1979 USAMO Problems/Problem 2

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.

1979 USAMO (Problems • Resources)
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1979 USAMO Problems/Problem 2

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Problem

Hint

Solution

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