Difference between revisions of "2009 AMC 12A Problems/Problem 16"
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Simplifying, we obtain <math>r^2 - 8r + 8 = 0</math>. By [[Vieta's formulas]] the sum of the two roots of this equation is <math>\boxed{8}</math>. | Simplifying, we obtain <math>r^2 - 8r + 8 = 0</math>. By [[Vieta's formulas]] the sum of the two roots of this equation is <math>\boxed{8}</math>. | ||
− | (We should actually solve for <math>r</math> to verify that there are two distinct positive roots. In this case we get <math>r=4\pm 2\sqrt 2</math>. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers.) | + | (We should actually solve for <math>r</math> to verify that there are two distinct positive roots. In this case we get <math>r=4\pm 2\sqrt 2</math>. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.) |
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{{AMC12 box|year=2009|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2009|ab=A|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:07, 16 January 2020
Problem
A circle with center is tangent to the positive
and
-axes and externally tangent to the circle centered at
with radius
. What is the sum of all possible radii of the circle with center
?
Solution
Let be the radius of our circle. For it to be tangent to the positive
and
axes, we must have
. For the circle to be externally tangent to the circle centered at
with radius
, the distance between
and
must be exactly
.
By the Pythagorean theorem the distance between and
is
, hence we get the equation
.
Simplifying, we obtain . By Vieta's formulas the sum of the two roots of this equation is
.
(We should actually solve for to verify that there are two distinct positive roots. In this case we get
. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.)
See Also
2009 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.