geometric-constructiongeometry
Is it possible to draw a triangle, if the length of its medians $(m_1, m_2, m_3)$ are given only?
Someone asked me this question, but I can not see it. Is it really possible?
UPDATE
Apart from the algebraic solution given by JimmyK4542, can anyone give me a direct construction? I mean, it should sound like:
Draw a line segment sufficiently long. Cut the length of $m_1$ from it. Then $\ldots$
I decided to upgrade my comment into an answer.
Angle $C$ is free to vary. That would change the length of $BD$ while preserving the given lengths, which is were you went astray (the length of $BD$ is not necessarily $4$).
Now, in general, if we know two side lengths of a triangle (call them $x$ and $y$) and the angle measure between them (call it $\theta$), then the area of the triangle is given by $A = \frac{1}{2}xy\sin(\theta)$.
Applying this to our scenario, since the angle between the two given sides can vary, then so too can the area of the triangle. This lack of uniqueness is evident in the figure shown below: Every triangle shown has the prescribed side lengths, but all have different areas. In short, we cannot determine the area with the given information.
Best Answer
From JimmyK4542's formulas it follows that $a={2\over3}\sqrt{2s^2-m_a^2}\>$, where $s:=\sqrt{m_b^2+m_c^2}$. From this one derives the following construction of ${3\over2}a$:
Construct $s$ as hypotenuse of a right triangle with legs $m_b$, $m_c$; then draw a square with side $s$ and find $\sqrt{2}\>s$ as length of the diagonal $d$. Draw a Thales semicircle with diameter $d$; then construct a right triangle with hypotenuse $d$ and one leg $m_a$. The other leg then is ${3\over2}a$. The rest should be easy.