Solution to Problem 144


Congratulations to this week's winner

Nathan Pauli

Correct solutions were also received from Kristian Chelstrom.  Further correct solutions were submitted by Nancy Schwarzkopf, Ron Welch, Paul Botham, Tim Riehle, Mark Foster, Lou Cairoli, Philippe Fondanaiche, Bill Webb, Ahmed Aboadham, Yves Thiry, Fransesc Suñol y Esquirol.


The required limit is 1/2.

The area of the sector of the circle OAB is q/2, while the area of the triangle OAB is sin(q/2)cos(q/2).  Triangle OAP is similar to triangle OEA so the length of AP is tan(q/2).  Therefore

b(q) = q/2 - sin(q/2)cos(q/2)  and r(q) = tan(q/2) - q/2.
The remaining limit can be calculated either by means of a sequence of applications of L'Hôpital's Rule (messy!) or by resorting to a Taylor approximation of the functions:
b(q) = q/2 - (q/2  - 1/3!(q/2)3 + 1/5!(q/2)5- ...)(1 - 1/2!(q/2)2 + 1/4!(q/2)4 - ...)
» 2/3(q/2)3 for small values of q,

r(q) = (q/2  + 1/3(q/2)3 + 2/15(q/2)5 + ...) - q/2 » 1/3(q/2)3 for small values of q,

from which the result follows.

You are visitor number 2553 to this page.