2005 AIME I Problems/Problem 11

Problem

A semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution

We note that aligning the base of the semicircle with a side of the square is certainly non-optimal. If the semicircle is tangent to only one side of the square, we will have "wiggle-room" to increase its size. Once it is tangent to two adjacent sides of the square, we will maximize its size when it touches both other sides of the square. This can happen only when it is arranged so that the center of the semicircle lies on one diagonal of the square.

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Now, let the square be $ABCD$ , and let $E \in AB$ and $F \in DA$ be the points at which the "corners" of the semicircle touch the square. Then by the comments above, $AE = AF = a$ . By the Pythagorean Theorem, $d^2 = 2a^2$ .

Now, if we draw a line through the center, $O$ , of the semicircle and its point of tangency with $BC$ , we see that this line is perpendicular to $BC$ and so parallel to $AB$ . Thus, by triangle similarity it cuts $AF$ in half, and so by symmetry the distance from $O$ to $AD$ is $\frac{a}{2}$ and so the distance from $O$ to $BC$ is $8 - \frac a2$ . But this latter quantity is also the radius of the semicircle, so $d = 16 - a$ .

Our two previous paragraphs give $2a^2 = (16 - a)^2$ so $a^2 + 32a - 256 = 0$ and $a = 16\sqrt{2} - 16$ (where we discard the negative root of that quadratic) and so $d = a\sqrt{2} = 32 - 16\sqrt{2} = 32 - \sqrt{512}$ , so the answer is $32 + 512 = 544$ .

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2005 AIME I Problems/Problem 11

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