Question

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?

$T = R = H$

$H \lt R = T$

$H = R \lt T$

$H \lt R \lt T$

$H \lt T \lt R$

Solution:

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 Problem 12

Solution:

Problem 12 in Other Years