4th Putnam 1941

Problem A3

A circle radius a rolls in the plane along the x-axis the envelope of a diameter is the curve C. Show that we can find a point on the circumference of a circle radius a/2, also rolling along the x-axis, which traces out the curve C.



Consider a circle radius a/2 with center initially at (0,a/2) rolling along the x-axis. After rolling through an angle θ, the point initially at (0,a) is at a/2 sin θ, a/2 cos θ relative to the center and hence at P (aθ/2 + (a/2) sin θ, a/2 + (a/2) cos θ). The tangent at P is (y - (a/2)(1 + cos θ)/(x - (a/2)(θ + sin θ) = (-sin θ)/(1 + cos θ), or x sin θ + y(1 - cos θ) = (a/2)(θ sin θ + 2 cos θ + 2).

If we put θ = 2φ, then sin θ = 2 sin φ cos φ, 1 + cos θ = 2 cos2φ, so the tangent at P has equation, x sin φ + y cos φ = a(φ sin φ + cos φ).

Now consider the circle radius a with center initially at (0,a). When it has rolled through an angle φ, its center is at (aφ, a), so the diameter which is initially horizontal lies on the line (y-a)/(x-aφ) = -tan φ, or x sin φ + y cos φ = a(φ sin φ + cos φ). In other words, the diameter is tangent to the point P of the curve traced out by the point on the circumference of the circle radius a/2. Hence the envelope of the diameter is that curve.

A rather more sophisticated solution by Dave Rusin is as follows

Consider the large circle. When the center is at (aφ, a), a point initially at (u,a) has rotated to (u cos φ, -u sin φ) relative to the center and hence is at the point f(φ,u) = (aφ + u cos φ, a - u sin φ). The envelope can be computed as the singular points of f. In matrix form the derivative is

a - u sin φ    cos φ
  - u cos φ   - sin φ
Its determinant is zero when u = a sin φ. The image of such a point is (x,y) = f(φ,u) = (aφ + a sin φ cos φ, a - a sin φ sin φ) = (a/2)(2φ + sin 2φ, 1 + cos 2φ). As noted above, this is also the position of a point on the circumference of a circle radius a/2 rolling along the x-axis.



4th Putnam 1941

© John Scholes
19 January 2004
Last corrected/updated 19 Jan 04