Solution to Problem 58



Congratulations to this week's winner
 
Nathan Pauli

Partial solutions were also received from Tracy Thatcher, Michael Mitchell. Further correct solutions were also received from Steve Young, Daniel J. Statman, Burkart Venzke, William Webb.


A very large number of submissions assumed that the line segment between the center of the circle and point of tangency of the circle to the parabola forms a degree angle.  This is not so.  The angle is actually a bit over 30 degrees.

Six correct submissions were received all with different solution techniques, and all different from mine, too!  The following is that of Nathan Pauli, it's also the most trigonometric.

Let P = (x,y) be the point of tangency of the circle and the parabola.  The slope of the tangent line to the parabola at P is 2x.  This tangent line is tangent to the circle at that point, too, so is perpendicular to the radius joining the center of the circle and P.  The slope of that radius is therefore -1/2x.  Call the angle between that radius and the horizontal q, so q = arctan(1/2x).  (The change in sign comes from the orientation of q.)  Now we can see that

x2 = y = 1 + sin q = 1 + sin(arctan(1/2x)) = 1 + 1/ Ö(1+4x2).

After straightforward algebra, the only significant solution is x2 = (7 + Ö 17)/8, or x = 1.17914... With the center of the circle at (a,b) we have b = 1 and

a = x + cos q = x + 2x/ Ö(1 + 4x2) = 2.0997...

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Page last updated 15 March 1999.