Difference between revisions of "2015 AMC 12A Problems/Problem 21"
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Solving the above simply yields that <math>b = 8</math>, so our answer is <math>a + b = \sqrt{15} + 8 \textbf{ (D)}</math>. | Solving the above simply yields that <math>b = 8</math>, so our answer is <math>a + b = \sqrt{15} + 8 \textbf{ (D)}</math>. | ||
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+ | == See Also == | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=20|num-a=22}} | ||
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+ | {{MAA Notice}} |
Revision as of 22:22, 4 February 2015
Problem
A circle of radius r passes through both foci of, and exactly four points on, the ellipse with equation The set of all possible values of
is an interval
What is
$\textbf{(A)}\ 5\sqrt{2}+4\qquad\textbf{(B)}\ \sqrt{17}+7\qquad\textbf{(C)}\ 6\sqrt{2}+3\qquad\textbf{(D)}}\ \sqrt{15}+8\qquad\textbf{(E)}\ 12$ (Error compiling LaTeX. Unknown error_msg)
Solution
We can graph the ellipse by setting and finding possible values for
, and vice versa. The points where the ellipse intersects the coordinate axes are
, and
. Recall that the two foci lie on the major axis of the ellipse and are a distance of
away from the center of the ellipse, where
, with
being the length of the major (longer) axis and
being the minor (shorter) axis of the ellipse. We have that
. Hence, the coordinates of both of our foci are
and
. In order for a circle to pass through both of these foci, we must have that the center of this circle lies on the y-axis.
The minimum possible value of belongs to the circle whose diameter's endpoints are the foci of this ellipse, so
. The value for
is achieved when the circle passes through the foci and only three points on the ellipse, which is possible when the circle touches
or
. Which point we use does not change what value of
is attained, so we use
. Here, we must find the point
such that the distance from
to both foci and
is the same. Now, we have the two following equations.
Substituting for
, we have that
Solving the above simply yields that , so our answer is
.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.