Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

A circle $\omega_{1}$ of radius $15$ intersects a circle $\omega_{2}$ of radius $13$ at points $P$ and $Q$ . Point $A$ is on line $PQ$ such that $P$ is between $A$ and $Q$ . $R$ and $S$ are the points of tangency from $A$ to $\omega_{1}$ and $\omega_{2}$ , respectively, such that the line $AS$ does not intersect $\omega_{1}$ and the line $AR$ does not intersect $\omega_{2}$ . If $PQ = 24$ and $\angle RAS$ has a measure of $90^{\circ}$ , compute the length of $AR$ .

Solution

Solution:

Let $O_{1}$ be the center of $\omega_{1}$ and $O_{2}$ be the center of $\omega_{2}$ . Then $O_{1}O_{2}$ and $PQ$ are perpendicular. Let their point of intersection be $X$ . Using the Pythagorean theorem, the fact that $PQ = 24$ , and our knowledge of the radii of the circles, we can compute that $O_{1}X = 9$ and $O_{2}X = 5$ , so $O_{1}O_{2} = 14$ .

Let $SO_{1}$ and $RO_{2}$ meet at $Y$ . Then $SARY$ is a square, say of side length $s$ . Then $O_{1}Y = s - 15$ and $O_{2}Y = s - 13$ . So, $O_{1}O_{2}Y$ is a right triangle with sides $14$ , $s-15$ , and $s-13$ . By the Pythagorean theorem,
$(s-13)^2 + (s-15)^2 = 14^2.$
We can write this as $2s^2 - 4 \cdot 14 s + 198 = 0$ , or $s^2 - 28s + 99 = 0$ . The quadratic formula then gives
$s = \frac{28 \pm \sqrt{388}}{2} = 14 \pm \sqrt{97}.$
Since $14 - \sqrt{97} < 15$ and $YO_{1} > 15$ , we can discard the root of $14 - \sqrt{97}$ , and the answer is therefore $14 + \sqrt{97}$ .

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