#### 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.