Question

If ABC is a triangle with sides a, b , c and opposite angles alpha, beta and gamma. If alpha=3*beta, prove that (a-b)^2*(a+b)=b*c^2?

So, if `alpha=3*beta` , prove that `(a-b)^2*(a+b)=b*c^2`
Angle `alpha` is opposite of side a, angle `beta` is opposite of b, and `gamma` is opposite of c.

Answer 1

Using the Law of Sines, the Law of Cosines, the sine triple angle formula and a little Alpha help factoring, I've proven something close, which Is probably the correct formulation:

Given triangle ABC, `A=3B`, and `b(a + b) ne c^2` then

`(a-b)^2(a+b) = bc^2`

Explanation:

I think I can almost get there.

I'll write the angles A, B, and C to keep to standard notation.

`A=3B`

Let's write the Law of Sines:

`a/sin A = b/sin B`

`a sin B = b sin A = b sin(3B) = b (3 sin B - 4 sin^3 B)`

`a = b (3 - 4 sin^2B) = b(3 - 4(1-cos^2 B)) = b(4cos^2 B-1)`

Let's write the Law of Cosines:

`b^2 = a^2 + c^2 - 2 ac cos B`

`2 a c cos B = a^2+c^2-b^2`

`4 cos^2 B = (a^2 + c^2 - b^2)^2/{a^2 c^2}`

`4 cos^2 B - 1 = {(a^2 + c^2 - b^2)^2-a^2c^2}/{a^2 c^2}`

`= {(a^2 -ac + c^2 + b^2)(a^2 + ac + c^2 + b^2)}/{a^2 c^2}`

`a = b(4cos^2 B-1)`

`a^3 c^2 =b(a^2 + c^2 - b^2)^2-ba^2c^2`

`0 = a^4b - a^3 c^2 +b^5+bc^4 + ba^2c^2 - 2b^3a^2 - 2b^3 c^2`

Yikes. I'll ask Alpha to factor.

`0 = (a b + b^2 - c^2) (a^3 - a^2 b - a b^2 + b^3 - b c^2)`

Nice. It's really a shame Euler or Gauss didn't have a computer.

So if `ab + b^2 ne c^2` then we know

`a^3 - a^2 b - a b^2 + b^3 - b c^2 = 0`

`a^3 - a^2 b + b^3 - a b^2 = b c^2`

`a^2(a - b) - b^2(a- b) = b c^2`

`(a-b)^2(a+b) = bc^2 quad sqrt`