27 March 2014

• 20:4820:48, 27 March 2014 diff hist +899 N 2014 AIME II Problems/Problem 12 Created page with "Obviously, <math>A+B+C=180</math> in degrees. Note that <math>\cos{3C}=-\cos{(3A+3B)}</math>. Thus, our expression is of the form <math>\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1</math>. ..."