Maximum triangle area

Question

I have a small problem. Consider I have a triangle. Which maximum area can it cover if two of his medians are 3 and 8? I think I'll need to use derivative here, but firstly I need to find a function of an area which it covers. I actually tried to use some sorts of formulas but didn't succeed. Could anyone give me a hint at least? Thanks

Answer 1

If the lengths of two medians of a triangle are $m_1$ and $m_2$ and the angle formed by these two medians is $\theta$, then the area of the triangle is $$K_\triangle=\frac{2}{3}m_1m_2\sin\theta.$$ Since the maximum value of $\sin\theta$ is $1$, the maximum area of your triangle is $\frac{2}{3}\cdot3\cdot8\cdot1=16$.

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edit The formula above is probably not obvious. Suppose we have $\triangle ADE$ with $B$ and $C$ being the midpoints of $\overline{AE}$ and $\overline{AD}$, respectively (more because that's what I happened to draw than anything else).

The area of any quadrilateral with diagonals $d_1$ and $d_2$ and angle between then $\theta$ is $\frac{1}{2}d_1d_2\sin\theta$ (to derive this, the diagonals split the quadrilateral into 4 triangles, each with sides that are parts of the diagonals and included angles $\theta$ or $\pi-\theta$, the area of a triangle with sides $x$ and $y$ and included angle $\phi$ is $\frac{1}{2}xy\sin\phi$, and do some algebra). This gives the area of quadrialteral (trapezoid) $BCDE$ as $\frac{1}{2}m_1m_2\sin\theta$.

Now, $\triangle ABC$ is a dilation image of $\triangle AED$ by a factor of $\frac{1}{2}$ centered at $A$ (because of the midpoints, etc.), so it has $\frac{1}{4}$ of the area of the larger triangle. That is, $K_{\triangle ABC}=\frac{1}{4}K_{\triangle ADE}$ and $$K_{\text{quad }BCDE}=\frac{3}{4}K_{\triangle ADE},$$ so $$K_{\triangle ADE}=\frac{4}{3}\frac{1}{2}m_1m_2\sin\theta=\frac{2}{3}m_1m_2\sin\theta.$$

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edit 2 Here is a picture of a triangle with medians with lengths in the ratio $8:3$ that are perpendicular: