Start with a square in which there is a point.  Join the point to each
of the corners with a line.  The length of each line is 3", 4", 5",
and "X".  What is the length of "X"?  Show the algebraic and/or
trigometric steps to the solution.

Draw 2 perpendicular lines through the point P, parallel to the sides
of the square. You can then set up a series of equations using the
known lengths (3, 4, 5, and X) as the diagonals of the rectangles
you've created.

Also, to determine the size of the square, labeling the 4 corners of
the square A, B, C and D and the point P

Let AP=a, BP=b ,CP=c and DP=d. AB=x. (AB = the length of one side of the square)

It is easy to see that a^2 + c^2 = b^2 + d^2, but I don't need this
result in my proof.

Let angle(ABP)= Beta, so angle(CBP)=90-Beta.

Looking at Triangle ABP and the cosine-rule:

(1)	a^2 = b^2 + x^2 -2*b*x*cos(Beta)

Looking at Triangle CBP and the cosine-rule:

(2)	c^2 = b^2 + x^2 -2*b*x*cos(90-Beta)

Because cos(90-Beta) = sin(Beta):

(3)	c^2 = b^2 + x^2 -2*b*x*sin(Beta)

Rewrite (1) and (3):

(4)	2*b*x*cos(Beta) = b^2 - a^2 + x^2
(5)	2*b*x*sin(Beta) = b^2 - c^2 + x^2

And thus:

(6)	4*b^2*x^2*cos^2(Beta) = (b^2 - a^2 + x^2)^2
(7)	4*b^2*x^2*sin^2(Beta) = (b^2 - c^2 + x^2)^2
	-------------------------------------------   +
(8)	4*b^2*x^2 = (b^2 - a^2 + x^2)^2 + (b^2 - c^2 + x^2)^2

(Because cos^2 + sin^2 = 1 for each possible angle)

So formula (8) is a quadratic formula in x^2
Rewriting this formula results in:

x =
0.5*sqrt(2*a^2+2*c^2+sqrt((2*a^2+2*c^2)^2-8*(b^2-a^2)^2-8*(b^2-c^2)^2))
(If P is inside the square)

AND

x =
0.5*sqrt(2*a^2+2*c^2-sqrt((2*a^2+2*c^2)^2-8*(b^2-a^2)^2-8*(b^2-c^2)^2))
(If P is outside the square)

This is a general solution for the size of the square. 
Source: http://ken.duisenberg.com/potw/archive/arch96/960730sol.html

To langenu-ga, thanks so much for the detailed and eloquently
described answer I was seeking.  Google should have you on their
staff.

Al