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Difference between revisions of "1996 AHSME Problems/Problem 20"
m (→‎Solution: edit typo)
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Let <math>O</math> be the center of the circle at <math>(6,8)</math>.
 
Let <math>O</math> be the center of the circle at <math>(6,8)</math>.
  
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<math>OA = 10</math> and <math>OB = 5</math> since <math>B</math> is on the circle.  Since <math>\trianlge OAB</math> is a right triangle with right angle <math>B</math>, we find that <math>AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}</math>.  This means that <math>\triangle OAB</math> is a <math>30-60-90</math> triangle with sides <math>5:5\sqrt{3}:10</math>.
+
<math>OA = 10</math> and <math>OB = 5</math> since <math>B</math> is on the circle.  Since <math>\triangle OAB</math> is a right triangle with right angle <math>B</math>, we find that <math>AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}</math>.  This means that <math>\triangle OAB</math> is a <math>30-60-90</math> triangle with sides <math>5:5\sqrt{3}:10</math>.
  
 
Notice that <math>OAD</math> is a line, since all points are on <math>y = 2x</math>.  In fact, it is a line that makes a <math>60^\circ</math> angle with the positive x-axis.  Thus, <math>\angle DOC = 60^\circ</math>, and <math>\angle AOB = 60^\circ</math>.  These are two parts of the stright line <math>OAD</math>.  The third angle is <math>\angle BOC</math>, which must be <math>60^\circ</math> as well.  Thus, the arc that we travel is a <math>60^\circ</math> arc, and we travel <math>\frac{C}{6} = \frac{2\pi r}{6} = \frac{2\pi \cdot 5}{6} = \frac{5\pi}{3}</math> around the circle.
 
Notice that <math>OAD</math> is a line, since all points are on <math>y = 2x</math>.  In fact, it is a line that makes a <math>60^\circ</math> angle with the positive x-axis.  Thus, <math>\angle DOC = 60^\circ</math>, and <math>\angle AOB = 60^\circ</math>.  These are two parts of the stright line <math>OAD</math>.  The third angle is <math>\angle BOC</math>, which must be <math>60^\circ</math> as well.  Thus, the arc that we travel is a <math>60^\circ</math> arc, and we travel <math>\frac{C}{6} = \frac{2\pi r}{6} = \frac{2\pi \cdot 5}{6} = \frac{5\pi}{3}</math> around the circle.

Revision as of 16:39, 2 August 2016

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

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 $\boxed{C}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AHSME Problems and Solutions

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