Difference between revisions of "2005 AMC 12A Problems/Problem 16"
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Revision as of 16:25, 1 January 2021
Contents
[hide]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