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

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Then <math>TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h</math>. So:
 
Then <math>TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h</math>. So:
  
−
<math>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</math>
+
<math>Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2</math>
  
 
Let <math>L</math> be the area of the side of line <math>l</math> containing regions <math>X, Y, Z</math>. Then
 
Let <math>L</math> be the area of the side of line <math>l</math> containing regions <math>X, Y, Z</math>. Then

Revision as of 17:27, 5 March 2017

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

[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label("$A$",(-16.43287,-9.3374),NE/2); dot((252,0)); label("$B$",(255.242,5.00321),NE/2); dot((0,-154.31785)); label("$D$",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label("$C$",(255.242,-149.55669),NE/2); dot((126,0)); label("$O$",(129.36332,5.00321),NE/2); dot((63,109.1192)); label("$N$",(44.91307,108.57427),NE/2); label("$126$",(28.18236,40.85473),NE/2); dot((84,0)); label("$U$",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label("$T$",(116.61611,-149.55669),NE/2); dot((63,0)); label("$N'$",(66.42398,5.00321),NE/2); label("$84$",(41.72627,-12.5242),NE/2); label("$168$",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); label("$T'$",(87.13819,-149.55669),NE/2); dot((252,0)); label("$I$",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle); [/asy]

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 $TUT'$ are similar. Thus:

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

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

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

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

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

See Also

  • <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (Problems • Answer Key • Resources)
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