Geometry+trigonometry exercise

Question

I can't get ahead with it. Could you possibly give me some help on how to get started?

So this is the task:

The side lengths of a triangle are a, b, and c. We know this a+b=3c. Let us prove that cot(α/2)⋅cot(β/2)=2.

I tried to use the sine and cosine rule, also the law of tangents. I tried the inscribed circle and angle bisectors. I got some equations but I couldn't do anything with them.

Any help would be greatly appreciated!

Answer 1

Answer 2

need to be able to prove cot(A/2)=√[s(s-a)/(s-b)(s-c)]

and cot(B/2)=√[s(s-b)/(s-a)(s-c)]

where s=(a+b+c)/2

then cot(A/2)cot(B/2)=√[s²/(s-c)²]

problem given has s=2c

cot(A/2)cot(B/2)=√[(4c²/c²]=2

Answer 3

a+b=3c

cot(A/2)cot(B/2) =2

A = angle opposite side a, B=angle opposite side b, c= side opposite angle C

A+B+C=180=pi for any triangle

a/sinA = b/sinB= c/sinC

c^2= a^2+b^2 -2abCosC

b^2=c^2+a^2- 2acCosB

a^2=c^2+b^2 -2bcCosA

about 6 unknowns, 3 angles 3 sides, but once you determine 3 sides, the 3 angles are fixed, so call it3 unknowns and 3 basic equations, a+b=3c, the cotangent equation and A+B+C=180

possibly no solution,as it may be an overdetermined system, or infinite solutions or 1 unique solution

use substitution elimination, if it leads to a contradiction, DNE,No solution

possibly A=B and a=b because they appear in the equations symmetrically

A=B=70.4,C=38.2

140.8+38.2 = 180

a=b=3c/2

a+b=3c

is a possibility, an isosceles triangle where a=b, and A=B

cot(35.2)cot(35.2) = cot^2(35.2) = close to 2

but an infinite number of solutions for a,b and c

such as c=n,a=b=3n/2 where n=any positive number

specifically

c=2, a=b=3

c=4, a=b=6

etc.

A/2=B/2=35.265 degrees is closer

A=B=about 70.53

C= 180-141.06 = 38.94