AoPS - Problem

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

Tags: geometry, mixtilinear, fixed, fixed angle, semicircle

parmenides51

23.11.2020 01:30

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)

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