Let ABC be a triangle with integral side lengths such that angle A=3 * angle B. Find the minimum value of its perimeter.
I decided to see what would happen if sin(B)=3/5.
sin(A)=sin(3B)=117/125
sin(C)=sin(1804B)=sin(4B)=336/625
by the law of sines
14000a=21840b=24375c
and the minimum integer solution to this is
a=195, b=125, c=112
for a perimeter of 432.
I don't know if this is a minimum. I haven't even tried sin(B)=4/5.

Posted by Jer
on 20150304 09:47:08 