2023 SSMO Accuracy Round Problems/Problem 7

Problem

Concentric circles $\omega$ and $\omega_1$ are drawn, with radii $3$ and $5,$ respectively. Chords $AB$ and $CD$ of $\omega_1$ are both tangent to $\omega$ and intersect at $P.$ If $PA=PC = 3,$ then the sum of all possible distinct values of $[PAD]$ can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

There are two cases. First, consider the case where $P$ lies outside the circle.

Using the Pythagorean theorem, we find that $AB = CD = 2\sqrt{5^2 - 3^2} = 8$ . We also have $OP = \sqrt{3^2 + 7^2} = \sqrt{58}$ .

Let $T$ be the midpoint of $\overline{CA}$ , which lies on $\overline{PA}$ by symmetry. Then $\triangle NOP \sim \triangle CTP$ , so since $[NOP] = \frac{1}{2} \cdot 3 \cdot 7 = \frac{21}{2},$ it follows that $[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{21}{2} \cdot \frac{9}{58} = \frac{189}{116},$ and thus $[PAC] = 2 \cdot [CTP] = \frac{189}{58}.$

Note that $PA = 3$ and $PD = 3 + 8 = 11$ , so $[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.$

We now proceed with the other case.

We solve to get $OP = \sqrt{10}$ and $PM = PN = 1$ .

Once again, $\triangle NOP \sim \triangle CTP$ , and since $[NOP] = \frac{1}{2} \cdot 3 \cdot 1 = \frac{3}{2},$ it follows that $[CTP] = [NOP] \cdot \left(\frac{CP}{OP}\right)^2 = \frac{3}{2} \cdot \frac{9}{10} = \frac{27}{20},$ so $[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.$

The sum of the areas is thus $\frac{477}{29}$ and the answer is $\boxed{506}$ .

~SMO_Team

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2023 SSMO Accuracy Round Problems/Problem 7

Problem

Solution