# Optimization: square inscribed in a square

1. Dec 5, 2007

### enian

1. The problem statement, all variables and given/known data
Each edge of a square has length L. Prove that among all squares inscribed in the given square, the one of minimum area has edges of length $$\frac{1}{2}L\sqrt{2}$$

2. Relevant equations

3. The attempt at a solution
I started by drawing a square of sides L. Then labeled the vertices: (0,0) (L,0) (0,L) (L,L) then drew an inscribed square with variable x and the vertices were: (x,0) (L,x) (L-x,L) (0,L-x)
Then from this I set the distance of all the lines between each of these inscribed triangles vertices equal to one anotehr, to determine what values of X would work for the equation.
I determined that x=x. Is this true? Could you rotate a square 360 degrees while still being inscribed within a square?

I was stumped at this part, but if it's true. Then area of the inscribed square is going to be the distance of one of the inscribed squares length squared.
such as.. $$((x-L)^{2}+(x)^{2})^{2}$$
Then i would take a derivative of this and determine the minimum point?
Is this right?

2. Dec 5, 2007

### Shooting Star

"Could you rotate a square 360 degrees while still being inscribed within a square?"

No, you cannot, if the four corners are still touching the sides. You must have made some small mistake.

Just find the area of the inscribed square as the sum of two triangles, by considering the vertices in proper order. Or you can find the dist between the parallel pairs of lines to get the area in terms of x. This one'd be easier. Whatever you do, you should be able to show the given result.

(The result shows that the least square is obtained by joining the mid-points of the bigger square.)