Difference between revisions of "2018 AMC 12B Problems/Problem 23"
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== Diagram == | == 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 | + | == Solution 1 (Tetrahedron) == |
+ | 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</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; | ||
− | == Solution 3 ( | + | 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 loss of generality, let <math>AC=BC=1.</math> For tetrahedron <math>ABCD:</math> | ||
+ | <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>In <math>\triangle ACD,</math> we apply the Law of Cosines to get <math>AD=\sqrt{AC^2+CD^2-2\cdot AC\cdot CD\cdot\cos\angle ACD}=\frac{\sqrt{10}}{2}.</math></li><p> | ||
+ | <li>In right <math>\triangle ABD,</math> we apply the Pythagorean Theorem to get <math>AB=\sqrt{AD^2+BD^2}=\sqrt{3}.</math></li><p> | ||
+ | <li>In <math>\triangle ABC,</math> we apply the Law of Cosines to get <math>\cos\angle ACB=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=-\frac12,</math> so <math>\angle ACB=\boxed{\textbf{(C) }120}</math> degrees.</li><p> | ||
+ | </ol> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 (Cartesian Coordinates and Vectors) == | ||
+ | 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</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 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> | ||
+ | /* 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 <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</cmath> by the dot product, so <math>\angle ACB=\boxed{\textbf{(C) }120}</math> degrees. | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | Without 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 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. | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 4 (It's not that deep) == | ||
+ | This solution refers to the diagram in Solution 2. Let the radius of the sphere be <math>r</math>. | ||
+ | |||
+ | Form <math>\triangle{ABC}</math> and notice that it is isosceles, with <math>AC=BC</math>. Draw the circle with radius equal to the sphere (great circle) with <math>0^{\circ}</math> latitude. | ||
+ | |||
+ | Drop a perpendicular from B onto this circle. Let the foot of this perpendicular be called <math>D</math>. Since <math>\angle{DCA} = 135^{\circ}</math>, by Law of Cosines on <math>\triangle{ACD}</math>, <math>(AD)^2 = \frac{5}{2}r^2</math>. Since <math>BD</math> is a perpendicular, we can form right triangle <math>\triangle{BDA}</math>. By Pythag, <math>AB=\sqrt{\frac{1}{2}r^2 + \frac{5}{2}r^2} = r\sqrt{3}</math>. Since <math>\triangle{ABC}</math> has side lengths <math>r, r, </math> and <math>r\sqrt{3}</math>, <math>\angle{BCA} = \boxed{\textbf{(C) }120^{\circ}}</math>. | ||
+ | |||
+ | -skibbysiggy | ||
==See Also== | ==See Also== |
Latest revision as of 02:48, 25 September 2024
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 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 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 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
Solution 4 (It's not that deep)
This solution refers to the diagram in Solution 2. Let the radius of the sphere be .
Form and notice that it is isosceles, with . Draw the circle with radius equal to the sphere (great circle) with latitude.
Drop a perpendicular from B onto this circle. Let the foot of this perpendicular be called . Since , by Law of Cosines on , . Since is a perpendicular, we can form right triangle . By Pythag, . Since has side lengths and , .
-skibbysiggy
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.