2024 SSMO Relay Round 4 Problems/Problem 2

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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2024 SSMO Relay Round 4 Problems/Problem 2

Problem

Solution