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Difference between revisions of "2015 AMC 12A Problems/Problem 21"

(Created page with "==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 ...")
 
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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 ==
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{{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 $x^2+16y^2=16.$ The set of all possible values of $r$ is an interval $[a,b).$ What is $a+b?$

$\textbf{(A)}\ 5\sqrt{2}+4\qquad\textbf{(B)}\ \sqrt{17}+7\qquad\textbf{(C)}\ 6\sqrt{2}+3\qquad\textbf{(D)}}\ \sqrt{15}+8\qquad\textbf{(E)}\ 12$ (Error compiling LaTeX. Unknown error_msg)

Solution

We can graph the ellipse by setting $x = 0$ and finding possible values for $y$, and vice versa. The points where the ellipse intersects the coordinate axes are $(0, 1), (0, -1), (4, 0)$, and $(-4, 0)$. Recall that the two foci lie on the major axis of the ellipse and are a distance of $c$ away from the center of the ellipse, where $c^2 = a^2 - b^2$, with $a$ being the length of the major (longer) axis and $b$ being the minor (shorter) axis of the ellipse. We have that $c^2 = 4^2 - 1^2 \implies$ $c^2 = 15 \implies c = \pm \sqrt{15}$. Hence, the coordinates of both of our foci are $(0, \sqrt{15})$ and $(0, -\sqrt{15})$. 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 $r$ belongs to the circle whose diameter's endpoints are the foci of this ellipse, so $a = \sqrt{15}$. The value for $b$ is achieved when the circle passes through the foci and only three points on the ellipse, which is possible when the circle touches $(0, 1)$ or $(0, -1)$. Which point we use does not change what value of $b$ is attained, so we use $(0, -1)$. Here, we must find the point $(0, y)$ such that the distance from $(0, y)$ to both foci and $(-1, 0)$ is the same. Now, we have the two following equations. \[(\sqrt{15})^2 + (y)^2 = b^2\] \[y + 1 = b \implies y = b - 1\] Substituting for $y$, we have that \[15 + (b - 1)^2 = b^2 \implies -2b + 16 = 0.\]

Solving the above simply yields that $b = 8$, so our answer is $a + b = \sqrt{15} + 8 \textbf{ (D)}$.

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


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