By Waclaw Sierpinski, I. N. Sneddon, M. Stark

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An . We get another such polygon D with vertices B1 , B2 , . . , Bn on a sphere (of the same radius) on Q at the corresponding vertex Q . Figure 27 By the isometry assumption corresponding edges of C and D have equal length (the edges are, up to a factor r , the interior angles of faces of P and Q at P and Q ). The present hypothesis z = 0 means that the angle of C at any Ai is greater than or equal to the angle of D at Bi . Since also A1 An equals B1 Bn , Cauchy’s Lemma implies that the two polygons are congruent.

The Isoperimetric Inequality Occasionally this inequality will be referred to as “the I-I”. An old version is “Dido’s Problem”: What route should a man take to walk from sunup to sundown and to enclose as much land as possible? The answer is: he should walk on a circular path. — We look at simple closed curves in the plane, of a given length L (some differentiability assumptions) and ask: Which curve encloses the largest area? The answer is that it is the circle, of radius L/2π ; it encloses an area F = L2 /4π .

The first one requires Dz to be independent of s , from our integral inequality and the relation Q (z) = Dz / 1 + Dz2 Dz ds. This means that the third component is constant along each circle z = const , or that the angle of the normal to F and the z axis is constant. One verifies that then the centers of the horizontal circles making up F must lie on a vertical line. Then F is a surface of revolution, √ and we are done. (What needs to be done to extend the result to any n ?