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Problem 12 Easy

The figure below shows an equilateral triangle, a rhombus with a 6 0 ^ { \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?

  • A.

    T = R = H

  • B.

    H < R = T

  • C.

    H = R < T

  • D.

    H < R < T

  • E.

    H < T < R

Answer:C
Solution

 

Let a be the length of a side of an equilateral triangle, r_1 be the radius in the 3-tangent-circle case, and r_2 be the one-tangent-circle case.

It can be checked that both R and H equal the ratio of the r_2 case.

Then: a = 2\sqrt{3} r_1 + 2r_1 = 2(\sqrt{3}+1)r_1 a = 2\sqrt{3} r_2

\frac{T}{R} = \frac{3\pi r_1^2}{\pi r_2^2} = \frac{3 \times 3}{(\sqrt{3}+1)^2} = \frac{9}{4+2\sqrt{3}} > 1

Thus T > R = H.

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