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Show that it is not possible to have a triangle with sides $a,b,c$ whose medians have length $\frac{2a}{3},\frac{2b}{3},\frac{4c}{5}$. Source ISI entrance exam sample questions

I could solve it as follows

It is well known that in a triangle with sides of lengths $a,b,c$ and medians $m_a,m_b,m_c$ respectively then the following identity holds true.

$m_a^{2}+m_b^{2}+m_c^{2}=\frac{3(a^2+b^2+c^2)}{4}$

Evaluating this we get $275(a^2+b^2)+99c^2=0$ which is obviously not true. Hence, proved. But I want some variety in the technique. I want to prove this with manipulation and not with some identity. Please help!

tryst with freedom
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Hawk
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1 Answers1

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HINT:

Use this to find the lengths of the sides.

We know for valid triangle, the sum of length of any two sides $>$ the other

lab bhattacharjee
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    -The solution I used is the effect of the solution you have recommended. I have specifically said in the question that I do not want to use any identity rather some manipulation or any elegant ideas to solve it. – Hawk Dec 03 '13 at 17:49