Difference between revisions of "2018 AMC 12B Problems/Problem 23"
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Tetrahedron)) |
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Cartesian Coordinates and Vectors)) |
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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 1 (Tetrahedron) == | == Solution 1 (Tetrahedron) == | ||
This solution refers to the <b>Diagram</b> section. | 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 | + | 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> | ||
<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} | + | <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} | + | <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} | + | <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> | + | <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> | </ol> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 2 ( | + | == 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 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> | ||
+ | /* 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 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 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 | ||
==See Also== | ==See Also== |
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.