2015 AMC 12A 第 21 题

Question

一个半径为

r

的圆经过椭圆

x^2 + 16y^2 = 16

的两个焦点，并且恰好经过该椭圆上的四个点。所有可能的

r

组成区间

[a, b)

。求

a + b

？

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

答案：D

知识点：椭圆圆坐标几何

难度评级：2170

小提示：

椭圆为

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

，焦点在

(\pm\sqrt{15}, 0)

The ellipse is

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

with foci at

(\pm\sqrt{15}, 0)

大提示：

经过两个焦点的圆心在

(0, k)

；要求它的最低点在椭圆外部

A circle through both foci is centered at

(0, k);

require its bottom point to lie outside the ellipse

解答：

椭圆

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

的半轴长为

4

和

1

，所以

c^2 = 16 - 1 = 15

，焦点为

(\pm\sqrt{15}, 0)

。

经过两个焦点的圆的圆心在

y

-轴上，设为

(0, k)

，半径为

\sqrt{k^2 + 15}

。它的最高点总在椭圆外。要有四个交点，它的最低点

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

必须低于

y = -1

，这恰好在

0 \le k \lt 7

时发生。

当

k

在

[0, 7)

中变化时，半径

\sqrt{k^2 + 15}

的取值范围为

[\sqrt{15}, 8)

，所以

a + b = \sqrt{15} + 8

。

因此，正确答案是 D。

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}