Question

Six regular hexagonal blocks of side length $1$ unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?

$\dfrac{13\sqrt{3}}{3}$

$\dfrac{216\sqrt{3}}{49}$

$\dfrac{9\sqrt{3}}{2}$

$\dfrac{14\sqrt{3}}{3}$

$\dfrac{243\sqrt{3}}{49}$

Solution:

The uncovered region is the frame's area minus the six unit blocks. A regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2,$ so each unit block is $\tfrac{3\sqrt3}{2}.$ The spacing rule, that each frame corner sits $\tfrac37$ from the nearest block vertex, pins the frame's side length at $3.$ So the uncovered area is $\tfrac{3\sqrt3}{2}\cdot 3^2 - 6\cdot\tfrac{3\sqrt3}{2} = \tfrac{27\sqrt3}{2} - 9\sqrt3 = \tfrac{9\sqrt{3}}{2}.$ Therefore, the answer is C.

2023 AMC 10A Problem 24

Solution:

Problem 24 in Other Years