Difference between revisions of "1974 USAMO Problems/Problem 3"
(New page: ==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 hemi...) |
(Is this completely rigorous?) |
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==Solution== | ==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 <math>EF</math> parallel to the chord but not on it. | |
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| + | <geogebra>e6603728dd656bd9b9e39f2b656ced562f94332c</geogebra> | ||
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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. | ||
==See also== | ==See also== | ||
Revision as of 14:31, 11 February 2009
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.
See also
| 1974 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
