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

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

Revision as of 10:50, 6 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 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 $(0, \sqrt{15})$ 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 $(0, -1)$ 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 $a + b = \sqrt{15} + 8 \textbf{ (D)}$.

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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