Difference between revisions of "2005 AMC 12A Problems/Problem 16"
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<div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div> | <div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div> | ||
− | Quite obviously <math>r > 1</math>, so <math>r = 9 | + | Quite obviously <math>r > 1</math>, so <math>r = 9</math>. |
== See also == | == See also == |
Revision as of 15:30, 30 March 2024
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
Set so that we only have to find
. 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:
![$(r-3)^2 + (r-1)^2 = (r+1)^2$](http://latex.artofproblemsolving.com/a/2/3/a234f93c17da5e55cc6d61dcb2dd96c94b967a8b.png)
![$r^2 - 10r + 9 = 0$](http://latex.artofproblemsolving.com/c/f/5/cf5947e2aa4bf9c1f4187105b5fcc854f8bddee5.png)
![$r = 1, 9$](http://latex.artofproblemsolving.com/4/8/2/482f67dafc8ed1baa5cf823440067133e5dc45fa.png)
Quite obviously , so
.
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 |
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