Bicentric quads formed by cevians

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Bicentric quads formed by cevians

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Source: India TST 2018, D4 P1

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#1 h Jul 18, 2018, 4:11 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Let $ABC$ be a triangle and $AD,BE,CF$ be cevians concurrent at a point $P$ . Suppose each of the quadrilaterals $PDCE,PEAF$ and $PFBD$ has both circumcircle and incircle. Prove that $ABC$ is equilateral and $P$ coincides with the center of the triangle.

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#3 h Jul 18, 2018, 2:14 PM • 2 Y

Y by Adventure10, Mango247

Observe $\angle AEF=\angle APF=\angle FBC$ so $BCEF$ is cyclic. Thus, $\angle PBA=\angle PCA$ . Likewise $\angle PAC=\angle PBC$ and $\angle PAB=\angle PCB$ . Let $Q$ be the isogonal conjugate of $P$ in $\triangle ABC$ . Then $QA=QB=QC$ hence $P=H$ is the orthocenter of $\triangle ABC$ . Now $AE-HE=AF-HF$ by Pithot's Theorem in $AEPF$ . Also $AE^2+HE^2=AF^2+HF^2=AH^2$ by Pythagoras Theorem. Solving yields that $\overline{AH}$ bisects angle $EAF$ . Now apply this cyclically: $H$ coincides with the incenter so $ABC$ is equilateral and $P$ is its center.

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#4 h Jul 18, 2018, 2:52 PM • 2 Y

Y by Adventure10, Mango247

My solution: Note that $\angle BFC = \angle ADC = 180^{\circ}-\angle BEC$ and $\angle AEF = \angle APF = \angle ABD$

$\Rightarrow BCEF$ is cyclic $\Rightarrow \angle BFC = \angle BEC = 90^{\circ} \Rightarrow P$ is the orthocenter of $\triangle ABC$ .

Now, Let $AP \cap EF = K$ and $T$ be the center of $\odot (AEPF) \Rightarrow K, T, I$ is collinear, where $I$ is the incenter of $AEPF$ .

$\Rightarrow I$ lies on $AP \Rightarrow AP$ bisects $\angle EAF$ and $\angle EPF$ .

Thus, $\angle AEF = \angle APF = \angle APE = 90^{\circ}-\angle EAP \Rightarrow AK \perp EF \Rightarrow EF \parallel BC$

But, $EF$ is antiparallel to $BC \Rightarrow \triangle ABC$ is isosceles with $AB = AC$ .

As the situation is symmetric, we get that $\triangle ABC$ is equilateral with $P$ as its orthocenter (i.e. its center).

This post has been edited 1 time. Last edited by math_pi_rate, Jul 18, 2018, 2:54 PM

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#5 h Jul 18, 2018, 7:28 PM • 2 Y

Y by Adventure10, Mango247

Just note that by PoP, we have $BCEF$ cyclic, and that $\angle AFP = \angle PDB = \angle PEC$ , so $\angle BEC = 90^\circ$ . So $P$ is the orthocenter. Also, by Pitot on $ABHC$ , we have that the orthocenter is an equal detour point of $ABC$ , which happens only if $ABC$ is equilateral. The rest is clear.

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#6 h Mar 23, 2023, 11:07 AM

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How is this a TST problem??

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