Difference between revisions of "1996 AHSME Problems/Problem 20"
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+ | ==Problem 20== | ||
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+ | In the xy-plane, what is the length of the shortest path from <math>(0,0)</math> to <math>(12,16)</math> that does not go inside the circle <math> (x-6)^{2}+(y-8)^{2}= 25 </math>? | ||
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+ | <math> \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 </math> | ||
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+ | ==Solution== | ||
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+ | The pathway from <math>A(0,0)</math> to <math>D(12,16)</math> will consist of three segments: | ||
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+ | 1) <math>\overline{AB}</math>, where <math>AB</math> is tangent to the circle at point <math>B</math>. | ||
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+ | 2) <math>\overline{CD}</math>, where <math>CD</math> is tangent to the circle at point <math>C</math>. | ||
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+ | 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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+ | 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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+ | 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 ADC = 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>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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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=19|num-a=21}} | {{AHSME box|year=1996|num-b=19|num-a=21}} |
Revision as of 20:33, 19 August 2011
Problem 20
In the xy-plane, what is the length of the shortest path from to that does not go inside the circle ?
Solution
The pathway from to will consist of three segments:
1) , where is tangent to the circle at point .
2) , where is tangent to the circle at point .
3) , where is an arc around the circle.
The actual path will go , so the acutal segments will be in order .
Let be the center of the circle at .
and since is on the circle. Since $\trianlge OAB$ (Error making remote request. No response to HTTP request) is a right triangle with right angle , we find that . This means that is a triangle with sides .
Notice that is a line, since all points are on . In fact, it is a line that makes a angle with the positive x-axis. Thus, , and . These are two parts of the stright line . The third angle is , which must be as well. Thus, the arc that we travel is a arc, and we travel around the circle.
Thus, , , and . The total distance is , which is option .
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 |