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Difference between revisions of "1996 AHSME Problems/Problem 20"

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3)  <math>\widehat {BC}</math>, where <math>BC</math> is an arc around the circle.
 
3)  <math>\widehat {BC}</math>, where <math>BC</math> is an arc around the circle.
  
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The actual path will go <math>A \rightarrow B \rightarrow C \rightarrow D</math>, so the acutal segments will be in order <math>1, 3, 2</math>.
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The actual path will go <math>A \rightarrow B \rightarrow C \rightarrow D</math>, so the actual segments will be in order <math>1, 3, 2</math>.
  
 
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>.
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<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>.
  
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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.
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Thus, <math>\angle DOC = 60^\circ</math>, and <math>\angle AOB = 60^\circ</math>.  Since <math>\angle AOB</math>, <math>\angle DOC</math> and <math>\angle BOC</math> lie on a straight line, <math>\angle BOC</math> 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.
  
 
Thus, <math>AB = 5\sqrt{3}</math>, <math>\widehat {BC} = \frac{5\pi}{3}</math>, and <math>{CD} = 5\sqrt{3}</math>.  The total distance is <math>10\sqrt{3} + \frac{5\pi}{3}</math>, which is option <math>\boxed{C}</math>.
 
Thus, <math>AB = 5\sqrt{3}</math>, <math>\widehat {BC} = \frac{5\pi}{3}</math>, and <math>{CD} = 5\sqrt{3}</math>.  The total distance is <math>10\sqrt{3} + \frac{5\pi}{3}</math>, which is option <math>\boxed{C}</math>.
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==Solution 2 (by elimination of options)==
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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
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<cmath>a + b\pi,</cmath>
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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 \(\boxed{C}\).
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 +
~Voidling
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=19|num-a=21}}
 
{{AHSME box|year=1996|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 21:24, 9 August 2025

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

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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