Difference between revisions of "1995 AIME Problems/Problem 4"
Federernadal (talk | contribs) (→Solution) |
(→Solution) |
||
Line 9: | Line 9: | ||
== 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{ | + | 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 12:12, 8 September 2011
Problem
Circles of radius and are externally tangent to each other and are internally tangent to a circle of radius . The circle of radius has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
Solution
We label the points as following: the centers of the circles of radii are respectively, and the endpoints of the chord are . Let be the feet of the perpendiculars from to (so are the points of tangency). Then we note that , and . Thus, (consider similar triangles). Applying the Pythagorean Theorem to , we find that
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 |