Solution to Problem 90


Congratulations to this week's winner
Ray Kremer

Correct solutions were submitted by Ariel Flat, Ravi Subramanian, Philippe Fondanaiche, Aaron David Kahn.


The sum of the distances from the wheel to the two points on the tree at which the rope is attached is constant.  Thus the locus of points traced out by the wheel is an arc of an ellipse with the foci at the two attachment points, A and B!  (For those of you who have forgotten your analytic geometry, this is the defining property of an ellipse.)

Now for a little algebra.  In order to write the equation of the ellipse, it's a good idea to place the coordinate system carefully.  The coordinate system of choice has the x-axis through the points A and B, and the y-axis halfway between these points.  See the diagram on the left.  The general form of an ellipse is

x2 / a2 + y2 / b2 =1;
using the fact that y = 0 at A and B, you'll find that a2 = 3600, b2= 475.

To find the lowest point, T, it's easiest to argue from physical principles.  See the diagram on the right where I've used a more standard coordinate system; the ellipse described by the trajectory of the wheel is in black and the rope is ATB.  At the lowest point of the path, the tangent to the ellipse is horizontal and the angles made to the normal by the rope, XTB and XTA,  must be equal and opposite; this will ensure that both the forward and returning force cancel.  A little geometry should convince you that the triangles BTD and CTD are equivalent, and by Pythagoras

AC 2  = AY 2  + YC 2
120 2  = 100 2  + (YB + BD + DC) 2
120 2  = 100 2  + (150 - 2n) 2
n = 75 - 10 Ö11 = 41.8337...
from which, using a little similar triangle geometry, it's easy to find that m = 12.311...

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Page last updated 5 May 2000.
ã2000 Alberto L. Delgado