1979 USAMO Problems/Problem 2

Revision as of 09:58, 31 July 2020 by Airplanes2007 (talk | contribs)

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Let SA, SB, SN be the great circles through A and C, B and C, and N and C respectively. Let C' be the point directly opposite C on the sphere. Then any great circle through C also goes through C'. So, in particular, SA, SB and SN go through C'.

Two great circles through C meet at the same angle at C and at C', so the spherical angles ACN and AC'N are equal. Now rotate the sphere through an angle 180o about the diameter through N. Then great circles through N map into themselves, so C and C' change places (C is on the equator). Also A and B change places (they are equidistant from N). SA must go into another great circle through C and C'. But since A maps to B, it must be SB. Hence the spherical angle AC'N = angle BCN (since one rotates into the other). Hence ACN and BCN are equal.


See Also

1979 USAMO (ProblemsResources)
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