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Difference between revisions of "1995 AIME Problems/Problem 4"

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== Solution ==
 
== Solution ==
−
We label the points as following: the centers of the circles of radii <math>3,6,9</math> are <math>O_3,O_6,O_9</math> respectively, and the endpoints of the chord are <math>P,Q</math>. Let <math>A_3,A_6,A_9</math> be the feet of the [[perpendicular]]s from <math>O_3,O_6,O_9</math> to <math>\overline{PQ}</math> (so <math>A_3,A_6</math> are the points of [[tangent (geometry)|tangency]]). Then we note that <math>\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}</math>, and <math>O_6O_9 : O_9O_3 = 3:6 = 1:2</math>. Thus, <math>O_9A_9 = \frac{1 \cdot O_6A_6 + 2 \cdot O_3A_3}{3} = 5</math> (consider similar triangles). Applying the [[Pythagorean Theorem]] to <math>\triangle O_9A_9P</math>, we find that  
+
We label the points as following: the centers of the circles of radii <math>3,6,9</math> are <math>O_3,O_6,O_9</math> respectively, and the endpoints of the chord are <math>P,Q</math>. Let <math>A_3,A_6,A_9</math> be the feet of the [[perpendicular]]s from <math>O_3,O_6,O_9</math> to <math>\overline{PQ}</math> (so <math>A_3,A_6</math> are the points of [[tangent (geometry)|tangency]]). Then we note that <math>\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}</math>, and <math>O_6O_9 : O_9O_3 = 3:6 = 1:2</math>. Thus, <math>O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5</math> (consider similar triangles). Applying the [[Pythagorean Theorem]] to <math>\triangle O_9A_9P</math>, we find that  
 
<cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath>
 
<cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath>
 
<center><asy>
 
<center><asy>

Revision as of 11:12, 8 September 2011

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

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]

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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