Difference between revisions of "2018 AMC 12B Problems/Problem 23"
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MRENTHUSIASM (talk | contribs) m (→Solution 2 (Cartesian Coordinates and Vectors)) |
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== Diagram == | == Diagram == | ||
<asy> | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
size(250); | size(250); | ||
import graph3; | import graph3; | ||
Line 34: | Line 35: | ||
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> | 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> | ||
+ | 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 <math>AC=BC=1.</math> For tetrahedron <math>ABCD:</math> | 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;"> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 2 ( | + | == Solution 2 (Cartesian Coordinates and Vectors) == |
This solution refers to the <b>Diagram</b> section. | This solution refers to the <b>Diagram</b> section. | ||
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Without the loss of generality, let <math>AC=BC=1.</math> As shown below, we place Earth in the <math>xyz</math>-plane with <math>C=(0,0,0)</math> such that the positive <math>x</math>-axis runs through <math>A,</math> the positive <math>y</math>-axis runs through <math>0^\circ</math> latitude and <math>160^\circ \text{ W}</math> longitude, and the positive <math>z</math>-axis runs through the North Pole. | Without the loss of generality, let <math>AC=BC=1.</math> As shown below, we place Earth in the <math>xyz</math>-plane with <math>C=(0,0,0)</math> such that the positive <math>x</math>-axis runs through <math>A,</math> the positive <math>y</math>-axis runs through <math>0^\circ</math> latitude and <math>160^\circ \text{ W}</math> longitude, and the positive <math>z</math>-axis runs through the North Pole. | ||
<asy> | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
size(300); | size(300); | ||
import graph3; | import graph3; | ||
Line 76: | Line 100: | ||
It follows that <math>A=(1,0,0)</math> and <math>D=(-t,t,0)</math> for some positive number <math>t.</math> Since <math>\triangle BCD</math> is an isosceles right triangle, we have <math>B=\left(-t,t,\sqrt{2}t\right).</math> By the Distance Formula, we get <math>(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,</math> from which <math>t=\frac12.</math> | It follows that <math>A=(1,0,0)</math> and <math>D=(-t,t,0)</math> for some positive number <math>t.</math> Since <math>\triangle BCD</math> is an isosceles right triangle, we have <math>B=\left(-t,t,\sqrt{2}t\right).</math> By the Distance Formula, we get <math>(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,</math> from which <math>t=\frac12.</math> | ||
− | As <math>\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}</math> and <math>\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},</math> we obtain <cmath>\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12 | + | As <math>\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}</math> and <math>\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},</math> we obtain <cmath>\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12</cmath> by the dot product, so <math>\angle ACB=\boxed{\textbf{(C) }120}</math> degrees. |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 3 ( | + | == Solution 3 (Spherical Coordinates and Vectors) == |
+ | This solution refers to the diagram in Solution 2. | ||
+ | |||
In spherical coordinates <math>(\rho,\theta,\phi),</math> note that <math>\rho,\theta,</math> and <math>\phi</math> represent the radial distance, the polar angle, and the azimuthal angle, respectively. | In spherical coordinates <math>(\rho,\theta,\phi),</math> note that <math>\rho,\theta,</math> and <math>\phi</math> represent the radial distance, the polar angle, and the azimuthal angle, respectively. | ||
Without the loss of generality, let <math>AC=BC=1.</math> As shown in Solution 2, we place Earth in the <math>xyz</math>-plane with origin <math>C</math> such that the positive <math>x</math>-axis runs through <math>A,</math> the positive <math>y</math>-axis runs through <math>0^\circ</math> latitude and <math>160^\circ \text{ W}</math> longitude, and the positive <math>z</math>-axis runs through the North Pole. | Without the loss of generality, let <math>AC=BC=1.</math> As shown in Solution 2, we place Earth in the <math>xyz</math>-plane with origin <math>C</math> such that the positive <math>x</math>-axis runs through <math>A,</math> the positive <math>y</math>-axis runs through <math>0^\circ</math> latitude and <math>160^\circ \text{ W}</math> longitude, and the positive <math>z</math>-axis runs through the North Pole. | ||
− | In spherical coordinates, we have <math>A=(1,90^\circ,0^\circ)</math> and <math>B=(1,45^\circ,135^\circ).</math> Now, we | + | In spherical coordinates, we have <math>A=(1,90^\circ,0^\circ)</math> and <math>B=(1,45^\circ,135^\circ).</math> Now, we express <math>\vec{A}</math> and <math>\vec{B}</math> in Cartesian coordinates: <cmath>\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}.</cmath> |
We continue with the last paragraph of Solution 2 to get the answer <math>\angle ACB=\boxed{\textbf{(C) }120}</math> degrees. | We continue with the last paragraph of Solution 2 to get the answer <math>\angle ACB=\boxed{\textbf{(C) }120}</math> degrees. | ||
Latest revision as of 22:39, 3 December 2021
Contents
Problem
Ajay is standing at point near Pontianak, Indonesia, latitude and longitude. Billy is standing at point near Big Baldy Mountain, Idaho, USA, latitude and longitude. Assume that Earth is a perfect sphere with center What is the degree measure of
Diagram
~MRENTHUSIASM
Solution 1 (Tetrahedron)
This solution refers to the Diagram section.
Let be the orthogonal projection of onto the equator. Note that and Recall that longitude is the same as longitude, so
We obtain the following diagram: Without the loss of generality, let For tetrahedron
- Since is an isosceles right triangle, we have
- In we apply the Law of Cosines to get
- In right we apply the Pythagorean Theorem to get
- In we apply the Law of Cosines to get so degrees.
~MRENTHUSIASM
Solution 2 (Cartesian Coordinates and Vectors)
This solution refers to the Diagram section.
Let be the orthogonal projection of onto the equator. Note that and Recall that longitude is the same as longitude, so
Without the loss of generality, let As shown below, we place Earth in the -plane with such that the positive -axis runs through the positive -axis runs through latitude and longitude, and the positive -axis runs through the North Pole. It follows that and for some positive number Since is an isosceles right triangle, we have By the Distance Formula, we get from which
As and we obtain by the dot product, so degrees.
~MRENTHUSIASM
Solution 3 (Spherical Coordinates and Vectors)
This solution refers to the diagram in Solution 2.
In spherical coordinates note that and represent the radial distance, the polar angle, and the azimuthal angle, respectively.
Without the loss of generality, let As shown in Solution 2, we place Earth in the -plane with origin such that the positive -axis runs through the positive -axis runs through latitude and longitude, and the positive -axis runs through the North Pole.
In spherical coordinates, we have and Now, we express and in Cartesian coordinates: We continue with the last paragraph of Solution 2 to get the answer degrees.
~MRENTHUSIASM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.