Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1979 USAMO Problems/Problem 2

Revision as of 17:25, 15 September 2023 by Tomasdiaz (talk | contribs) (→‎Solution)

Problem

$N$ is the north pole. $A$ and $B$ are points on a great circle through $N$ equidistant from $N$. $C$ is a point on the equator. Show that the great circle through $C$ and $N$ bisects the angle $ACB$ in the spherical triangle $ABC$ (a spherical triangle has great circle arcs as sides).

Hint

Draw a large diagram. A nice, large, and precise diagram. Note that drawing a sphere entails drawing a circle and then a dashed circle (preferably of a different color) perpendicular (in the plane) to the original circle.

Solution

Since $N$ is the north pole, we define the Earth with a sphere of radius one in space with $N=(0,0,1)$ and sphere center $O=(0,0,0)$ We then pick point $N$ on the sphere and define the $xz-plane$ as the plane that contains great circle points $A$ , $B$, and $N$ with the $x-axis$ perpendicular to the $z-axis$ and in the direction of $A$.

Using this coordinate system and $x$, $y$, and $z$ axes $A=(cos(\phi),0,sin(\phi))$ where $\phi$ is the angle from the $xy-plane$ to $A$ or latitude on this sphere with $\frac{-\pi}{2} < \phi < \frac{\pi}{2}$

Since $A$ and $B$ are points on a great circle through $N$ equidistant from $N$, then $B=(-cos(\phi),0,sin(\phi))$

Since $C$ is a point on the equator, then $C=(cos(\theta),sin(\theta),0)$ where $\theta$ is the angle on the $xy-plane$ from the origin to $C$ or longitude on this sphere with $-\pi < \phi \le \pi$

We note that vectors from the origin to points $N$, $A$, $B$, and $C$ are all unit vectors because all those points are on the unit sphere.

So, we're going to define points $N$, $A$, $B$, and $C$ as unit vectors with their coordinates.

We also define the following vectors as follows:

Vector $\overrightarrow{V_{CN}}$ is the unit vector in the direction of arc $CN$ and tangent to the great circle of $CN$ at $C$

Vector $\overrightarrow{V_{CA}}$ is the unit vector in the direction of arc $CA$ and tangent to the great circle of $CA$ at $C$

Vector $\overrightarrow{V_{CB}}$ is the unit vector in the direction of arc $CB$ and tangent to the great circle of $CB$ at $C$

To calculate each of these vectors we shall use the cross product as follows:

$\overrightarrow{V_{CN}}=(\overrightarrow{C}\times\overrightarrow{N})\times\overrightarrow{C}$

$\overrightarrow{V_{CN}}=(\left\langle cos(\theta),sin(\theta),0 \right\rangle\times\left\langle 0,0,1 \right\rangle)\times \left\langle cos(\theta),sin(\theta),0 \right\rangle$

$\overrightarrow{V_{CN}}=(\left\langle cos(\theta),sin(\theta),0 \right\rangle\times\left\langle 0,0,1 \right\rangle)\times \left\langle cos(\theta),sin(\theta),0 \right\rangle$

$\overrightarrow{V_{CN}}=\left\langle sin(\theta),-cos(\theta),0 \right\rangle\times \left\langle cos(\theta),sin(\theta),0 \right\rangle$

$\overrightarrow{V_{CN}}=\left\langle 0,0,1 \right\rangle$


$\left\langle \right\rangle$

~Tomas Diaz

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1979 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1979_USAMO_Problems/Problem_2&oldid=198258"