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 "2004 AIME I Problems/Problem 10"

(nicer solution)
(→‎Solution 3)
 
(4 intermediate revisions by 4 users not shown)
Line 14: Line 14:
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
[[Image:2004_I_AIME-10a.png]]
+
[[Image:2004_I_AIME-10.png]]
  
 
Let the bisector of <math>\angle CAD</math> be <math>AE</math>, with <math>E</math> on <math>CD</math>. By the angle bisector theorem, <math>DE = 36/5</math>. Since <math>\triangle AOR \sim \triangle AED</math> (<math>O</math> is the center of the circle), we find that <math>AR = 5</math> since <math>OR = 1</math>. Also <math>AT = 35</math> so <math>RT = OQ = 30</math>.
 
Let the bisector of <math>\angle CAD</math> be <math>AE</math>, with <math>E</math> on <math>CD</math>. By the angle bisector theorem, <math>DE = 36/5</math>. Since <math>\triangle AOR \sim \triangle AED</math> (<math>O</math> is the center of the circle), we find that <math>AR = 5</math> since <math>OR = 1</math>. Also <math>AT = 35</math> so <math>RT = OQ = 30</math>.
  
−
We can apply the same principle again to find that <math>PT = 27/2</math>, and since <math>QT = 1</math>, we find that <math>PQ = 27/2 - 1 = 25/2</math>. The locus of all possible centers of the circle on this "half" of the rectangle is triangle <math>\triangle OPQ</math>. There exists another congruent triangle that is symmetric over <math>AC</math> that has the same area as triangle <math>\triangle OPQ</math>. <math>\triangle APQ</math> has area <math>\frac {1}{2}\cdot OP \cdot PQ = \frac {1}{2}\cdot 30\cdot \frac {25}{2}</math>, since <math>\angle OQP</math> is right. Thus the total area that works is <math>30\cdot \frac {25}{2} = 375</math>, and the area of the locus of all centers of any circle with radius 1 is <math>34\cdot 14 = 442</math>. Hence, the desired probability is <math>\frac {375}{442}</math>, and our answer is <math>\boxed {817}</math>.
+
We can apply the same principle again to find that <math>PT = 27/2</math>, and since <math>QT = 1</math>, we find that <math>PQ = 27/2 - 1 = 25/2</math>. The locus of all possible centers of the circle on this "half" of the rectangle is triangle <math>\triangle OPQ</math>. There exists another congruent triangle that is symmetric over <math>AC</math> that has the same area as triangle <math>\triangle OPQ</math>. <math>\triangle APQ</math> has area <math>\frac {1}{2}\cdot OP \cdot PQ = \frac {1}{2}\cdot 30\cdot \frac {25}{2}</math>, since <math>\angle OQP</math> is right. Thus the total area that works is <math>30\cdot \frac {25}{2} = 375</math>, and the area of the locus of all centers of any circle with radius 1 is <math>34\cdot 13 = 442</math>. Hence, the desired probability is <math>\frac {375}{442}</math>, and our answer is <math>\boxed {817}</math>.
  
 
=== Solution 3 ===
 
=== Solution 3 ===
Line 32: Line 32:
 
The two values of <math>c</math> correspond to the triangle on top and below the diagonal. We are considering <math>A'B'C'</math> which is below, so <math>c = -\frac{13}{12}</math>. Then the equation of <math>\overline{A'C'}</math> is <math>y = \frac{5}{12}x - \frac{13}{12}</math>. Solving for its intersections with the lines <math>y = 1, x = 35</math> (boundaries of the internal rectangle), we find the coordinates of <math>A'B'C'</math> are at <math>A'\ (5,1)\ B'\ (35,1)\ C'\ (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>.
 
