# Maximum triangle area

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

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Can you find a formula for the area in terms of the three medians? That's not a rhetorical question - I don't know that I've ever seen such a formula myself, but if there is one, it would be a logical place to start. EDIT: see jwilson.coe.uga.edu/emt725/Medians.Triangle/… – Gerry Myerson Jan 15 '12 at 21:10
There is a formula to find one side of a triangle. But it uses three medians – user1131662 Jan 15 '12 at 21:19
Well, thanks, but I still do not know how to use this. Unfortunately – user1131662 Jan 15 '12 at 21:23

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 $\frac{2}{3}\cdot3\cdot8\cdot1=16$.

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 $$K_{\triangle ADE}=\frac{4}{3}\frac{1}{2}m_1m_2\sin\theta=\frac{2}{3}m_1m_2\sin\theta.$$

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

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Is that formula easy to find? Is it obvious that there is a triangle with the given medians, with the medians being orthogonal? – Gerry Myerson Jan 15 '12 at 21:14
Wow, I've never seen this formula. I've tried to count the area as I said already, and then the derivative. The cos of this angle was zero. But I thought it'd be rather stupid, because You'll never get a triangle like that. So I decided to ask. – user1131662 Jan 15 '12 at 21:16
Oh yes, now it's bright and clear. But again, how could the angle between medians be 90 grad? – user1131662 Jan 15 '12 at 21:28
@GerryMyerson: I don't know that I'd seen that formula somewhere before; I derived it from the area of a quadrilateral in terms of its diagonals and the angle formed by them (and have edited this into my answer above). As to whether such a triangle exists, the medians have to meet at a point that is $\frac{1}{3}$ of the way from one endpoint to the other of each median, with the farther endpoints of the medians being vertices of the triangle. From that and any arbitrary angle between the medians, I am fairly confident (though I don't have a proof offhand), a triangle can be constructed. – Isaac Jan 15 '12 at 21:29
@user1131662: The angles that I'm saying have measure 90° are the ones at the intersection point of the diagonals (which is unlabeled in the earlier diagram that has points $A$, $B$, $C$, $D$, and $E$ labeled), which doesn't make $BCDE$ a rectangle. – Isaac Jan 15 '12 at 21:36