3rd Putnam 1940 Problem A4 p is a positive constant. Let R is the curve y 2 = 4px. Let S be the mirror image of R in the y-axis (y 2 = - 4px). R remains fixed and S rolls around it without slipping. O is the point of S initially at the origin. Find the equation for the locus of O as S rolls. Solution Straightforward.
: x(x 2 + y 2 ) + 2p y 2 = 0. Take the point of contact as (X, Y), so Y 2 = 4pX. The tangent at to R at this point has gradient 2p/Y, and hence has equation (y - Y) = 2p/Y (x - X). The perpendicular to the tangent through the origin has equation x = - 2p/Y y. If (x, y) is their point of intersection, then (2x, 2y) is the point O (since S in its new position is the reflection of R in the tangent). Solving for x, y: y = Y 3 /(2(Y 2 + 4p 2 ) ), x = - 2p/Y = - pY/(Y 2 + 4p 2 ). So the point O is (-2pY 2 /(Y 2 + 4p 2 ), Y 3 /(Y 2 + 4p 2 ) ) (*). Using Y = -2py/x, we get x = -2py 2 /(x 2 + y 2 ), or x(x 2 + y 2 ) + 2p y 2 = 0 (**). We have shown that the locus is given by (*) and that all points on (*) are on (**). However, we must check that (**) does not include additional points. Writing (**) as y 2 = - x 3 /(x + 2p), shows that for each value of x in the range (-2p, 0) there are exactly two possible values of y, and the only other point on (**) is the origin. Inspection shows that the same is true for (*), so the two expressions are equivalent. 3rd Putnam 1940 © John Scholes jscholes@kalva.demon.co.uk 15 Sep 1999