Difference between revisions of "2004 AIME I Problems/Problem 10"
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=== Solution 2 === | === Solution 2 === | ||
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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 | + | 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 === | ||
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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 = 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 == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:46, 28 March 2016
Problem
A circle of radius 1 is randomly placed in a 15-by-36 rectangle so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal is where and are relatively prime positive integers. Find
Solution
Solution 1
The location of the center of the circle must be in the rectangle that is one unit away from the sides of rectangle . We want to find the area of the right triangle with hypotenuse one unit away from . Let this triangle be .
Notice that and share the same incenter; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of extend one unit farther than those of . From , we note that . Thus , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, .
The probability is , and .
Solution 2
Let the bisector of be , with on . By the angle bisector theorem, . Since ( is the center of the circle), we find that since . Also so .
We can apply the same principle again to find that , and since , we find that . The locus of all possible centers of the circle on this "half" of the rectangle is triangle . There exists another congruent triangle that is symmetric over that has the same area as triangle . has area , since is right. Thus the total area that works is , and the area of the locus of all centers of any circle with radius 1 is . Hence, the desired probability is , and our answer is .
Solution 3
Again, the location of the center of the circle must be in the rectangle that is one unit away from the sides of rectangle . We want to find the area of the right triangle with hypotenuse one unit away from .
Let be at the origin, , , . The slope of is . Let be the right triangle with sides one unit inside . Since , they have the same slope, and the equation of is . Manipulating, . We need to find the value of , which can be determined since is one unit away from . Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:
The two values of correspond to the triangle on top and below the diagonal. We are considering which is below, so . Then the equation of is . Solving for its intersections with the lines (boundaries of the internal rectangle), we find the coordinates of are at . The area is .
Finally, the probability is , and .
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:
This yields .
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 |
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