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

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== Solution 2 (Coordinate Geometry) ==
 
== Solution 2 (Coordinate Geometry) ==
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This solution refers to the <b>Diagram</b> section.
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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>
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Without the loss of generality, let <math>AC=BC=1.</math> We place the diagram in the <math>xyz</math>-plane with the equator in the <math>xy</math>-plane. Let <math>C=(0,0,0),A=(1,0,0),</math> and <math>B</math> be above the equator. It follows that <math>D=(-r,r,0)</math> and <math>B=(-r,r,s)</math> for some positive numbers <math>r</math> and <math>s.</math>
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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

Revision as of 02:35, 5 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 $\angle BDA = \angle BDC = 90^\circ, \angle BCD = 45^\circ,$ 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)

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 the equator in the $xy$-plane. Let $C=(0,0,0),A=(1,0,0),$ and $B$ be above the equator. It follows that $D=(-r,r,0)$ and $B=(-r,r,s)$ for some positive numbers $r$ and $s.$






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 $\boxed{\textbf{C}}.$

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

Solution 3 (Coordinate Geometry)

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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