1996 AHSME Problems/Problem 20

Problem 20

In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$ ?

$\text{(A)}\ 10\sqrt 3\qquad\text{(B)}\ 10\sqrt 5\qquad\text{(C)}\ 10\sqrt 3+\frac{ 5\pi}{3}\qquad\text{(D)}\ 40\frac{\sqrt{3}}{3}\qquad\text{(E)}\ 10+5\pi$

Solution

The pathway from $A(0,0)$ to $D(12,16)$ will consist of three segments:

1) $\overline{AB}$ , where $AB$ is tangent to the circle at point $B$ .

2) $\overline{CD}$ , where $CD$ is tangent to the circle at point $C$ .

3) $\widehat {BC}$ , where $BC$ is an arc around the circle.

The actual path will go $A \rightarrow B \rightarrow C \rightarrow D$ , so the acutal segments will be in order $1, 3, 2$ .

Let $O$ be the center of the circle at $(6,8)$ .

$OA = 10$ and $OB = 5$ since $B$ is on the circle. Since $\trianlge OAB$ (Error compiling LaTeX. Unknown error_msg) is a right triangle with right angle $B$ , we find that $AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}$ . This means that $\triangle OAB$ is a $30-60-90$ triangle with sides $5:5\sqrt{3}:10$ .

Notice that $OAD$ is a line, since all points are on $y = 2x$ . In fact, it is a line that makes a $60^\circ$ angle with the positive x-axis. Thus, $\angle ADC = 60^\circ$ , and $\angle AOB = 60^\circ$ . These are two parts of the stright line $OAD$ . The third angle is $\angle BOC$ , which must be $60^\circ$ as well. Thus, the arc that we travel is a $60^\circ$ arc, and we travel $\frac{C}{6} = \frac{2\pi r}{6} = \frac{2\pi \cdot 5}{6} = \frac{5\pi}{3}$ around the circle.

Thus, $AB = 5\sqrt{3}$ , $\widehat {BC} = \frac{5\pi}{3}$ , and ${CD} = 5\sqrt{3}$ . The total distance is $10\sqrt{3} + \frac{5\pi}{3}$ , which is option $\boxed{C}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1996 AHSME Problems/Problem 20

Problem 20

Solution

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