ABCD is a tetrahedron. The three edges at B are mutually perpendicular. O is the midpoint of AB and K is the foot of the perpendicular from O to CD.

Given that AC = x, BD = y, AB = z, and:
2*x*y = z²

Determine the ratio: volume(KOAC)/volume(KOBD) in terms of x, y and z.

Since the larger tetrahedra KBAC and KABD have the same bases and are twice the height of KOAC and KOBD, their volume ratio is the same.

Since these tetrahedra have the same height and their bases together cover triangle CBD, the ratio of their bases is the ratio of their volumes.

Since BK is splits this triangle, the ratio of areas CBK and DBK is the same as the ratio of the lengths CK to DK.

To find this ratio, focus on triangle COD, of which OK is an altitude.  The formula for the ratio of the parts CD is split into (which I had to derive) would be

CD²+OC²-OD²
-----------
CD²-OC²+OD²

These lengths can be found via pythagorean theorem as

CD²=x²+y²-z² 
OC²=x²-.75z² 
OD²=y²+.25z² 

Substituting these into the formula, the result simplifies to

x²-z²
-----
 y²

Posted by Jer on 2015-01-22 08:28:06