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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 8"

(Created page with "==Problem== There is a quadrilateral <math>ABCD</math> inscribed in a circle <math>\omega</math> with center <math>O</math>. In quadrilateral <math>ABCD</math>, diagonal <math...")
 
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==Solution==
 
==Solution==
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Since <math>EF = 3</math> and <math>\triangle EOF</math> is a <math>30^\circ</math>–<math>60^\circ</math>–<math>90^\circ</math> triangle, it follows that <cmath>EO = 2\sqrt{3},</cmath> so <cmath>BC = 4\sqrt{3}</cmath> and <cmath>AC = 8\sqrt{3}.</cmath>
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This means the radius of <math>\omega</math> is <math>4\sqrt{3}</math>. Since <math>\angle BOD</math> is a right angle by the inscribed angle theorem, it follows that <cmath>BD = 4\sqrt{6}.</cmath>
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Thus, <cmath>AC \cdot BD = 8\sqrt{3} \cdot 4\sqrt{6} = 96\sqrt{2},</cmath> and the answer is <cmath>96 + 2 = \boxed{98}.</cmath>

Revision as of 21:11, 9 September 2025

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 \[96 + 2 = \boxed{98}.\]

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