Difference between revisions of "2005 AMC 12A Problems/Problem 16"
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− | Three [[circle]]s of [[radius]] <math>s</math> are drawn in the first [[quadrant]] of the <math>xy</math>-[[plane]]. The first circle is tangent to both axes, the second is [[tangent (geometry)|tangent]] to the first circle and the <math>x</math>-axis, and the third is tangent to the first circle and the <math>y</math>-axis. A circle of radius <math>r > 1</math> is tangent to both axes and to the second and third circles. What is <math>r</math>? | + | Three [[circle]]s of [[radius]] <math>s</math> are drawn in the first [[quadrant]] of the <math>xy</math>-[[plane]]. The first circle is tangent to both axes, the second is [[tangent (geometry)|tangent]] to the first circle and the <math>x</math>-axis, and the third is tangent to the first circle and the <math>y</math>-axis. A circle of radius <math>r > 1</math> is tangent to both axes and to the second and third circles. What is <math>r</math>? |
<asy> | <asy> |
Revision as of 00:52, 10 January 2021
Contents
Problem
Three circles of radius are drawn in the first quadrant of the -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the -axis, and the third is tangent to the first circle and the -axis. A circle of radius is tangent to both axes and to the second and third circles. What is ?
Solution
Solution 1
Draw the segment between the center of the third circle and the large circle; this has length . We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs and hypotenuse . The Pythagorean Theorem yields:
Quite obviously , so .
Solution 2
duh
Solution by im bad