, 2018

Geometry Difficulty 8.1 Shortlist Prove it Saudi Arabia

Let $C$ be a point lies outside the circle $(O)$ and $CS, CT$ are tangent lines of $(O)$ . Take two points $A, B$ on $(O)$ with $M$ is the midpoint of the minor $\operatorname{arc} AB$ such that $A, B, M$ differ from $S, T$ . Suppose that $MS, MT$ cut line $AB$ at $E, F$ . Take $X \in OS$ and $Y \in OT$ such that $EX, FY$ are perpendicular to $AB$ . Prove that $XY$ and $CM$ are perpendicular.

Solution

First, note that $OM \perp AB$ then $OM \parallel XE$ . But $OMS$ is isosceles triangle implies that triangle $XES$ is also isosceles, or $XE = XS$ . Similarly, $YE = YT$ . Denote $\left(\omega_{1}\right), \left(\omega_{2}\right)$ as the circle of center $X$ , radius $XS$ and center $Y$ , radius $YT$ . Since $CS \perp XS$ , we have $CS$ is tangent to $\left(\omega_{1}\right)$ so $\mathscr{P}_{C /\left(\omega_{1}\right)} = CS^{2}$ . On the other hand, $\mathscr{P}_{C /\left(\omega_{2}\right)} = CT^{2}$ and $CS = CT$ imply that $C$ belongs to radical axis of two circles $\left(\omega_{1}\right), \left(\omega_{2}\right)$ .

By similar triangles, we get $MA^{2} = MS \cdot ME$ and $MB^{2} = MF \cdot MT$ , but $MA = MB$ then $MS \cdot ME = MF \cdot MT$ , thus $\mathscr{P}_{M /\left(\omega_{1}\right)} = \mathscr{P}_{M /\left(\omega_{2}\right)}$ . By combining these results, we obtain $CM$ is the radical axis of two circles $\left(\omega_{1}\right), \left(\omega_{2}\right)$ ; hence, $CM \perp XY$ .

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