Congratulations to this week's winner
Shaun Lewis
A partial solution was submitted by Scott Baker. Further solutions
were submitted by Philippe Fondanaiche, Sudipta Das, Steve Schaefer, Paul
Botham, Bill Soudriette, Alejandro Vellano, Bill Webb, Nancy Schwarzkopf,
Steven Young, Bill Evans, Iñigo Picaza, Lou Cairoli, Richard Bischoff.
By
symmetry, we can restrict out attention to points with positive coordinates.
Consider an arbitrary point T = (t,t2),
t > 0, on the graph of the parabola. The tangent
line at T has slope 2t so the normal will have slope
-1/2t.
The equation of the normal line is y - t2
= (-1/2t)(x - t). This line intersects
the parabola at the further point S = (s,s2)
which satisfies s2 - t2
= (-1/2t)(s - t), or s = -t
- 1/2t. The area of the shaded region is given by
Carrying out the integration (with respect to x) and
differentiating (with respect to t) gives the equation for an extreme
value (4t2 - 1)(4t2 + 1)/16t4
= 0 which has the positive solution t = 1/2. Either
a second derivative test (sheer madness!) or recourse to the geometry of
the situation will ensure that this does yield a mininmum area of 4/3 square
unit.
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