Difference between revisions of "2015 AIME II Problems/Problem 15"
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− | + | Then we can find the coordinates of <math>D</math> by finding the point <math>(x,y)</math> other than <math>A = (0,0)</math> where the circle <math>\mathcal{P}</math> intersects <math>\ell</math>. <math>\mathcal{P}</math> can be represented with the equation <math>(x + 1)^2 + y^2 = 1</math>, and substituting <math>y = -\frac{3}{2}x</math> into this equation yields <cmath>x^2 + 2x + 1 + \frac{9}{4}x^2 = 1 \implies \frac{13}{4}x^2 + 2x = 0 \implies x = 0, -\frac{8}{13}.</cmath> Discarding <math>x = 0</math>, the <math>y</math>-coordinate of <math>D</math> is <math>-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}</math>. The distance from <math>D</math> to <math>A</math> is then <math>\sqrt{\left(\frac{-8}{13}\right)^2 + \left(\frac{12}{13}\right)^2} = \frac{4}{\sqrt{13}}.</math> The perpendicular distance from <math>B</math> to <math>AD</math> or the height of <math>\triangle{DBA}</math> is <math>\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.</math> Finally, the common area is <math>\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}</math>, and <math>m + n = 64 + 65 = \boxed{129}</math>. | |
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:49, 28 March 2015
Problem
Circles and have radii and , respectively, and are externally tangent at point . Point is on and point is on so that line is a common external tangent of the two circles. A line through intersects again at and intersects again at . Points and lie on the same side of , and the areas of and are equal. This common area is , where and are relatively prime positive integers. Find .
Solution
Call and the centers of circles and , respectively, and call and the feet of the altitudes from to and to , respectively. Extend and to meet at point . Using the fact that and setting , we have that . We can do some more length chasing using triangles similar to to get that , , and . Now, consider the circles and on the coordinate plane, where is the origin. If the line through intersects at and at , then . To verify this, notice that from the fact that both triangles are isosceles with , which are corresponding angles. Since , we can conclude that .
Hence, we need to find the slope of line such that the perpendicular distance from to is four times the perpendicular distance from to . This will mean that the product of the bases and heights of triangles and will be equal, which in turn means that their areas will be equal. Let the line have the equation . Then, the coordinates of are , and the coordinates of are . Using the point-to-line distance formula and the fact that , we have Since takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have Thus, the equation of is .
Then we can find the coordinates of by finding the point other than where the circle intersects . can be represented with the equation , and substituting into this equation yields Discarding , the -coordinate of is . The distance from to is then The perpendicular distance from to or the height of is Finally, the common area is , and .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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