Difference between revisions of "2004 AIME I Problems/Problem 10"
(analytical geometry solution, let's see if I can find a non-coordinate solution..) |
(revise my solution, solution 1 credit to chess64) |
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== Solution == | == Solution == | ||
− | [[Image:2004_I_AIME-10.png | + | [[Image:2004_I_AIME-10.png]] |
=== Solution 1 === | === Solution 1 === | ||
+ | [[Image:2004_I_AIME-10a.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>. | ||
+ | |||
+ | 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>. | ||
=== Solution 2 === | === Solution 2 === | ||
− | [[Image:2004_I_AIME-10b.png | + | [[Image:2004_I_AIME-10b.png]] |
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>. | 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>. | ||
− | Let <math>A</math> be at the origin, <math>B | + | Let <math>A</math> be at the origin, <math>B (36,0)</math>, <math>C (36,15)</math>, <math>D (0,15)</math>. The slope of <math>\overline{AC}</math> is <math>\frac{15}{36} = \frac{5}{12}</math>. Let <math>\triangle A'B'C'</math> be the right triangle with sides one unit inside <math>\triangle ABC</math>. Since <math>\overline{AC} || \overline{A'C'}</math>, they have the same slope, and the equation of <math>A'C'</math> 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 since <math>\overline{AC}</math> is one unit away from <math>\overline{A'C'}</math>. Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin: |
− | <cmath>\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1 | + | <cmath>\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</cmath> |
− | |||
<cmath>c = \pm \frac{13}{12}</cmath> | <cmath>c = \pm \frac{13}{12}</cmath> | ||
− | + | 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>. |
Revision as of 20:07, 13 October 2007
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
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 2
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 .
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 |