Question

Equilateral triangle $T$ is inscribed in circle $A,$ which has radius $10.$ Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T.$ Circles $C$ and $D,$ both with radius $2,$ are internally tangent to circle $A$ at the other two vertices of $T.$ Circles $B,$ $C,$ and $D$ are all externally tangent to circle $E,$ which has radius $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Place the center of circle $A$ at the origin with the triangle's vertices at $(0, 10)$ and $\left(\pm 5\sqrt{3}, -5\right).$ A circle internally tangent to $A$ at a vertex has its center on the radius to that vertex, so circle $B$ has center $(0, 7)$ and circles $C$ and $D$ have centers $\left(\mp 4\sqrt{3}, -4\right)$ (at distance $10 - 2 = 8$ from the origin).

By symmetry the center of circle $E,$ of radius $r,$ lies on the $y$ -axis at $(0, y).$ External tangency to $B$ gives $7 - y = r + 3,$ so $y = 4 - r.$ External tangency to $C$ gives $\left(4\sqrt{3}\right)^2 + (4 - r + 4)^2 = (r + 2)^2,$ that is, $48 + (8 - r)^2 = (r + 2)^2,$ which simplifies to $112 - 16r = 4r + 4,$ so $r = \frac{27}{5}.$

Then $m + n = 27 + 5 = 32.$

2009 AIME II Problem 5

Solution:

Problem 5 in Other Years