Only to the best mathematicians- repost at the request of math prof.?

Question

Given a triangle ABC. M is on BC , P is on AC and N is on AB. BP, AM,
and CN intersect at G. GM=GN=GP.
AG+BG+CG = 43.Find (AG)(BG)(CG)
it is NOT given the triangle is equilateral.

yes alexander, i have posted the question before and the title of my quetion now did say repost. Somebody told me that they can prove the triangle must be equilateral and asked me to repost it. thanx anyway.

Answer 1

This problem is incorrectly stated.

Lets denote
R = GM=GN=GP
a = AG
b = BG
c = CG.

Areas of triangles:
S(AGB) + S(BGC) + S(CGA) = S(ABC)
therefore

R/(a + R) + R/(b+R) + R/(c+R) = 1

R[(b+R)(c+R) + (a+R)(c+R) + (a+R)(c+R)] = (a+R)(b+R)(c+R)

R²(a+b+c) + 2R³ = abc


As it turns out, the answer depends on the value of R.
It's easy to show that R can be chosen rather arbitrarily,
even if we consider only isoceles triangles.




Edit:
I just noticed, that you already asked this question
many times, and many people already told you
that data provided is insufficient to obtain unique answer.

Answer 2

Maybe it is NOT given that the triangle ABC
is equilateral, but that hypothesis doesn't contradict
any of the other givens.

So if there is a numerical solution that works for all such
triangles, it will work also for the equilateral case.

In that case, let AG=BG=CG, so AG=BG=CG=43/3,
and (AG)(BG)(CG) = (43/3)^3.