It's more than 5 lines but it works (and I think it rather pretty). Nice problem!If we can find a non-negative function f(x,y) such that the integral of f(x,y) over any strip intersected with the circle is equal to A, while f(x,y) integrated over the whole circle is B with B > 9A, we are done.How can we find such a function f? Note that if the function is such that the line integral of f(x,y) over any chord of the circle is the same (call this number A), then this function will have the property above (with A being the integral as well). This is because the integral over a strip intersected with the circle can be broken up (using essentially Fubini) into an integral along the radial direction and integrals along the chords. Since the strips are of width 1, this ends up being the integral from 0 to 1 of A, i.e. just A. (minor note: I assume here that the strip is such that along the radial direction you do get at least one whole unit in this integral; if not, it doesn't really affect the argument, as in that case the integral is less tjhan A, so 9 of them are still less than B).I find it easier to rescale everything here so that the circle has radius 1, and the strips have width 1/5.What specific function should we use. Try f(x,y) = 1/sqrt(1-x^2-y^2). This function is circularly symmetric, so we only need to evaluate its integral along a chord where we fix on the variables, say x. Note:This last term is a standard trig sub integral (u=sin), and evaluates to pi/2. This is half of a chord, so the whole chord is pi. Thus, A is pi/5 (remember the rescaling above).What is the integral over the whole circle? Put it in polar coordinates and you get 2pi * (integral 0 to 1 of 1/sqrt(1-r^2)). Put in polar coords and simplify, you get 2*pi*(integral of sin(theta) from 0 to pi/2). This is 2pi, which is our B.