Difference between revisions of "2018 AMC 12B Problems/Problem 23"
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Suppose that Earth is a unit sphere with center <math>(0,0,0).</math> We can let | Suppose that Earth is a unit sphere with center <math>(0,0,0).</math> We can let | ||
<cmath>A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).</cmath>The angle <math>\theta</math> between these two vectors satisfies <math>\cos\theta=A\cdot B=-\frac{1}{2},</math> yielding <math>\theta=120^{\circ},</math> or <math>\boxed{\textbf{C}.}</math> | <cmath>A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).</cmath>The angle <math>\theta</math> between these two vectors satisfies <math>\cos\theta=A\cdot B=-\frac{1}{2},</math> yielding <math>\theta=120^{\circ},</math> or <math>\boxed{\textbf{C}.}</math> | ||
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+ | Note (not by author): Alternatively, to find the angle without dot products, one may compute the distance from <math>A</math> to <math>B</math> as | ||
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+ | <math>\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}=\sqrt{3}</math>. | ||
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+ | From the Law of Cosines, <math>3=1^2+1^2-2\cos{\theta}</math>, so <math>\cos{\theta}=-\frac{1}{2},</math> from which the desired conclusion follows. | ||
==See Also== | ==See Also== |
Revision as of 18:09, 26 January 2020
Problem
Ajay is standing at point near Pontianak, Indonesia,
latitude and
longitude. Billy is standing at point
near Big Baldy Mountain, Idaho, USA,
latitude and
longitude. Assume that Earth is a perfect sphere with center
. What is the degree measure of
?
Solution
Suppose that Earth is a unit sphere with center We can let
The angle
between these two vectors satisfies
yielding
or
Note (not by author): Alternatively, to find the angle without dot products, one may compute the distance from to
as
.
From the Law of Cosines, , so
from which the desired conclusion follows.
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.