Solution to Problem 85


No Bradley student solved this week's problem.   From outside Bradley University, correct solutions were submitted by Monty Gray, Bill Webb, Philippe Fondanaiche.  A partial solution was sent in by WC Soudriette.

This problem is very rich in ideas from both calculus and plane geometry.  The solution I offer is purely analytic, but Monty Gray submitted a solution which is entirely geometric.

Consider the figure on the left.  I orient the slice of pizza and choose coordinates for that the pizza lies in the first quadrant with one edge along the y-axis between the origin and the point at (0,1) and with the bug at B = (a,b).  Since the cut may not eliminate any of the tasty crust, we must find where the line through BS intersects the y-axis; a little algebra later we'll see that k must lie in the feasible interval

The line through OC has equation y = (cot q) x, while that through AC has equation

they intersect at the point

The area, A, of the discarded piece as a function of k is

We want to minimize A, so we differentiate and find only one critical point in the feasible interval, at

k* = 2(b - a cot q).

The minimum area therefore occurs at k = k*, t, or 1 with the respective areas being

But when does each occur?  We first find when k*lies in the feasible interval by solving two inequalities (a good exercise for the soul) and find the conditions

b £ a cot q + 1/2,    a £ (sinq)/2.

These conditions break up the slice of pizza into three regions, see the figure on the left, with the red region forming a quadrilateral in which k* yields the minimum.  It's a straightforward task to show that for points in the blue region the minimum occurs at k = t, and for points in the green region the minimum occurs for k = 1.

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Page last updated 3 April 2000.

ã2000 Alberto L. Delgado