2010 AIME I Problems/Problem 13

Problem

Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ , $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ , $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .

Solution

The center of the semicircle is also the midpoint of $AB$ . Let this point be O. Let $h$ be the length of $AD$ .

Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$ . Then $AB = 6$ so $OA = OB = 3$ .

Since $ON$ is a radius of the semicircle, $ON = 3$ . Thus $OAN$ is an equilateral triangle.

Let $X$ , $Y$ , and $Z$ be the areas of triangle $OUN$ , sector $ONB$ , and trapezoid $UBCT$ respectively.

$X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}$

$Y = \frac {1}{3}\pi(3)^2 = 3\pi$

To find $Z$ we have to find the length of $TC$ . Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$ . Notice that $UNN'$ and $UTT'$ are similar. Thus:

$\frac {UT'}{TT'} = \frac {UN'}{NN'} \implies \frac {UT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies UT' = \frac {\sqrt {3}}{9}h$ .

Then $TC = T'B = UB - UT' = 4 - \frac {\sqrt {3}}{9}h$ . So:

$Z = \frac {1}{2}(BX + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2$

Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$ . Then

$L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2$

Obviously, the $L$ is greater than the area on the other side of line $l$ . This other area is equal to the total area minus $L$ . Thus:

$\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L$ .

Now just solve for $h$ .

$ $\begin{aligned} 12 h + 9 π & = \frac{9}{4} \sqrt{3} + 9 π + 12 h - \frac{\sqrt{3}}{6} h^{2} \\ 0 & = \frac{9}{4} \sqrt{3} - \frac{\sqrt{3}}{6} h^{2} \\ h^{2} & = \frac{9}{4} (6) \\ h & = \frac{3}{2} \sqrt{6} \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

Don't forget to un-rescale at the end to get $AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}$ .

Finally, the answer is $63 + 6 = \boxed{069}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2010 AIME I Problems/Problem 13

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