Difference between revisions of "1974 USAMO Problems/Problem 3"
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Now take the plane through EF perpendicular to AG. This creates two hemispheres, and the curve is in one and only one of them. | Now take the plane through EF perpendicular to AG. This creates two hemispheres, and the curve is in one and only one of them. | ||
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==See Also== | ==See Also== | ||
Revision as of 14:09, 17 September 2012
Problem
Two boundary points of a ball of radius 1 are joined by a curve contained in the ball and having length less than 2. Prove that the curve is contained entirely within some hemisphere of the given ball.
Solution
Draw a Great Circle containing the two points and the curve. Then connect the two points with a chord in the circle. The circle has radius 1, so the circle has diameter 2, so the two points are all but diametrically opposite. Therefore we can draw a diameter 
parallel to the chord but not on it.
<geogebra>e6603728dd656bd9b9e39f2b656ced562f94332c</geogebra>
Now take the plane through EF perpendicular to AG. This creates two hemispheres, and the curve is in one and only one of them.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1974 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
