Is there a solution to this problem???Please do not waste my time with dumb answers.?

Question

I have posted this question twice already but I have yet to find a solution that convinces me.

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.

Is there enough information to solve this one?
Is it (43/3)^3 or is there not enough info?
if there isn't enough info, can you convince me?
i don't know if i am just not finding it or if there is insufficient info, so your help will be appreciated.
thankyou.

Note: 1- the triangle is not equilateral.
I do not think as one of the people who answerd that G is the incenter. If it was, then GM, GN and GP would have to be perpendicular to AB, BC and AC.
can someone help??
dutch professor and all you smart people out there????

wise bloke you have not answered the question.
so?
what is the product? can we find it or is there insufficient info???

thankyou zanti.
at least now i know that some of the smart guys actually read the question.
i was not sure if i was asking when all the smart guys were not on and the question went unnoticed by the people i would be interested in their feedback.
this was a question on a computer exam
you have to enter the answer onto computer [not really multiple choice but a computer test, where the computer tell you correct or wrong when you enter.
I gave the answer (43/3)^3 but it returned wrong.
I think if it is incenter then GM, GN and GP would have to be perpendicular to the sides.

thankyou zanti. That was my logic for getting [43/3]^3. I thought that if there is a unique solution, then we can construct an equilateral triangle that satisfies the conditions of the problem. I was not sure if my reasoning was wrong.

Answer 1

We're not ignoring you. This is just really hard. I've been looking at the diagram for maybe an hour, and I *think* the fact that points A, G and N are collinear (the same can be said of C, G, and N) put strong limitations on the possible placements of points M, N, and P -- in other words, you can't just select the points arbitrarily and still get the diagram to work. Proving that AG*BG*CG is a constant (if it is a constant) and what that constant is, however, will take some work...

* * * * *

OK, thinking about this a little more (sorry, had to leave this for awhile to do some actual work :) ), here is an underhanded way to answer this one:

Let's assume the answer is a constant. We don't know that, of course, but the fact that this is a question on a computer quiz is a pretty strong hint. So, given that assumption, how about we just find a figure that fits all the requirements and find AG * BG * CG for that figure.

All right, then, here is a figure that works nicely: Make ABC an equilateral triangle with M, N and P the midpoints of the sides. In this figure, AG = BG = CG, and we know AG + BG + CG = 43. So, AG* BG*CG = (43/3)³.

If the answer is a constant, (43/3)³ must be correct. If that is not correct, it seems to me "answer varies" is the only other possibility.

Answer 2

From what you describe, G is indeed the incenter of the circle, because GM, GN, and GP are radii. If you examine the segment AG, the circle with its center at G (the incenter) bisects AG (no, I don't have the details of the proof at this time but it is still true). The same holds true for BG and CG. This means that AG = BG = CG.
Since AG = BG = CG and AG + BG + CG = 43, then (AG)(BG)(CG) = (43/3)^3.

Answer 3

If ABC is not an equalateral triangle, then M, N, and P are not all in the middle of their sides ...
... but one of them could be if it is an isosoles triangle (2 sides equal) That would make it easier to find equal lengths for MG, NG, and PG.
. . . . . . .A
. . . . . . . /\
. . . . . ../ .. \
. . . . .. /. . . .\
. . N ./ . .G . . \ P
. B./_______\ C
. . . . . . .M

In this case, BCG is also an isosoles triangle with BG = CG. CGM is perpendicular.
I still can't find a way to work out AG without knowing at least MB (MB^2 + MG^2 = BG^2)

0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0
I agree with Zani that if it is equalateral, then your formula (43/3) = GA = BG = CG is enough.
But you said it is not equalateral. In that case AG > BG = CG and you can't work out what those lengths are.
so BG and CG are 14 or less

Answer 4

7