Difference between revisions of "1979 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
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| + | 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'. | ||
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| + | 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. | ||
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==See Also== | ==See Also== | ||
Revision as of 10:58, 31 July 2020
Contents
Problem
is the north pole. 
and 
are points on a great circle through equidistant from . 
is a point on the equator. Show that the great circle through and bisects the angle 
in the spherical triangle 
(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 (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. 
