2011 AMC 12A 第 24 题

Question

考虑所有满足

AB = 14

、

BC = 9

、

CD = 7

且

DA = 12

的四边形

ABCD

。在这样的四边形内部或边界上能放入的最大圆的半径是多少？

Consider all quadrilaterals

ABCD

such that

AB = 14,

BC = 9,

CD = 7,

and

DA = 12.

What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

\sqrt{15}

\sqrt{21}

2\sqrt{6}

5

2\sqrt{7}

答案：C

知识点：内切圆、内心与内切圆半径婆罗摩笈多公式圆内接四边形最优化

难度评级：2460

小提示：

若半径为

r

的圆能放入，从圆心向四边分割可知该四边形的面积至少为

21r

If a circle of radius

r

fits, splitting from its center shows the quadrilateral’s area is at least

21r

大提示：

布雷特施奈德不等式用圆内接情形给出面积上界；取等的四边形也有内切圆，因为两组对边之和相等

Bretschneider’s inequality bounds the area by the cyclic case; the equality case is also tangential because opposite side sums are equal

解答：

设以

X

为圆心、半径为

r

的圆能放入其中一个四边形。如果

h_1,h_2,h_3,h_4

是

X

到四条边所在直线的距离，那么每个

h_i\ge r

。将四边形分成四个三角形，得到

\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r \end{aligned}\text{。}

布雷特施奈德不等式表明，给定这些边长的四边形面积不超过圆内接情形：

\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6 \end{aligned}\text{。}

因此

r\le \frac{K}{21}\le2\sqrt6

。等号可以达到：具有这些边长的圆内接四边形也有内切圆，因为

14+7=9+12

，其内切圆半径为

\frac{K}{21}=2\sqrt6

。

因此，正确答案是 C。

Suppose a circle of radius

r

centered at

X

fits in one of the quadrilaterals. If

h_1,h_2,h_3,h_4

are the distances from

X

to the four side lines, then each

h_i\ge r.

Splitting the quadrilateral into four triangles gives

\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned}

Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case:

\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned}

Therefore

r\le \frac{K}{21}\le2\sqrt6.

Equality is attainable: the cyclic quadrilateral with these sides is also tangential because

14+7=9+12,

and its incircle has radius

\frac{K}{21}=2\sqrt6.

Thus, the correct answer is C.

2011 AMC 12A 第 24 题

答案：C

难度评级：2460

小提示：

大提示：

解答：

其他年份的第 24 题