Putnam 1998, Problem A-2

Let \( s \) be any arc of the unit circle lying entirely in the first quadrant. Let \( A \) be the area of the region lying below \( s \) and above the \( x \)-axis and let \( B \) be the area of the region lying to the right of the \( y \)-axis and to the left of \( s \). Prove that \( A + B \) depends only on the arc length, and not on the position, of \( s \).

Solution:
First solution: to fix notation, let \( A \) be the area of region \( DEFG \), and \( B \) be the area of \( DEIH \); further let \( C \) denote the area of sector \( ODE \), which only depends on the arc length of \( s \). If \( [XYZ] \) denotes the area of triangle \( [XYZ] \), then we have \( A = C + [OEG] - [ODF] \) and \( B = C + [ODH] - [OEI] \). But clearly \( [OEG] = [OEI] \) and \( [ODF] = [ODH] \), and so \( A + B = 2C \).

Second solution: We may parametrize a point in \( s \) by any of \( x, y \), or \( \theta = \tan^{-1}(y/x) \). Then \( A \) and \( B \) are just the integrals of \( y \, dx \) and \( x \, dy \) over the appropriate intervals; thus \( A + B \) is the integral of \( x \, dy - y \, dx \) (minus because the limits of integration are reversed). But \( d\theta = x \, dy - y \, dx \), and so \( A + B = \Delta\theta \) is precisely the radian measure of \( s \). (Of course, one can perfectly well do this problem by computing the two integrals separately. But what's the fun in that?)