Difference between revisions of "2005 AMC 12A Problems/Problem 16"
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== Solution == | == Solution == | ||
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+ | ===Solution 1=== | ||
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Quite obviously <math>r > s = 1</math>, so <math>r = 9</math> and <math>\frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}</math>. | Quite obviously <math>r > s = 1</math>, so <math>r = 9</math> and <math>\frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | Don't do this unless really really desperate. But I actually solved this with a ruler (try and see!!). Let <math>s=1</math> and find <math>r</math> in terms of <math>s</math>. The rest is easy. | ||
+ | |||
+ | Solution by franzliszt | ||
== See also == | == See also == |
Revision as of 19:07, 11 July 2020
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
Without loss of generality, let . 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
and
.
Solution 2
Don't do this unless really really desperate. But I actually solved this with a ruler (try and see!!). Let and find
in terms of
. The rest is easy.
Solution by franzliszt
See also
2005 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.