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Difference between revisions of "2024 SSMO Relay Round 4 Problems/Problem 2"

(Created page with "==Problem== Let <math>T = TNYWR.</math> Regular octagon <math>OLYMPIAD</math> is perfectly inscribed within Circle <math>Q</math>. Circle <math>Q</math> has area <math>T\pi</...")
 
 
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==Solution==
 
==Solution==
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Since the area of circle <math>Q</math> is <math>100\pi,</math> the radius of circle <math>Q</math> is <math>10.</math> Let the center of circle <math>Q</math> be <math>Q_1.</math> From the Law of Cosines on <math>OQ_1L,</math> we have <cmath>OL^2 = 10^2+10^2-2\cdot\left(\frac{\sqrt{2}}{2}\right)(10)(10) = 200-100\sqrt{2}.</cmath> Now, since the area of an octagon with sidelength <math>s</math> is <math>s^2(2+2\sqrt{2}),</math> we have <cmath>[OLYMPIAD] = (200-100\sqrt{2})(2+2\sqrt{2}) = 200\sqrt{2}\implies 200+2 = \boxed{202}.</cmath>
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~SMO_Team

Latest revision as of 14:47, 10 September 2025

Problem

Let $T = TNYWR.$ Regular octagon $OLYMPIAD$ is perfectly inscribed within Circle $Q$. Circle $Q$ has area $T\pi$. If the area of octagon $OLYMPIAD$ is $a\sqrt{b},$ for squarefree $b,$ find $a+b.$

Solution

Since the area of circle $Q$ is $100\pi,$ the radius of circle $Q$ is $10.$ Let the center of circle $Q$ be $Q_1.$ From the Law of Cosines on $OQ_1L,$ we have \[OL^2 = 10^2+10^2-2\cdot\left(\frac{\sqrt{2}}{2}\right)(10)(10) = 200-100\sqrt{2}.\] Now, since the area of an octagon with sidelength $s$ is $s^2(2+2\sqrt{2}),$ we have \[[OLYMPIAD] = (200-100\sqrt{2})(2+2\sqrt{2}) = 200\sqrt{2}\implies 200+2 = \boxed{202}.\]

~SMO_Team

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