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

m (→‎Solution 1 (Tetrahedron))
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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>
 
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, <math>AC=BC=1.</math> For tetrahedron <math>ABCD:</math>
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Without the loss of generality, let <math>AC=BC=1.</math> For tetrahedron <math>ABCD:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li>Since <math>\triangle BCD</math> is an isosceles right triangle, we have <math>BD=CD=\frac{\sqrt2}{2}.</math></li><p>
 
   <li>Since <math>\triangle BCD</math> is an isosceles right triangle, we have <math>BD=CD=\frac{\sqrt2}{2}.</math></li><p>

Revision as of 00:55, 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

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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)

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)

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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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