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

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

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

因此 $r\le K/21\le2\sqrt6.$ 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 $K/21=2\sqrt6.$

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 $r\le K/21\le2\sqrt6.$ Equality is attainable: the cyclic quadrilateral with these sides is also tangential because $14+7=9+12,$ and its incircle has radius $K/21=2\sqrt6.$

Thus, the correct answer is C.

2011 AMC 12A 第 24 题

答案：C

难度评级：2460

解答：

其他年份的第 24 题