4th Putnam 1941

Problem A7

Do either (1) or (2):

(1)   Do either (1) or (2):

(1)   Let A be the 3 x 3 matrix

1+x2-y2-z2     2(xy+z)    2(zx-y)

  2(xy-z)    1+y2-z2-x2   2(yz+x)

  2(zx+y)     2(yz-x)   1+z2-x2-y2

Show that det A = (1 + x2 + y2 + z2)3.

(2)   A solid is formed by rotating about the x-axis the first quadrant of the ellipse x2/a2 + y2b2 = 1. Prove that this solid can rest in stable equilibrium on its vertex (corresponding to x = a, y = 0 on the ellipse) iff a/b ≤ √(8/5).



(1) subtract z times row 2 from row 1 and add y times row 3 to row 1. After taking out the common factor 1+x2+y2+z2 from row 1 we get:

     1           z         -y

  2(xy-z)    1+y2-z2-x2   2(yz+x)

  2(zx+y)     2(yz-x)   1+z2-x2-y2

Subtract z times col 1 from col 2 and add y times col 1 to col 3. We get:
     1           0            0

  2(xy-z)   1+y2+z2-x2-2xyz   2x(1+y2)

  2(zx+y)     -2x(1+z2)   1+z2-x2+y2+2xyz

Multiplying this out, we get (1-x2+y2+z2)2 - 4x2y2z2 + 4x2(1+y2+z2+y2z2) = (1+x2+y2+z2)2. Hence with the additional factor we took out, we get the result.

(2) We first have to find the position of the centre of mass on the axis. The moment about the y-axis of the solid is ∫0a πy2 x dx = πb20a (x - x3/a2) dx = πb2a2/4. The volume is ∫0a πy2 dx = πb20a (1 - x2/a2) dx = 2/3 πb2a. Hence the centre of mass is a distance 3a/8 from the flat surface or 5a/8 from the point of contact.

Now suppose the point of contact is at (a cos t, b sin t). The tangent has gradient - b/a cot t, so the normal has gradient a/b tan t. So the equation of the normal is y - b sint = a/b tan t (x - a cos t). This meets the x-axis at a(1 - b2/a2) cos t. For stability we want this to be closer to the origin than the centre of mass, in other words we want (1 - b2/a2) cos t < 3/8. The point of contact is at cos t = 1, so we require (1 - b2/a2) < 3/8 or b/a > √(5/8).



4th Putnam 1941

© John Scholes
5 Mar 2002