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.

1 Answer
Dean R.
Apr 24, 2018

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

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