The length of each of the sides of a triangle ABC is a positive integer with: ∠ BAC = 2* ∠ ABC and, ∠ ACB is obtuse.
Find the minimum length of the perimeter.
I don't know if this is really a D2 problem but here is what I did.
Let AC = b = 1, then if I can make the other sides rational I can just multiply through to make them integers. [And the smallest perimeter should come from the fraction with least denominator.]
Call angle ABC=x so angle BAC=2x and angle ACB=1803x
By the law of sines
sin(x)/1 = sin(2x)/a = sin (3x)/c
a=sin(2x)/sin(x) = 2cos(x)
c=sin(3x)/sin(x) = [3sin(x)4sin(x)^3]/sin(x) = 4cos(x)^2  1
So we require cos(x) to be rational. However, there are limitations on cos(x). Since angle ACB is obtuse 0<x<30 so
sqrt(3)/2 < cos(x) < 1
The fraction with minimum denominator in this range is 7/8.
[This will be especially good because 8 is divisible by 2 so c will reduce further]
This gives a=2*7/8=14/8, b=1, c=4(7/8)^21=33/16
Multiplying all by 16 produces
28, 16, 33 for a perimeter of 77. for comparison the next few fractions [8/9, 9/10 etc] give
400, 126, 620, 187, 888, 260, ...
the even denominators tend to be smaller but are increasing.
Edited on December 10, 2010, 4:23 pm

Posted by Jer
on 20101210 16:12:22 