Some Ways to Solve

First Way

The problem asks for the minimum of the sum of two squares as shown below. If the sides of the squares are extended in our sketch to form a square of length AB on each side, four regions are formed:

Therefore, the minimum sum of the squares occurs when the two rectangles have maximum area. But a rectangle has maximum area when it is a square or when AP = PB.




Second Way

Variations on the above approach include the following. Let AB = x and PB = y.
Then we want to minimize .

By the Arithmetic Mean-Geometric Mean inequality,

Therefore the sum of the two squares is always greater than the combined areas of the two rectangles except when x = y. So the minimum area occurs when P is the midpoint.

Third Way

Another approach is to formulate the area as a function of a single variable. Let AP = x and PB = AB - x. The the area is

This is a parabola with the following graph

where the vertex is at .




Fourth Way

Another approach is the following, using the arithmetic mean -- geometric mean inequality.

with equality iff x = AB - x

Another . . .

For the high school student with just enough cookbook calculus to take f(x), find its derivative f'(x) = 4x - 2(AB), and set f'(x) = 0, the result is that he can conclude the function reaches the above minimum without having to think about it.

Still Another . . .

Another approach is to particularize the length AB and compute a sequence of values for sum of the two squares as P is placed along points on the line. Let AB = 10 and x = AP. Then the following table can be generated quickly.

			  x  	0	1  	2	3	4	5	6	7	8 	9 	 10
       10 - x   10	9	8	7	6	5	4	3	2	1	  0
          sum  100  82	68	58	52	50	52	58	68	82	100

This provides good intuition that the desired location for P is at the midpoint of AB.

And another . . .

Another variation is to construct a GSP animation to see the areas change as P is moved along AB.

Return to the Problem Statement.