fixed point for circumcircles of APQ

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fixed point for circumcircles of APQ

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Source: 2006 Ukraine NMO 9.8

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#1 h Jul 11, 2021, 9:03 AM

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Given an acute triangle $ABC$ . Let $D$ be a point on the side $BC$ different from the vertices, and let the points $P$ and $Q$ be the centers of the circumcircles of the triangles $ABD$ and $ACD$ , respectively. Consider all sorts of triangles $APQ$ obtained for all such points $D$ . Prove that the circles circumscribed around all these triangles have a common point other than point $A$ .

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#2 h Oct 6, 2021, 8:07 AM • 1 Y

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Let $A(0,a),B(-b,0),C(c,0)$ .
Choose $D(\lambda,0)$ .

Circumcircle $\triangle ABD\ :\ x^{2}+y^{2}+(b-\lambda)x+\frac{b\lambda-a^{2}}{a}y-b\lambda=0$ , midpoint $P(\frac{\lambda-b}{2},\frac{a^{2}-b\lambda}{2a})$ .
Circumcircle $\triangle ACD\ :\ x^{2}+y^{2}-(c+\lambda)x-\frac{c\lambda+a^{2}}{a}y+c\lambda=0$ , midpoint $Q(\frac{c+\lambda}{2},\frac{a^{2}+c\lambda}{2a})$ .

Circumcircle $\triangle APQ\ :\ x^{2}+y^{2}+\frac{a^{2}(b-c-\lambda)-bc\lambda}{2a^{2}} \cdot x-\frac{3a^{2}-b(c+\lambda)+c\lambda}{2a} \cdot y+\frac{a^{2}-b(c+\lambda)+c\lambda}{2}=0$ .

Fixed points $A(0,a)$ and $V(\frac{c-b}{2},\frac{a^{2}-bc}{2a})$ , the circumcenter of $\triangle ABC$ .

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