Solution to Problem #1

Solutions were submitted by Fred W. Helenius of Athens GA, Keith Roberts of Glen Burnie MD, Derek May of Southwest Missouri State University, Ross Millikan of San Mateo CA, Jud McCranie of Lawrenceville GA, Robin Stokes of the University of New England (Australia), Susan Hoover of Houston TX, Phil Kenny of Santa Clara CA, Ruth Hauptman of Ames IA, and Ken Duisenberg of Roseville CA. Fred W. Helenius' solution was:

Let the equation of the common tangent line be y = mx + b.  Perform a
linear transformation, taking x -> u and y -> u + mv + b.  The tangent
line becomes the u-axis, and the equation of the new curve is
v = u^4 - 4u^2 - (m + 3)u - b.  Since each tangent point represents a
double root, this polynomial must be of the form c(u - r)^2 (u - s)^2.
Clearly c = 1, and for the polynomial to be a square it must be
(u^2 - 2)^2.  Thus m = -3 and b = -4, so the original tangent line
is y = -3x - 4.

[and the two points are (-Sqrt[2],-4+3*Sqrt[2]) and (Sqrt[2],-4-3*Sqrt[2])]

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