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
Maximum triangle area
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0Can 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 http://jwilson.coe.uga.edu/emt725/Medians.Triangle/Area.Medians.Tri.html – 2012-01-15
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0There is a formula to find one side of a triangle. But it uses three medians – 2012-01-15
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0Well, thanks, but I still do not know how to use this. Unfortunately – 2012-01-15
1 Answers
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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0Is that formula easy to find? Is it obvious that there is a triangle with the given medians, with the medians being orthogonal? – 2012-01-15
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0Wow, 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. – 2012-01-15
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0Oh yes, now it's bright and clear. But again, how could the angle between medians be 90 grad? – 2012-01-15
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0@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. – 2012-01-15
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0@user1131662: I've edited in a picture of a triangle with medians whose lengths are in the ratio $8:3$ and perpendicular (form 90° angle). – 2012-01-15
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0If we look at a picture above and consider that one angle between the medians is 90 grad. Then vertical angle is 90 grad too. Two other ones are 180 - 90 = 90 grad. Then we can see that DCBE is a rectangle. Which is not appropriate. Am I right? – 2012-01-15
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0@Isaac: Sorry, could you explain again? – 2012-01-15
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1@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. – 2012-01-15
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0Oops, my bad just thinking very simple. Thanks a lot. – 2012-01-15
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0I've edited the first diagram to show where $\theta$ is (or it could be the supplementary angle as $\sin\theta=\sin(\pi-\theta)$). – 2012-01-15
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0Yes, I understand where this angle is, just thought that it can't be 90. I got this answer by myself and was not happy. – 2012-01-15