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2018 AMC 12B Problems/Problem 23

Revision as of 21:22, 2 September 2024 by Mineric (talk | contribs) (→‎Solution 4 (It's not that deep))

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

[asy] /* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); draw(unitsphere,white,light=White); dot(A^^B^^C,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--C--B); label("$A$",A,3*dir(A)); label("$B$",B,3*dir(B)); label("$C$",C,3*(0,0,-1)); [/asy] ~MRENTHUSIASM

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 $\angle ACD=135^\circ.$

We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); draw(surface(A--B--C--cycle),yellow); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D); label("$A$",A,3*dir(A)); label("$B$",B,3*dir(B)); label("$C$",C,3*(0,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); [/asy] 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 (Cartesian Coordinates and Vectors)

This solution refers to the Diagram section.

Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ 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 $AC=BC=1.$ 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. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); [/asy] 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 \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.

~MRENTHUSIASM

Solution 3 (Spherical Coordinates and Vectors)

This solution refers to the diagram in Solution 2.

In spherical coordinates $(\rho,\theta,\phi),$ note that $\rho,\theta,$ and $\phi$ represent the radial distance, the polar angle, and the azimuthal angle, respectively.

Without the loss of generality, let $AC=BC=1.$ As shown in Solution 2, we place Earth in the $xyz$-plane with origin $C$ 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.

In spherical coordinates, we have $A=(1,90^\circ,0^\circ)$ and $B=(1,45^\circ,135^\circ).$ Now, we express $\vec{A}$ and $\vec{B}$ in Cartesian coordinates: \[\vec{A} = \begin{pmatrix}\sin90^\circ \cos0^\circ \\ \sin90^\circ \sin0^\circ \\ \cos90^\circ \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \text{ and } \vec{B} = \begin{pmatrix}\sin45^\circ \cos135^\circ \\ \sin45^\circ \sin135^\circ \\ \cos45^\circ \end{pmatrix} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix}.\] We continue with the last paragraph of Solution 2 to get the answer $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.

~MRENTHUSIASM


Solution 4 (It's not that deep)

This solution refers to the diagram in Solution 2. Let the radius of the sphere be $r$.


Form $\triangle{ABC}$ and notice that it is isosceles, with $AC=BC$. Draw the circle with radius equal to the sphere (great circle) with $0^{\circ}$ latitude.


Drop a perpendicular from B onto this circle. Let the foot of this perpendicular be called $D$. Since $\angle{DCA} = 135^{\circ}$, by Law of Cosines, $(AD)^2 = \frac{5}{2}r^2$. Since $BD$ is a perpendicular, we can form right triangle $\triangle{BDA}$. By Pythag, $AB=\sqrt{\frac{1}{2}r^2 + \frac{5}{2}r^2} = r\sqrt{3}$. Since $\triangle{ABC}$ has side lengths $r, r,$ and $r\sqrt{3}$, $\angle{BCA} = \boxed{D) 120^{\circ}}$.

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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