Difference between revisions of "2004 AIME I Problems/Problem 10"

(should have a solution soon)
(analytical geometry solution, let's see if I can find a non-coordinate solution..)
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== Problem ==
 
== Problem ==
A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely within the rectangle. Given that the p[[robability]] that the circle will not touch diagonal <math> AC </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
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A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely within the rectangle. Given that the [[probability]] that the circle will not touch diagonal <math> AC </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m + n. </math>
  
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__TOC__
 
== Solution ==
 
== Solution ==
[[Image:2004_I_AIME-10.png]]
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[[Image:2004_I_AIME-10.png|500px]]
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=== Solution 1 ===
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=== Solution 2 ===
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[[Image:2004_I_AIME-10b.png|500px]]
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The location of the center of the circle must be in the <math>34 \times 13</math> rectangle that is one unit away from the sides of rectangle <math>ABCD</math>. We want to find the area of the [[right triangle]] with [[hypotenuse]] one unit away from <math>\overline{AC}</math>.
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Let <math>A</math> be at the origin, <math>B</math> at <math>(36,0)</math>, <math>C</math> at <math>(36,15)</math>, <math>D</math> at <math>(0,15)</math>. The slope of diagonal <math>\overline{AC}</math> is <math>\frac{15}{36} = \frac{5}{12}</math>. Since the hypotenuse is parallel to the diagonal, it has the same slope, and its equation is <math>y = \frac{5}{12}x + c</math>. Manipulating, <math>5x - 12y + 12c = 0</math>. We need to find the value of <math>c</math>, which can be determined using the fact that the hypotenuse is one unit away from the diagonal. Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:
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<cmath>\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1</cmath>
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<cmath>\left|\frac{(5)(0) + (-12)(0) + 12c}{\sqrt{5^2 + (-12)^2}}\right| = 1</cmath>
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<cmath>c = \pm \frac{13}{12}</cmath>
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It makes sense that we have two values of <math>c</math>, one for the triangle on top of the diagonal, and one for the bottom. We will just consider the bottom triangle, so <math>c = -\frac{13}{12}</math>. Then the equation of the line is <math>y = \frac{5}{12}x - \frac{13}{12}</math>. Solving for its intersections with the lines <math>y = 1, x = 35</math> we find the coordinates of the triangles are at <math>(5,1)(35,1)(35,\frac{27}{2})</math>. The area is <math>\frac{1}{2}bh = \frac{1}{2}(35-5)\left(\frac{27}{2} - 1\right) = \frac{375}{2}</math>.
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Finally, the probability is <math>\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}</math>, and <math>m + n = 817</math>. 
  
 
== See also ==
 
== See also ==
{{AIME box|year=2004|num-b=9|num-a=11}}
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{{AIME box|year=2004|num-b=9|num-a=11|n=I}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 19:52, 13 October 2007

Problem

A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

2004 I AIME-10.png

Solution 1

Solution 2

2004 I AIME-10b.png

The location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$. We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$.

Let $A$ be at the origin, $B$ at $(36,0)$, $C$ at $(36,15)$, $D$ at $(0,15)$. The slope of diagonal $\overline{AC}$ is $\frac{15}{36} = \frac{5}{12}$. Since the hypotenuse is parallel to the diagonal, it has the same slope, and its equation is $y = \frac{5}{12}x + c$. Manipulating, $5x - 12y + 12c = 0$. We need to find the value of $c$, which can be determined using the fact that the hypotenuse is one unit away from the diagonal. Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:

\[\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1\] \[\left|\frac{(5)(0) + (-12)(0) + 12c}{\sqrt{5^2 + (-12)^2}}\right| = 1\] \[c = \pm \frac{13}{12}\]

It makes sense that we have two values of $c$, one for the triangle on top of the diagonal, and one for the bottom. We will just consider the bottom triangle, so $c = -\frac{13}{12}$. Then the equation of the line is $y = \frac{5}{12}x - \frac{13}{12}$. Solving for its intersections with the lines $y = 1, x = 35$ we find the coordinates of the triangles are at $(5,1)(35,1)(35,\frac{27}{2})$. The area is $\frac{1}{2}bh = \frac{1}{2}(35-5)\left(\frac{27}{2} - 1\right) = \frac{375}{2}$.

Finally, the probability is $\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}$, and $m + n = 817$.

See also

2004 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions
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