Difference between revisions of "2015 AMC 12A Problems/Problem 21"
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As above, we can show that the foci of the ellipse are <math>(\sqrt{15},0), (-\sqrt{15},0).</math> | As above, we can show that the foci of the ellipse are <math>(\sqrt{15},0), (-\sqrt{15},0).</math> | ||
− | To obtain the lower bound, note that the smallest circle is when the diameter is on the line segment formed by the two foci. We can check that this indeed passes through four points on the ellipse since <math>\sqrt{15}>1 | + | To obtain the lower bound, note that the smallest circle is when the diameter is on the line segment formed by the two foci. We can check that this indeed passes through four points on the ellipse since <math>\sqrt{15}>1,</math> so <math>a=\sqrt{15}.</math> |
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 | 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 |
Revision as of 22:43, 4 December 2017
Contents
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
.
Solution 2
As above, we can show that the foci of the ellipse are
To obtain the lower bound, note that the smallest circle is when the diameter is on the line segment formed by the two foci. We can check that this indeed passes through four points on the ellipse since so
To get the upper bound, note that the circle must go through either or
WLOG, let let the circle go through
We know that the circle must go through the foci of the ellipse
So we can apply power of a point to find the diameter. Let
denote the length of the line segment from the origin to the lower point on the circle. Note that
lies on the diameter. Then by POP, we have
yielding
, and so the radius of the circle is
so
Thus
.
~ ccx09 (Roy Short)
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.