The two values of <math>c</math> correspond to the triangle on top and below the diagonal. We are considering <math>A'B'C'</math> which is below, so <math>c = -\frac{13}{12}</math>. Then the equation of <math>\overline{A'C'}</math> is <math>y = \frac{5}{12}x - \frac{13}{12}</math>. Solving for its intersections with the lines <math>y = 1, x = 35</math> (boundaries of the internal rectangle), we find the coordinates of <math>A'B'C'</math> are at <math>A'\ (5,1)\ B'\ (35,1)\ C'\ (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>.
  
−
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>.
+
Finally, the probability is <math>\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}</math>, and <math>m + n = \boxed{817}</math>.
 +
 
 +
For this solution, if you didn't know the formula for the distance from a point to a line, you can use similar triangles to get the ratio:
 +
 
 +
<math>\frac{1}{36}=\frac{c}{39}</math>
 +
 
 +
This yields <math>c=\frac{13}{12}</math>.
  
 
== See also ==
 
== See also ==
Line 38: Line 44:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 06:53, 15 October 2023

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

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 this triangle be $A'B'C'$.

Notice that $ABC$ and $A'B'C'$ share the same incenter; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of $\triangle ABC$ extend one unit farther than those of $\triangle A'B'C'$. From $A = rs$, we note that $r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6$. Thus $r_{A'B'C'} = r_{ABC} - 1 = 5$, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, $[A'B'C'] = [ABC] \cdot \left(\frac{r_{A'B'C'}}{r_{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}$.

The probability is $\frac{2[A'B'C']}{34 \times 13} = \frac{375}{442}$, and $m + n = \boxed{817}$.

Solution 2

2004 I AIME-10.png

Let the bisector of $\angle CAD$ be $AE$, with $E$ on $CD$. By the angle bisector theorem, $DE = 36/5$. Since $\triangle AOR \sim \triangle AED$ ($O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$. Also $AT = 35$ so $RT = OQ = 30$.

We can apply the same principle again to find that $PT = 27/2$, and since $QT = 1$, we find that $PQ = 27/2 - 1 = 25/2$. The locus of all possible centers of the circle on this "half" of the rectangle is triangle $\triangle OPQ$. There exists another congruent triangle that is symmetric over $AC$ that has the same area as triangle $\triangle OPQ$. $\triangle APQ$ has area $\frac {1}{2}\cdot OP \cdot PQ = \frac {1}{2}\cdot 30\cdot \frac {25}{2}$, since $\angle OQP$ is right. Thus the total area that works is $30\cdot \frac {25}{2} = 375$, and the area of the locus of all centers of any circle with radius 1 is $34\cdot 13 = 442$. Hence, the desired probability is $\frac {375}{442}$, and our answer is $\boxed {817}$.

Solution 3

2004 I AIME-10b.png

Again, 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, $B (36,0)$, $C (36,15)$, $D (0,15)$. The slope of $\overline{AC}$ is $\frac{15}{36} = \frac{5}{12}$. Let $\triangle A'B'C'$ be the right triangle with sides one unit inside $\triangle ABC$. Since $\overline{AC} || \overline{A'C'}$, they have the same slope, and the equation of $A'C'$ is $y = \frac{5}{12}x + c$. Manipulating, $5x - 12y + 12c = 0$. We need to find the value of $c$, which can be determined since $\overline{AC}$ is one unit away from $\overline{A'C'}$. Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:

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

The two values of $c$ correspond to the triangle on top and below the diagonal. We are considering $A'B'C'$ which is below, so $c = -\frac{13}{12}$. Then the equation of $\overline{A'C'}$ is $y = \frac{5}{12}x - \frac{13}{12}$. Solving for its intersections with the lines $y = 1, x = 35$ (boundaries of the internal rectangle), we find the coordinates of $A'B'C'$ are at $A'\ (5,1)\ B'\ (35,1)\ C'\ (35,\frac{27}{2})$. 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 $m + n = \boxed{817}$.

For this solution, if you didn't know the formula for the distance from a point to a line, you can use similar triangles to get the ratio:

$\frac{1}{36}=\frac{c}{39}$

This yields $c=\frac{13}{12}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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=2004_AIME_I_Problems/Problem_10&oldid=199662"