AIME 1995 · 第 4 题

Solution 1

We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}$, and $O_6O_9 : O_9O_3 = 3:6 = 1:2$. Thus, $O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5$ (consider similar triangles). Applying the Pythagorean Theorem to $\triangle O_9A_9P$, we find that

$$PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}$$

Solution 2 (Analytic Geometry)

Let $A$ be defined as the origin of a coordinate plane with the $y$-axis running across the chord and $C(6\sqrt{2},0)$ by the Pythagorean Theorem. Then we have $D(0,-6)$ and $F(6\sqrt{2},-3)$, and since $\frac{DE}{DF}=\frac{1}{3}$, the point $E$ is one-third of the way from $D$ to $F$, so point $E$ has coordinates $(2\sqrt{2},-5)$. $E$ is the center of the circle with radius $9$, so the equation of this circle is $(x-2\sqrt{2})^2+(y+5)^2=81$. Since the chord's equation is $y=0$, we must find all values of $x$ satisfying the equation of the circle such that $y=0$. We find that $x-2\sqrt{2}=\pm\sqrt{56}$, so the chord has length $|\sqrt{56}+2\sqrt{2}-(-\sqrt{56}+2\sqrt{2})|=2\sqrt{56}$ and the answer is $(2\sqrt{56})^2=\boxed{224}$.

~eevee9406