2023 SSMO Speed Round Problems/Problem 10

Problem

In a circle centered at $O$ with radius $7,$ we have non-intersecting chords $AB$ and $DC.$ $O$ is outisde of quadrilateral $ABCD$ and $AB<CD.$ Let $X = AO\cup CD$ and $Y = BO\cup CD.$ Suppose that $XO+YO = 7$ . If $YC-DX=2$ and $XY = 3$ , then $AB = \frac{a\sqrt{b}}{c}$ for $\gcd(a,c) = 1$ and squareless $b.$ Find $a+b+c.$

Solution

Let $CX = x+2, DY=x$ .

Then, by power of the point we have that $(x+2)(x+3) = 49 - YO^2$ $(x+5)x = 49 - XO^2$ and subtracting gives that $XO^2 - YO^2 = 6$ . Since we know that $XO + YO = 7$ , dividing gives that $XO - YO = \frac{6}{7}$ so $XO = \frac{55}{14}$ and $YO = \frac{43}{14}$ .

Then, by law of cosines, it follows that $OY^2 + OX^2 - 2 \cdot OX \cdot OY \cos\angle YOX = YX^2$ which implies that $\cos\angle YOX = \frac{311}{473}$ .

Then, $AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cos\angle YOX$ which implies that $AB = \frac{126 \sqrt{473}}{473}$ so the answer is then $\boxed{1072}$ .

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2023 SSMO Speed Round Problems/Problem 10

Problem

Solution