1974 USAMO Problems/Problem 3


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.


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 $EF$ parallel to the chord but not on it.


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 $Q$ be the plane through the center O of the sphere perpendicular to $BB'$ and passing through the midpoint of $BB'$. We claim that the curve must be in the hemisphere divided by $Q$ containing points A and B.

Assume the contrary and suppose that the curve intersects $Q$ 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 $Q$ 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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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