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
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
As a further challenge, you may like to prove that the position
of X is given by: ( C X/
X B) = ( C