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This is my model:

enter image description here

What I do know:

  1. A, B, C, which form an equilateral triangle
  2. Mab, Mbc, Mac which are the middle points
  3. Angles x and y, which are the angles formed by the segment from the unknown point to the medians' points and a perpendicular from the faces of the triangle (I put a square to represent it).
  4. If needed, I can do my algorithm find the same angle "z", which is not represented on the model, but is the same idea that x and y, but with middle point Mbc

What I don't know:

  1. The unknown point represented on the model

There is no restriction on where the unknown point can be in 2D space.

I couldn't find a formula to find the unknown point with just with internal angles of the triangle and the x and y angles. If you can provide some reference, subject or even some clue on how to solve this, I would be grateful.

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up vote 6 down vote accepted


Consider the Triangle with vertices $MAC$, $MAB$ and $U$ (the unknown). We know the length of the line from $MAC$ to $MAB$, and we know the angles $U-MAB-MAC$ and $U-MAC-MAB$. Hence we know everything about the triangle ($ASA$). We can then calculate the lengths using the sin law, from which you can find your point $U$.

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Thanks for your answer. My problem is to find the angles U - MAB - MAC and U - MAC - MAB. After I find these angles I can use the sin law that you mention, but X and Y are sometimes external angles and sometimes internal angles formed with the U points. Seems simple, but I am not seeing how to find these internal angles formed with point U. – JeanK May 28 '11 at 22:06
@JeanK: Well, notice that angle $X$ can be decomposed as the sum of angles $U-MAB-MAC$, $B-MAB-MAC$ and $90^o$. Since we can find $B-MAB-MAC$ (it is 120 degrees) we see that $X-210$ is equal to angle $U-MAB-MAC$. – Eric Naslund May 28 '11 at 22:11
Thank you, I am doing some tests, seems to work. Just to ensure, it is the same for y and "z" right? – JeanK May 28 '11 at 22:22
In all cases of U, the internal angles will be x-210 and y-210? – JeanK Jun 1 '11 at 14:31

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