1955 AMC 12 Problem 34

Question

A

6

-inch-diameter pole and an

18

-inch-diameter pole are placed together and bound together with wire. The length of the shortest wire that will go around them is:

12\sqrt3+16\pi

12\sqrt3+7\pi

12\sqrt3+14\pi

12+15\pi

24\pi

The wire consists of two common external tangent segments and one exposed arc on each circle

Use radii

3

and

9

; each tangent segment has length

\sqrt{12^2-6^2}

The centers are

12

inches apart and their radii differ by

6.

Thus each common external tangent segment has length

\sqrt{12^2-6^2}=6\sqrt3,

contributing

12\sqrt3

in all. The geometry of the tangent lines leaves a

120^\circ

exposed arc on the radius-

3

circle and a

240^\circ

exposed arc on the radius-

9

circle. Their lengths total

\begin{aligned} &\frac{120}{360}(2\pi\cdot3)\\ &\quad{}+\frac{240}{360}(2\pi\cdot9)\\ &\qquad=2\pi+12\pi\\ &\qquad=14\pi. \end{aligned}

Hence the wire length is

12\sqrt3+14\pi.

Thus, the correct answer is C.

Problem 34 in other years