2018 AMC 12B Problems/Problem 23
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
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
,
.
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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