2023 SSMO Accuracy Round Problems/Problem 8

Problem

There is a quadrilateral $ABCD$ inscribed in a circle $\omega$ with center $O$ . In quadrilateral $ABCD$ , diagonal $AC$ is a diameter of the circle, $\angle BAC = 30^\circ,$ and $\angle DAC = 15^\circ.$ Let $E$ be the base of the altitude from $O$ onto side $BA$ . Let $F$ be the base of the altitude from $E$ onto $BO$ . Given that $EF = 3,$ and that the product of the lengths of the diagonals of $ABCD$ is $a\sqrt{b},$ for some squarefree $b,$ find $a+b.$

Solution

Since $EF = 3$ and $\triangle EOF$ is a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle, it follows that $EO = 2\sqrt{3},$ so $BC = 4\sqrt{3}$ and $AC = 8\sqrt{3}$ .

This means the radius of $\omega$ is $4\sqrt{3}$ . Since $\angle BOD$ is a right angle by the inscribed angle theorem, it follows that $BD = 4\sqrt{6}.$

Thus, $AC \cdot BD = 8\sqrt{3} \cdot 4\sqrt{6} = 96\sqrt{2},$ and the answer is $96 + 2 = \boxed{98}.$

~SMO_team

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2023 SSMO Accuracy Round Problems/Problem 8

Problem

Solution