2015 AIME II Problems/Problem 15
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 .
Hint
This is a #15 on an AIME, so it must be difficult. Indeed, although there are several possible approaches, all of them are very computational.
Solution 1
Call and the centers of circles and , respectively, and extend and to meet at point . Call and the feet of the altitudes from to and to , respectively. 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 , and let be a positive real number so that the negative slope of is preserved. Setting , the coordinates of are , and the coordinates of are . Using the point-to-line distance formula and the condition , we have If , then clearly and would not lie on the same side of . Thus since , 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 as solutions. 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 .
Solution 2
By homothety, we deduce that . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of and to .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from to is four times that from to . Let the distance from be and the distance from be .
Let and be the centers of their respective circles. Then dropping a perpendicular from to creates a right triangle, from which and, if , that . Then , and the Law of Cosines on triangles and gives and
Now, using the Pythagorean Theorem to express the length of the projection of onto line gives Squaring and simplifying gives and squaring and solving gives
By the Law of Sines on triangle , we have But we know , and so a small computation gives The Pythagorean Theorem now gives and so the common area is The answer is
Alternate Path to x
Call the intersection of lines and .You can use similar triangles to find that the distance from to is four times the distance from to . Then draw a perpendicular from to and call the point . and , so by the Pythagorean Theorem, . You can now use similar triangles to find that and continue on like in solution 2.
Solution 3
goes through , the point of tangency of both circles. So intercepts equal arcs in circle and : homothety. Hence, . We will use such similarity later.
The diagonal distance between the centers of the circles is . The difference in heights is . So .
The triangle connecting the centers with a side parallel to is a right triangle. Since , the height of is . Drop an altitude from to and call it : and . Since right , is a right triangle also; form a geometric progression .
Extend through to a point on the other side of . By homothety, . By angle chasing through right triangle , we deduce that is a right angle. Since is cyclic, is also right. So is a diameter of . Because of this, , the tangent line. is right and .
so and .
Since , the common area is . because the triangles are similar with a ratio of . So we only need to find now.
Extend through to intersect the tangent at . Because , the altitude from to is times the height from to . So and . We look at right triangle . and . is a right triangle. Hypotenuse intersects at a point, we call it . . So .
By Power of a Point, . So . The height from to is .
Thus, . The area of the whole cyclic quadrilateral is . Lastly, the common area is the area of the quadrilateral, or . So .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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