Is the sum of the squares of two opposite sloping edges of a rectangular based pyramid equal to the sum of the squares of the other two sloping edges?
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
ABCD is a square. P is the midpoint of AB and is joined to C. A line from D perpendicular to PC meets the line at the point Q. Prove AQ = AD.
The solution to this problem only uses two very simple facts:
Putting the two together, for any right angled triangle the length of P Q is a minimum when X is the foot of the perpendicular from A to B C. As the point X moves along the hypotenuse the rectangle A P X Q changes but P Q is always equal in length to A X.
As a further challenge, you may like to prove that the position of X is given by: ( C X/ X B) = ( C A/ A B)².