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1995 AIME Problems/Problem 4

Problem

Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$. The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(A,9)); D(CR(B,3)); D(CR(C,6)); D(P--Q); [/asy]

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

[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP("O_9",A)),9)); D(CR(D(MP("O_3",B)),3)); D(CR(D(MP("O_6",C)),6)); D(MP("P",P,NW)--MP("Q",Q,NE)); D((-9,0)--(9,0)); D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); [/asy]

Solution 2 (Analytic Geometry)

[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP("E",A)),9)); D(CR(D(MP("F",B)),3)); D(CR(D(MP("D",C)),6)); D((-9,0)--(9,0)); D(MP("",P,NW)--MP("",Q,NE)); D(A--MP("B",G,N)); D(B--MP("C",F,N)); D(C--MP("A",D,N)); D(rightanglemark(A,G,P,12)); D(rightanglemark(C,D,P,12)); D(rightanglemark(B,F,P,12)); [/asy]

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

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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