# Difference between revisions of "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 making remote request. No response to HTTP request) 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, $\angle DOC = 60^\circ$, 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 $\boxed{C}$.