Question

下图显示了一个等边三角形、一个有 $60^\circ$ 角的菱形和一个正六边形。每个图形中都放有一些两两相切的全等圆盘。分别令 $T$ 、 $R$ 、 $H$ 表示每种情况下圆盘总面积与外部多边形面积之比。

下列哪一项正确？

The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T,$ $R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon.

Which of the following is true?

$T = H = R$

$H \lt R = T$

$H = R \lt T$

$H \lt R \lt T$

$H \lt T \lt R$

答案：C

知识点：相切圆面积比正多边形

难度评级：1710

解答：

设三角形边长为 $s$ 。三个半径为 $r$ 的圆盘给出 $s = 2r(1 + \sqrt3)$ ，所以 $T = \dfrac{3\pi r^2}{(\sqrt3/4)s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73$ 。对于边长为 $a$ 的菱形，两个圆盘位于长对角线上，满足 $a\sqrt3 = 6r$ ，所以 $r = \tfrac{a}{2\sqrt3}$ 。对于边长为 $a$ 的正六边形，每个圆盘都在一条边中点处相切，同样有 $r = \tfrac{a}{2\sqrt3}$ ，且 $R = \dfrac{2\pi r^2}{(a^2\sqrt3/2)} = \dfrac{\pi\sqrt3}{9} \approx 0.60$ 。同时 $H = \dfrac{6\pi r^2}{(3\sqrt3/2)a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60$ 。因此 $H = R \lt T$ ，正确答案是 C。

Take the triangle with side $s.$ Three disks of radius $r$ give $s = 2r(1 + \sqrt3),$ so $T = \dfrac{3\pi r^2}{(\sqrt3/4)s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73.$ For the rhombus with side $a,$ the two disks sit on the long diagonal $a\sqrt3 = 6r,$ so $r = \tfrac{a}{2\sqrt3}$ and $R = \dfrac{2\pi r^2}{(a^2\sqrt3/2)} = \dfrac{\pi\sqrt3}{9} \approx 0.60.$ For the hexagon with side $a,$ each of the six disks touches a side at its midpoint, again giving $r = \tfrac{a}{2\sqrt3}$ and $H = \dfrac{6\pi r^2}{(3\sqrt3/2)a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60.$ So $H = R \lt T.$ Therefore, the answer is C.

2025 AMC 10B 第 12 题

答案：C

难度评级：1710

解答：

其他年份的第 12 题