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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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==Solution 2 (Basically Solution 1)==
 +
To obtain the lower bound, refer to the above solution.
 +
To get the upper bound, note that the circle must go through either <math>(0,1)</math> or <math>(0,-1).</math> WLOG, let let the circle go through <math>(0,1).</math> We know that the circle must go through the foci of the ellipse <math>(\sqrt{15},0), (-\sqrt{15},0},</math> So we can apply power of a point to find the diameter. Let <math>x</math> denote the length of the line segment from the origin to the lower point on the circle. Note that <math>x</math> lies on the diameter. Then by POP, we have
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<math>x * 1 = \sqrt{15} * \sqrt{15},</math> yielding <math>x=15</math>, and so the radius of the circle is <math>(15+1)/2=8.</math> So <math>b=8.</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:33, 2 December 2017

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 $a+b?$

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

Solution

We can graph the ellipse by seeing that the center is $(0, 0)$ and finding the ellipse's intercepts. 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 half the length of the major (longer) axis and $b$ being half 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 $(\sqrt{15}, 0)$ and $(-\sqrt{15}, 0)$. 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 $(0, -1)$ 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 $a + b = \sqrt{15} + 8 \textbf{ (D)}$.

Solution 2 (Basically Solution 1)

To obtain the lower bound, refer to the above solution. To get the upper bound, note that the circle must go through either $(0,1)$ or $(0,-1).$ WLOG, let let the circle go through $(0,1).$ We know that the circle must go through the foci of the ellipse $(\sqrt{15},0), (-\sqrt{15},0},$ (Error compiling LaTeX. Unknown error_msg) So we can apply power of a point to find the diameter. Let $x$ denote the length of the line segment from the origin to the lower point on the circle. Note that $x$ lies on the diameter. Then by POP, we have $x * 1 = \sqrt{15} * \sqrt{15},$ yielding $x=15$, and so the radius of the circle is $(15+1)/2=8.$ So $b=8.$

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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