025 10.2

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#1 h Mar 30, 2025, 12:41 PM

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Let $AB$ be an acute scalene triangle. A point $D$ varies on its side $BC$ . The points $P$ and $Q$ are the midpoints of the arcs $\widehat{AB}$ and $\widehat{AC}$ (not containing $D$ ) of the circumcircles of triangles $ABD$ and $ACD$ , respectively. Prove that the circumcircle of triangle $PQD$ passes through a fixed point, independent of the choice of $D$ on $BC$ .

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#2 h Mar 30, 2025, 5:35 PM • 1 Y

Y by Primeniyazidayi

Let $E$ be the midpoint of segment $BC$ . We claim that $(PQD)$ passes through $E$ . Let $F,G$ be midpoints of segment $AB$ and $AC$ respectively Clearly, $PF$ and $QG$ are the perpendicular bisector of $AB$ and $AC$ respectively. Let $O=PF$ $\cap$ $QG$ be the circumcenter of $\triangle ABC.$ Since $\angle BDP=\angle ADP,\angle ADQ=\angle CDQ$ and $\angle BDP+\angle ADP+\angle ADQ+\angle CDQ=180^{\circ}$ we obtain $\angle PDQ = 90^{\circ}$ . Let $H$ be antipode of $P$ w.r.t. $(ABD)$ and $I$ be antipode of $Q$ w.r.t. $(ACD)$ . $\angle PDH=90^{\circ}=\angle PDQ \Longrightarrow D,H,Q$ collinear. Leftwise, $P,I,D$ collinear.

$\angle APF = 90^{\circ}-\angle PAF=90^{\circ}-\angle PAB=90^{\circ}-\angle PDB=\angle QDC = \angle QAC = \angle QAG$ .
Hence $\triangle APF \sim \triangle QAG \Longrightarrow \frac{PF}{FA}=\frac{AG}{GQ} \Longrightarrow \frac{PF}{EG} = \frac{FE}{GQ} \Longrightarrow \frac{PF}{FE} = \frac{EG}{GQ}$ .

Notice that $\angle PFE = 90^{\circ}+\angle BFE = 90^{\circ}+\angle CGE =\angle EGQ$ which means $\triangle PFE \sim EGQ$ .

Hence $\angle PEQ = \angle POQ - \angle EPF-\angle EQG=180^{\circ}-\angle BAC- \angle EPF-\angle PEF = 180^{\circ}-\angle BAC -(90^{\circ}-\angle BAC)=90^{\circ}$ . Hence, $(PQD)$ passes through a fixed point, independent of the choices of $D$ on $BC$ .

This post has been edited 1 time. Last edited by E50, Mar 31, 2025, 2:59 AM
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#3 h Mar 30, 2025, 7:45 PM • 2 Y

Y by Primeniyazidayi, E50

Let $M$ be the midpoint of $BC$ . Invert around $D$ . We have $(M^*,D;B^*,C^*)=-1$ so $\frac{M^*C}{M^*B}.\frac{P^*B}{P^*A}.\frac{Q^*A}{Q^*C}=\frac{DC^*}{DB^*}.\frac{DB^*}{DA^*}.\frac{DA^*}{DC^*}=1$ . By Menelaus $M^*,P^*,Q^*$ are collinear thus, $(DPQ)$ passes through $M$ as desired. $\blacksquare$

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