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 $\ldots$
On
The formulas for the lengths of the medians of a triangle given the sidelengths are:
$m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$
$m_b^2 = \dfrac{2c^2+2a^2-b^2}{4}$
$m_c^2 = \dfrac{2a^2+2b^2-c^2}{4}$
Solving for $a,b,c$ in terms of $m_a,m_b,m_c$ gives:
$a^2 = \dfrac{8m_b^2+8m_c^2-4m_a^2}{9}$
$b^2 = \dfrac{8m_c^2+8m_a^2-4m_b^2}{9}$
$c^2 = \dfrac{8m_a^2+8m_b^2-4m_c^2}{9}$
This gives you the lengths of the three sides of the triangle.
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.