2015 AMC 12A Problem 21

Question

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?

5\sqrt{2} + 4

\sqrt{17} + 7

6\sqrt{2} + 3

\sqrt{15} + 8

12

The ellipse is

\dfrac{x^2}{16} + y^2 = 1,

with foci at

(\pm\sqrt{15}, 0)

A circle through both foci is centered at

(0, k);

require its bottom point to lie outside the ellipse

The ellipse

\dfrac{x^2}{16} + y^2 = 1

has semi-axes

4

and

1,

so

c^2 = 16 - 1 = 15

and the foci are

(\pm\sqrt{15}, 0).

A circle through both foci has its center on the

y

-axis, say

(0, k),

with radius

\sqrt{k^2 + 15}.

Its top point always lies outside the ellipse. For four intersection points, its bottom point

(0, k - \sqrt{k^2 + 15})

must be below

y = -1,

which happens exactly when

0 \le k \lt 7.

As

k

ranges over

[0, 7),

the radius

\sqrt{k^2 + 15}

ranges over

[\sqrt{15}, 8),

so

a + b = \sqrt{15} + 8.

Thus, the correct answer is D.

Problem 21 in other years