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

Difference between revisions of "1979 USAMO Problems/Problem 2"

(→‎Solution)
(→‎Solution)
Line 10: Line 10:
 
We then pick point <math>N</math> on the sphere and define the <math>xz-plane</math> as the plane that contains great circle points <math>A</math> , <math>B</math>, and <math>N</math> with the <math>x-axis</math> perpendicular to the <math>z-axis</math> and in the direction of <math>A</math>.
 
We then pick point <math>N</math> on the sphere and define the <math>xz-plane</math> as the plane that contains great circle points <math>A</math> , <math>B</math>, and <math>N</math> with the <math>x-axis</math> perpendicular to the <math>z-axis</math> and in the direction of <math>A</math>.
  
−
Using this coordinate system and <math>x</math>, <math>y</math>, and <math>z</math> axes <math>A=(cos(\phi),0,sin(\phi))</math> where <math>\phi</math> is the angle from the <math>xy-plane</math> to <math>A</math> or latitude on this sphere with <cmath>-\dfrac{\pi}{2} \lt \phi \lt \dfrac{\pi}{2}</cmath>
+
Using this coordinate system and <math>x</math>, <math>y</math>, and <math>z</math> axes <math>A=(cos(\phi),0,sin(\phi))</math> where <math>\phi</math> is the angle from the <math>xy-plane</math> to <math>A</math> or latitude on this sphere with <cmath>\frac{-\pi}{2}\lt \phi \lt \frac{\pi}{2}</cmath>
  
 
Since <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>, then <math>B=(-cos(\phi),0,sin(\phi))</math>  
 
Since <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>, then <math>B=(-cos(\phi),0,sin(\phi))</math>  

Revision as of 16:53, 15 September 2023

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 

\[\frac{-\pi}{2}\lt \phi \lt \frac{\pi}{2}\] (Error compiling LaTeX. Unknown error_msg)

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 $C=(cos(\theta),sin(\theta),0)$ ~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=198231"