Difference between revisions of "1974 USAMO Problems/Problem 3"
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| − | + | '''Question: Why is the curve necessarily on a great circle? The curve is arbitrary–in space. Also, there is only one great circle through the two points, as three points determine a plane–a counterexample to this claim would then be readily found.''' | |
| + | 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. | ||
| − | ==See | + | <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. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Let the curve intersect the sphere at points A and B. Let B' be the point diametrically opposite A, and let <math>Q</math> be the plane through the center O of the sphere perpendicular to <math>BB'</math> and passing through the midpoint of <math>BB'</math>. We claim that the curve must be in the hemisphere divided by <math>Q</math> containing points A and B. | ||
| + | |||
| + | Assume the contrary and suppose that the curve intersects <math>Q</math> at point C. Reflect the portion of the curve from C to B to obtain a same-length curve from C to B'. Then the length of the curve equals the length of the new curve from A to B', which is at least the distance from A to B', or 2. This is a contradiction to the claim that the length of the curve is less than 2, so the curve must be contained in the hemisphere divided by <math>Q</math> containing points A and B. | ||
| + | |||
| + | {{alternate solutions}} | ||
| + | |||
| + | ==See Also== | ||
{{USAMO box|year=1974|num-b=2|num-a=4}} | {{USAMO box|year=1974|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
| + | [[Category:3D Geometry Problems]] | ||
Latest revision as of 23:13, 18 July 2016
Contents
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
Question: Why is the curve necessarily on a great circle? The curve is arbitrary–in space. Also, there is only one great circle through the two points, as three points determine a plane–a counterexample to this claim would then be readily found.
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.
Solution 2
Let the curve intersect the sphere at points A and B. Let B' be the point diametrically opposite A, and let 
be the plane through the center O of the sphere perpendicular to 
and passing through the midpoint of . We claim that the curve must be in the hemisphere divided by containing points A and B.
Assume the contrary and suppose that the curve intersects at point C. Reflect the portion of the curve from C to B to obtain a same-length curve from C to B'. Then the length of the curve equals the length of the new curve from A to B', which is at least the distance from A to B', or 2. This is a contradiction to the claim that the length of the curve is less than 2, so the curve must be contained in the hemisphere divided by containing points A and B.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
