Difference between revisions of "2010 AIME I Problems/Problem 13"

Revision as of 16:36, 13 March 2015

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

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Solution 1

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:

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

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 $h$ .

$\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}$

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 $63 + 6 = \boxed{069}$ .

Solution 2

Let $O$ be the center of the semicircle. It follows that $AU + UO = AN = NO = 126$ , so triangle $ANO$ is equilateral.

Let $Y$ be the foot of the altitude from $N$ , such that $NY = 63\sqrt{3}$ and $NU = 21$ .

Finally, denote $DT = a$ , and $AD = x$ . Extend $U$ to point $Z$ so that $Z$ is on $CD$ and $UZ$ is perpendicular to $CD$ . It then follows that $ZT = a-84$ . Since $NYU$ and $UZT$ are similar,

$\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}$

Given that line $NT$ divides $R$ into a ratio of $1:2$ , we can also say that

$(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})$

where the first term is the area of trapezoid $AUTD$ , the second and third terms denote the areas of $\frac{1}{6}$ a full circle, and the area of $NUO$ , respectively, and the fourth term on the right side of the equation is equal to $R$ . Cancelling out the $\frac{126^2\pi}{6}$ on both sides, we obtain

$(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})$

By adding and collecting like terms, $\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})$

$\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})$ .

Since $a - 84 = \frac{x}{3\sqrt{3}}$ ,

$\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})$

$\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})$

$x^2 = (63)(126)(3) = (2)(3^5)(7^2)$

$x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}$ , so the answer is $\boxed{069}.$

Solution 3

Note that the total area of $\mathcal{R}$ is $252DA + \frac {126^2 \pi}{2}$ and thus one of the regions has area $84DA + \frac {126^2 \pi}{6}$

As in the above solutions we discover that $\angle AON = 60^\circ$ , thus sector $ANO$ of the semicircle has $\frac{1}{3}$ of the semicircle's area.

Similarly, dropping the $N'T'$ perpendicular we observe that $[AN'T'D] = 84DA$ , which is $\frac{1}{3}$ of the total rectangle.

Denoting the region to the left of $\overline {NT}$ as $\alpha$ and to the right as $\beta$ , it becomes clear that if $[\triangle UT'T] = [\triangle NUO]$ then the regions will have the desired ratio.

Using the 30-60-90 triangle, the slope of $NT$ , is ${-3\sqrt{3}}$ , and thus $[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}$ .

$[NUO]$ is most easily found by $\frac{absin(c)}{2}$ : $[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}$

Equating, $\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}$

Solving, $63 * 21 * 3 * 6 = DA^2$

$DA = 63 \sqrt{6} \longrightarrow \boxed {069}$

@@ Line 84: / Line 84: @@
 Now just solve for <math>h</math>.
-<math>\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\
+<cmath>\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\
 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\
 h^2 & = \frac {9}{4}(6) \\
-h & = \frac {3}{2}\sqrt {6} \end{align*}</math>
+h & = \frac {3}{2}\sqrt {6} \end{align*}</cmath>
 Don't forget to un-rescale at the end to get <math>AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}</math>.

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