Difference between revisions of "2018 AMC 12B Problems/Problem 23"

Revision as of 02:08, 23 November 2021

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 $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$

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

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

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

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

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 $AC=BC=1.$ As shown below, we place Earth 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.

DIAGRAM IN PROGRESS

It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ 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 $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain $\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12,$ so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.

~MRENTHUSIASM

Solution 3 (Coordinate Geometry: Spherical Coordinates)

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

@@ Line 15: / Line 15: @@
 This solution refers to the <b>Diagram</b> section.
-Let <math>D</math> be the orthogonal projection of <math>B</math> onto the equator. Note that <math>\angle BDA = \angle BDC = 90^\circ, \angle BCD = 45^\circ,</math> and <math>\angle ACD=135^\circ.</math>
+Let <math>D</math> be the orthogonal projection of <math>B</math> onto the equator. Note that <math>\angle BDA = \angle BDC = 90^\circ</math> and <math>\angle BCD = 45^\circ.</math> Recall that <math>115^\circ \text{ W}</math> longitude is the same as <math>245^\circ \text{ E}</math> longitude, so <math>\angle ACD=135^\circ.</math>
 Without the loss of generality, let <math>AC=BC=1.</math> For tetrahedron <math>ABCD:</math>

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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