Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
Line 9: Line 9:
 
We can graph the ellipse by 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 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.
  
−
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>(-1, 0)</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>
  
 
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>.

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 $(0, -1)$ 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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_21&oldid=67517"