Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"

Revision as of 21:10, 9 September 2025

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 $\boxed{506}$ .

@@ Line 12: / Line 12: @@
 Note that <math>PA = 3</math> and <math>PD = 3 + 8 = 11</math>, so
 <cmath>[PAD] = \frac{PD}{PC} \cdot [PAC] = \frac{11}{3} \cdot \frac{189}{58} = \frac{693}{58}.</cmath>
-<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.
@@ Line 59: / Line 24: @@
 <cmath>[PAC] = 2 \cdot [CTP] = \frac{27}{10} \quad \text{and} \quad [PAD] = \frac{9}{2}.</cmath>
-<asy>
+The sum of the areas is thus <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>.
-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 <math>\frac{477}{29}</math> and the answer is <math>\boxed{506}</math>

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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 7"

Revision as of 21:10, 9 September 2025

Problem

Solution