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 actual 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 $\triangle OAB$ 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 $5:5\sqrt{3}:10$ .

Thus, $\angle DOC = 60^\circ$ , and $\angle AOB = 60^\circ$ . Since $\angle AOB$ , $\angle DOC$ and $\angle BOC$ lie on a straight line, $\angle BOC$ 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 $\boxed{C}$ .

Solution 2 (by elimination of options)

The shortest path likely consists of two congruent line segments connected by an arc of the circle, so the total length can be expressed in the form $a + b\pi,$ where $a$ and $b$ are positive real numbers.

Using this observation, we can eliminate options A, B, and D, as they do not fit this form.

Next, the total circumference of the circle is $10\pi$. Since half the circumference is $5\pi$, option E can be ruled out because it corresponds to exactly half the circle's circumference, which is unlikely to be the shortest path involving line segments.

Therefore, by elimination of the other choices, the only remaining option is $10\sqrt{3} + \frac{5\pi}{3}$, which matches option $\boxed{C}$.

~Voidling

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

Contents

Problem 20

Solution

Solution 2 (by elimination of options)

See also