View source for Mrs. Miniver's problem

==Solution==
The problem can be solved by cutting the lune along the [[line segment]] between the two crossing points of the circles, into two [[circular segment]]s, and using the formula for the area of a circular segment to relate the distance between the crossing points to the total area that the problem requires the lune to have. This gives a [[transcendental equation]] for the distance between crossing points but it can be solved numerically.{{r|ingenious|icons}} There are two boundary conditions whose distances between centers can be readily solved: the farthest apart the centers can be is when the circles have equal radii, and the closest they can be is when one circle is contained completely within the other, which happens when the ratio between radii is <math>\sqrt2</math>. If the ratio of radii falls beyond these limiting cases, the circles cannot satisfy the problem's area constraint.{{r|icons}}

In the case of two circles of equal size, these equations can be simplified somewhat. The [[rhombus]] formed by the two circle centers and the two crossing points, with side lengths equal to the radius, has an angle <math>\theta\approx 2.605</math> radians at the circle centers, found by solving the equation <math display=block>\theta-\sin\theta=\frac{2\pi}{3},</math> from which it follows that the ratio of the distance between their centers to their radius is <math>2\cos\tfrac\theta2\approx0.529864</math>.{{r|icons}}