angle bisector wanted, 2 of 3 planes are perpendicular, 1 circle

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angle bisector wanted, 2 of 3 planes are perpendicular, 1 circle

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Source: 2017 Kurchatov Olympiad 11 p6 - Russia

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#1 h Mar 8, 2021, 8:14 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $A$ and $B$ be different points belonging to the line of intersection of the perpendicular planes $\pi_1$ and $\pi_2$ . Point $C$ belongs to the plane $\pi_2$ , but does not belong to $\pi_1$ . Let $P$ denote the intersection point of the bisector of the angle $ACB$ with line $AB$ and through $\omega$ a circle with diameter $AB$ in the plane $\pi_1$ . The plane $\pi_3$ containing $CP$ meets the circle $\omega$ at points $D$ and $E$ . Prove that $CP$ is the bisector of the angle $DCE$ .

This post has been edited 2 times. Last edited by parmenides51, Mar 8, 2021, 3:05 PM
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#2 h Jul 1, 2021, 7:47 PM

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Let the plane passing through $C$ and perpendicular to $CP$ intersect $AB$ and $DE$ at $Q$ and $R$ respectively. Obviously $QR$ is perpendicular to $AB$ and $QR$ is the polar of $P$ wrt $\omega$ . It means that the point $S=AD\cap BE$ lies on $QR$ so considering the central projection with center $S$ we get $\frac{DP}{PE}:\frac{DR}{RE}=\frac{AP}{PB}:\frac{AQ}{QB}=1$ . Combining this with the fact that $\angle PCR=90^\circ$ we conclude that $CP$ and $CR$ are the bisectors of the angle $DCE$ as desired.

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