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2023 SSMO Accuracy Round Problems/Problem 7

Revision as of 21:09, 9 September 2025 by Pinkpig (talk | contribs)

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 \[[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.\]

[asy] size(4cm); pair O = (0,0); pair P = (7.6158,0); filldraw(circle(O,3), lightgreen, green); filldraw(circle(O,5), lightgreen, green); pair A = (6.9, 2.6); pair C = (6.9, -2.6); pair B = (9.7, 3.8); pair D = (9.7, -3.8); pair M = (4.1, 1.6); pair N = (4.1, -1.6); pair T = (3.5, 0); draw(B--P--D, blue); draw(N--O--M, dashed+green); draw(C--A, dashed+blue); draw(O--P, dashed+blue); dot("$A$", A, dir(345)); dot("$C$", C, dir(55)); dot("$B$", B, dir(240)); dot("$D$", D, dir(135)); dot("$M$", M, dir(330)); dot("$N$", N, dir(60)); dot("$T$", T, dir(135)); dot("$P$", P, dir(45)); dot("$O$", O, dir(45)); clip((20,20)--(-20,20)--(-20,-20)--(20,-20)--cycle); [/asy]

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 \[[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.\]

[asy] size(4cm); pair O = (0,0); pair P = (3.16, 0); path outer = Circle(O, 5); path inner = Circle(O, 3); filldraw(outer, lightgreen, green); filldraw(inner, lightgreen, green); pair A = (2.2, 2.6); pair B = (5, 6); pair C = (2.2, -2.6); pair D = (5, -6); pair M = (1.3, 1.6); pair N = (1.3, -1.6); pair T = (1.1, 0); draw(A--B, blue); draw(C--D, blue); draw(N--O--M, dashed+green); draw(A--C, blue+dashed); draw(O--P, blue+dashed); dot("$A$", A, NE); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SE); dot("$M$", M, N); dot("$N$", N, S); dot("$T$", T, dir(0)); dot("$O$", O, dir(135)); dot("$P$", P, dir(45)); [/asy]

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

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