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2018 AMC 12B Problems/Problem 23

Revision as of 02:20, 5 November 2021 by MRENTHUSIASM (talk | contribs)

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 $\angle ACB?$

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

Diagram

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

Solution 1 (Tetrahedron)

This solution refers to the Diagram section.

Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ, \angle BCD = 45^\circ,$ and $\angle ACD=135^\circ.$

Without the loss of generality, let $AC=BC=1.$ For tetrahedron $ABCD:$

  1. Since $\triangle BCD$ is an isosceles right triangle, we have $BD=CD=\frac{\sqrt2}{2}.$

  2. In $\triangle ACD,$ we apply the Law of Cosines to get $AD=\sqrt{AC^2+CD^2-2\cdot AC\cdot CD\cdot\cos\angle ACD}=\frac{\sqrt{10}}{2}.$

  3. In right $\triangle ABD,$ we apply the Pythagorean Theorem to get $AB=\sqrt{AD^2+BD^2}=\sqrt{3}.$

  4. In $\triangle ABC,$ we apply the Law of Cosines to get $\cos\angle ACB=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=-\frac12,$ so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.

~MRENTHUSIASM

Solution 2 (Coordinate Geometry: Vectors)

This solution refers to the Diagram section.

Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ, \angle BCD = 45^\circ,$ and $\angle ACD=135^\circ.$

Without the loss of generality, let $AC=BC=1.$ We place the diagram in the $xyz$-plane with $C=(0,0,0)$ such that the positive $x$-axis runs through $A,$ the positive $y$-axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$-axis runs through the North Pole.

It follows that $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$

As $\overrightarrow{CA} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\overrightarrow{CB} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\overrightarrow{CA}\bullet\overrightarrow{CB}}{\left\lVert\overrightarrow{CA}\right\rVert\left\lVert\overrightarrow{CB}\right\rVert}=-\frac12,\] so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.

~MRENTHUSIASM

Solution 3 (Coordinate Geometry: Spherical Coordinates)

IN CONSTRUCTION ... NO EDIT PLEASE ... WILL FINISH BY TODAY ...

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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