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

Revision as of 09:58, 31 July 2020

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

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 ==Solution==
 {{solution}}
-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 (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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

Revision as of 09:58, 31 July 2020

Contents

Problem

Hint

Solution

See Also