Difference between revisions of "2004 AMC 10B Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
− | Using [[Descartes' Circle Formula]], we can assign curvatures to all the circles: <math>1</math>, <math>1</math>, <math>1</math>, and <math>-\frac{1}{r}</math> (b/c | + | Using [[Descartes' Circle Formula]], we can assign curvatures to all the circles: <math>1</math>, <math>1</math>, <math>1</math>, and <math>-\frac{1}{r}</math> (b/c the bigger circle is externally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve: |
<math>2(1^2+1^2+1^2+(-\frac{1}{r})^2) = (1+1+1-\frac{1}{r})^2</math> | <math>2(1^2+1^2+1^2+(-\frac{1}{r})^2) = (1+1+1-\frac{1}{r})^2</math> |
Revision as of 13:23, 29 May 2021
Contents
Problem
Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?
Solution 1
The situation is shown in the picture below. The radius we seek is . Clearly
. The point
is clearly the center of the equilateral triangle
, thus
is
of the altitude of this triangle. We get that
. Therefore the radius we seek is
.
Solution 2
Using Descartes' Circle Formula, we can assign curvatures to all the circles: ,
,
, and
(b/c the bigger circle is externally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve:
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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