Start with a rectangular solid, two ends of which are squares. Label the vertices of one square A, B, C, D. The square opposite this is labeled A', B', C', D', with A' connected to A by an edge of the solid, and so forth.
Make a cut through plane ACD', and another cut through A'C'D. This results in the block being divided into four pieces. Discard the smallest three pieces. The volume of the remaining piece is a twodigit number of cubic centimeters. Each of the two digits happens to be one of the dimensions of the original rectangular solid, in centimeters.
What is that volume of that largest piece?
Let x = [AA'] (length of prism)
Let y = [AD] & [DC] (length of sides of square)
Let [ADCD'] & [A'D'C'D] represent the two tetrahedrons.
Let [XYX] mark the triangle representing the intersection of the two tetrahedrons.
The volume of the biggest piece = xy²  vol[ADCD']  vol[A'D'C'D] +
vol[XYZD'] + vol[XYZD]. This equation can be written as follows:
Volume = xy²  (xy²/6)  (xy²/6) + (xy²/24) + (xy²/24)
Volume = 17xy²/24
By setting this volume to equal either 10x+y OR x+10y, two equations for x can be derived:
x = 240y / (17y²  24) OR x = 24y / (17y²  240)
Since y → {1,2,3,4,5,6,7,8,9}, trial and error can be used for each equation to find x such that x → {1,2,3,4,5,6,7,8,9}

x = 3, y = 4, volume = 34

Posted by hoodat
on 20070918 16:01:53 