1955 AHSME Problems/Problem 34

Problem 34

A $6$-inch and $18$-inch diameter poles are placed together and bound together with wire. The length of the shortest wire that will go around them is:

$\textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi$


[asy] draw(circle((0,0),3)); draw(circle((12,0),9)); draw((-1,2sqrt(2)) -- (9,6sqrt(2))); draw((-1,-2sqrt(2)) -- (9,-6sqrt(2))); draw((0,0) -- (-1,2sqrt(2))); draw((12,0) -- (9,6sqrt(2))); draw((0,0) -- (12,0)); draw((0,0) -- (10,3sqrt(3))); [/asy] \[\text{A diagram of the problem...aside from a few blips.}\] We can separate the total length of the wire into two sections: the straight and the curve.

In order to find out the length of the straight, we can combine the two radii (halved because $6$ and $18$ are diameters) to be the hypotenuse and the difference between said radii to be the short leg. With this, we can establish a $30$-$60$-$90$ triangle, which means that the remaining leg must be $6\sqrt{3}$ This is the length of one part of the straight. Since there are two parts, the total length is $12\sqrt{3}$.

Next, we find the curved part of the wire. We must first find the arc length of the curves, and the angle measurements to find the lengths. We can do this by doing two different things for two different circles. For the radius $3$ circle, we can subtract two right angles and two $30$ degree angles from the total $360$, which makes for a total of $360 - 240 = 120$ degree measure. That means that the curved section on the radius $3$ is $\frac{1}{3}6\pi=2\pi$. For the bigger circle, we can simply subtract twice the $60$ degree angle from the circle to get $360 - 120 = 240$ degrees. When plugged into the formula, then we get $\frac{2}{3}18\pi=12\pi$.

Add everything up and you get $\boxed{\textbf{(C)} 12\sqrt{3} + 14\pi}$

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