Solution to Problem 5

"Solution" by Mike Arsenault

[Editorial Note: Here is Mike's solution which relies on a computerprogram to find the optimal answer. Can anyone give a purely mathematicalsolution?]

I started with an attempt at trig and calculus,but I didn't get any where. So I wrote a simple program in C and came upwith the maximum result is attained when A=B=C=60 degrees and that valueis 0.75.

Here's the program:

#include #include #include #include double pi = 3.1415926535;void main() {   double a,b,c, maxVal = 0.0, value;   while (!kbhit()) {      do {         a = 90.0*rand()/RAND_MAX;         b = (90-a)*rand()/RAND_MAX + a;         c = 180.0-a-b;      } while (a > b || b > c || c > 90.0);      value = sin(a*2*pi/360.0)*tan(b*2*pi/360.0)*cos(c*2*pi/360.0);      if (value > maxVal) {         maxVal = value;         printf("a = %.3f, b = %.3f, c = %.3f, value = %.3f\n",a,b,c,value);      }   }}
Also solved by Brian Gregory.
Back to the Problem of the Week page
Back toMathBack to the Math home page