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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 17:09, 26 January 2020

Problem

Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C$. What is the degree measure of $\angle ACB$?

\[\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad\]

Solution

Suppose that Earth is a unit sphere with center $(0,0,0).$ We can let \[A=(1,0,0), B=\left(-\frac{1}{2},\frac{1}{2},\frac{\sqrt 2}{2}\right).\]The angle $\theta$ between these two vectors satisfies $\cos\theta=A\cdot B=-\frac{1}{2},$ yielding $\theta=120^{\circ},$ or $\boxed{\textbf{C}.}$


Note (not by author): Alternatively, to find the angle without dot products, one may compute the distance from $A$ to $B$ as

$\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}=\sqrt{3}$.

From the Law of Cosines, $3=1^2+1^2-2\cos{\theta}$, so $\cos{\theta}=-\frac{1}{2},$ 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

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