Difference between revisions of "1996 AHSME Problems/Problem 20"
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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>\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>. | ||
− | 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 | + | 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. |
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>. | ||
− | |||
==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 21:35, 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 compiling LaTeX. Unknown error_msg) 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 |