fixed angle wanted, 2 mixtlinears created by a perpendicular on semicircle

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fixed angle wanted, 2 mixtlinears created by a perpendicular on semicircle

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Source: 2020 Austrian Federal Competition For Advanced Students, Part 2, p5

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#1 h Nov 22, 2020, 10:30 PM

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Let $h$ be a semicircle with diameter $AB$ . Let $P$ be an arbitrary point inside the diameter $AB$ . The perpendicular through $P$ on $AB$ intersects $h$ at point $C$ . The line $PC$ divides the semicircular area into two parts. A circle will be inscribed in each of them that touches $AB, PC$ and $h$ . The points of contact of the two circles with $AB$ are denoted by $D$ and $E$ , where $D$ lies between $A$ and $P$ . Prove that the size of the angle $DCE$ does not depend on the choice of $P$ .

(Walther Janous)

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#2 h Oct 1, 2021, 12:07 PM • 1 Y

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Given $A(-a,0),B(a,0),O(0,0)$ ; choose $P(\lambda,0)$ and $C(\lambda,\sqrt{a^{2}-\lambda^{2}})$ .

Midpoints $O_{1}(\lambda-u,u)$ and $O_{2}(\lambda+v,v)$ , then $D(\lambda-u,0)$ and $E(\lambda+v,0)$ .
$OO_{1}=a-u \Rightarrow u=\lambda-a+\sqrt{2a^{2}-2a\lambda}$ .
$OO_{2}=a-v \Rightarrow v=-\lambda-a+\sqrt{2a^{2}+2a\lambda}$ .
$D(a-\sqrt{2a^{2}-2a\lambda},0)$ and $E(-a+\sqrt{2a^{2}+2a\lambda},0)$ .

Slope of the line $CD\ :\ m_{CD}=\frac{\sqrt{a^{2}-\lambda^{2}}}{\lambda-a+\sqrt{2a^{2}-2a\lambda}}$ and slope of the line $CE\ :\ m_{CE}=\frac{\sqrt{a^{2}-\lambda^{2}}}{\lambda+a-\sqrt{2a^{2}+2a\lambda}}$ .
$\tan \angle DCE=\frac{m_{CE}-m_{CD}}{1+m_{CE} \cdot m_{CD}}=1 \Rightarrow \angle DCE=\frac{\pi}{4}$ .

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