Solution to Problem 142


Congratulations to this week's winner

Nathan Pauli

Further correct solutions were submitted by Dane Brooke, Ritwik Chaudhuri, Nancy Schwarzkopf, Jayavel Sounderpandian, Ron Welch, Paul Botham, Philippe Fondanaiche, Sudipta Das, Yves Thiry, Francesc Suñol, Bill Webb, Alejandro Vellano, Andrew Newton, Ahmed Aboadham, Lilliana Ponce de Leon, Tim Riehle, Irene Theotokas, Bill Soudriette.


Label the points as on the left with the origin at o.  The point (P,Q) lies at the intersection of the circle of radius 1 centered at the origin, the circle of radius A centered at (0,A) and the circle of radius B centered at (B,0).  The three corresponding equations are
P2 + Q2 = 1
P2 + (Q -A)2 = A2
(P - B)2 + Q2 = B2
A little algebra gives 1/(2A)2 + 1/(2B)2 = 1.  Solving for B gives
B = A / Ö((2A)2- 1)
from which the area of the triangle, T, can be written as
T = A2/ 2Ö((2A)2- 1)
with A > 1/Ö3.  Differentiate, set equal to zero and solve to get the critical points A = 1, 1/Ö2.  The minimum value of T is 1/4 and the maximum value is 1/2Ö3.

There are a number of possible ways to set this up.  Different choices of independent variable  yield very different equations, of course.

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