A convex polygon has n sides. Each vertex is joined to a point P not in the same plane. If A, B, C are adjacent vertices of the polygon take the angle between the planes PBA and PBC. The sum of the n such angles equals the sum of the n angles subtended at P by the sides of the polygon (such as the angle APB). Show that n = 3.
Solution
n = 3 is certainly possible. For example, take ∠APB = ∠APC = ∠BPC = 90o (so that the lines PA, PB, PC are mutually perpendicular). Then the three planes through P are also mutually perpendicular, so the two sums are both 270o.
We show that n > 3 is not possible.
The sum of the n angles APB etc at P is less than 360o. This is almost obvious. Take another plane which meets the lines PA, PB, PC etc at A', B', C', ... and so that the foot of the perpendicular from P to the plane lies inside the n-gon A'B'C' ... then as we move P down the perpendicular the angles A'PB' etc all increase. But when it reaches the plane their sum is 360o. However, I do not immediately see how to make that rigorous. Instead, take any point O inside the n-gon ABC... . We have ∠PBA + ∠PBC > ∠ABC. Adding the n such equations we get ∑(180o - APB) > ∑ ABC = (n - 2) 180o. So ∑ APB < 360o.
The sum of the n angles between the planes is at least (n - 2) 180o. If we take a sphere center P. Then the lines PA, PB intersect it at A", B", ... which form a spherical polygon. The angles of this polygon are the angles between the planes. We can divide the polygon into n - 2 triangles. The angles in a spherical triangle sum to at least 180o. So the angles in the spherical polygon are at least (n - 2) 180o. So we have (n - 2) 180o < 360o and hence n < 4.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002