Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
 
(13 intermediate revisions by 6 users not shown)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
 +
===Diagram===
 
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
 
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
 
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500);
 
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500);
Line 66: Line 67:
 
<math>Y = \frac {1}{3}\pi(3)^2 = 3\pi</math>
 
<math>Y = \frac {1}{3}\pi(3)^2 = 3\pi</math>
  
−
To find <math>Z</math> we have to find the length of <math>TC</math>. Project <math>T</math> and <math>N</math> onto <math>AB</math> to get points <math>T'</math> and <math>N'</math>. Notice that <math>UNN'</math> and <math>UTT'</math> are similar. Thus:
+
To find <math>Z</math> we have to find the length of <math>TC</math>. Project <math>T</math> and <math>N</math> onto <math>AB</math> to get points <math>T'</math> and <math>N'</math>. Notice that <math>UNN'</math> and <math>TUT'</math> are similar. Thus:
  
−
<math>\frac {UT'}{TT'} = \frac {NN'}{UN'} \implies \frac {UT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies UT' = \frac {\sqrt {3}}{9}h</math>.
+
<math>\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h</math>.
  
−
Then <math>TC = T'B = UB - UT' = 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
Line 146: Line 147:
  
 
<math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math>
 
<math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math>
 +
 +
=== Solution 4 (Coordinates) ===
 +
Like above solutions, note that <math>ANO</math> is equilateral with side length <math>126,</math> where <math>O</math> is the midpoint of <math>AB.</math> Then, if we let <math>DA=a</math> and set origin at <math>D=(0,0),</math> we get <math>N=(63,a+63\sqrt{3}), U=(84,a).</math> Line <math>NU</math> is then <math>y-a=\sqrt{27}(x-84),</math> so it intersects <math>CA,</math> the <math>x</math>-axis, at <math>x=(a/\sqrt{27}+84),</math> giving us point <math>T.</math> Now the area of region <math>R</math> is <math>252a+\pi(126)^2 / 2,</math> so one third of that is <math>84a+\pi(126)^2 / 6.</math>
 +
 +
The area of the smaller piece of <math>R</math> is <math>[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}</math>
 +
<math>=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.</math>
 +
Setting this equal to <math>84a+\pi(126)^2 / 6</math> and canceling the <math>84a + \pi(126)^2</math> yields
 +
<math>\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},</math> so <math>a = 63 \sqrt{6}</math> and the anser is <math>\boxed{069}.</math>
 +
~ rzlng
 +
 +
=== Solution 5 (Minimal calculation) ===
 +
 +
Once we establish that <math>\Delta ANO</math> is equilateral, we have
 +
<cmath>[{\rm Sector } BON] = 2[{\rm Sector } AON],  [BCT'U]=2[ADT'U]</cmath>
 +
<cmath>\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]</cmath>
 +
On the other hand,
 +
<cmath>[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]</cmath>
 +
Therefore, <math>[UT'T]=[NUO]</math>.
 +
 +
Now, <math>UO=42, NU=21 \Rightarrow [UT'T]=[NUO]=2[NN'U]</math>. Also <math>\Delta UT'T \sim \Delta NN'U</math>. Therefore,
 +
<cmath>DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed {069}</cmath>
 +
 +
~asops
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:05, 20 October 2021

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

Diagram

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

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

Solution 4 (Coordinates)

Like above solutions, note that $ANO$ is equilateral with side length $126,$ where $O$ is the midpoint of $AB.$ Then, if we let $DA=a$ and set origin at $D=(0,0),$ we get $N=(63,a+63\sqrt{3}), U=(84,a).$ Line $NU$ is then $y-a=\sqrt{27}(x-84),$ so it intersects $CA,$ the $x$-axis, at $x=(a/\sqrt{27}+84),$ giving us point $T.$ Now the area of region $R$ is $252a+\pi(126)^2 / 2,$ so one third of that is $84a+\pi(126)^2 / 6.$

The area of the smaller piece of $R$ is $[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}$ $=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.$ Setting this equal to $84a+\pi(126)^2 / 6$ and canceling the $84a + \pi(126)^2$ yields $\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},$ so $a = 63 \sqrt{6}$ and the anser is $\boxed{069}.$ ~ rzlng

Solution 5 (Minimal calculation)

Once we establish that $\Delta ANO$ is equilateral, we have \[[{\rm Sector } BON] = 2[{\rm Sector } AON], [BCT'U]=2[ADT'U]\] \[\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]\] On the other hand, \[[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]\] Therefore, $[UT'T]=[NUO]$.

Now, $UO=42, NU=21 \Rightarrow [UT'T]=[NUO]=2[NN'U]$. Also $\Delta UT'T \sim \Delta NN'U$. Therefore, \[DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed {069}\]

~asops

See Also

  • <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_13&oldid=163829"