Difference between revisions of "2015 AMC 12A Problems/Problem 21"
m (→Solution) |
m (fixed latex) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | A circle of radius r passes through both foci of, and exactly four points on, the ellipse with equation <math>x^2+16y^2=16.</math> The set of all possible values of <math>r</math> is an interval <math>[a,b).</math> What is <math>a+b?</math> | + | A circle of radius <math>r</math> passes through both foci of, and exactly four points on, the ellipse with equation <math>x^2+16y^2=16.</math> The set of all possible values of <math>r</math> is an interval <math>[a,b).</math> What is <math>a+b?</math> |
<math> \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 </math> | <math> \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 </math> |
Revision as of 01:45, 30 January 2016
Problem
A circle of radius 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
Solution
We can graph the ellipse by seeing that the center is and finding the ellipse's intercepts. 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 half the length of the major (longer) axis and
being half 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.