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2005 AMC 12A Problems/Problem 16

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Problem

Three circles of radius s are drawn in the first quadrant of the xy-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the x-axis, and the third is tangent to the first circle and the y-axis. A circle of radius r > s is tangent to both axes and to the second and third circles. What is r/s?

(\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10

Image:2005_12A_AMC-16.png

Solution

Image:2005_12A_AMC-16b.png

Without loss of generality, let s = 1. Draw the segment between the center of the third circle and the large circle; this has length r+1. 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 r-3,r-1 and hypotenuse r+1. The Pythagorean Theorem yields:

(r-3)^2 + (r-1)^2 = (r+1)^2
r^2 - 10r + 9 = 0
r = 1, 9

Quite obviously r > s = 1, so r = 9 and \frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}.

See also

2005 AMC 12A (ProblemsResources)
Preceded by
Problem 15
Followed by
Problem 17
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