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In triangle $\triangle ABC$, let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$, and $AB$ respectively and let $G$ be the intersection of $AD$ and $BE$. If $AG = 15$, $BG = 13$, and $F G = 7$, what is the area of triangle $\triangle ABC$?

So there is a problem that I solves to $a^2+b^2+c^2=1770$ by Stewart's theorem , and I need to find the area of the triangle. I thought of am-gm inequality but it seems doesn't work out. Is there a better way to solve this?

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    @SteamyRoot it seems that u know everything in math... can u just give me a hint this time? – 2017-01-02
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    I just know that I must use the definition of centroid and the ratio of segments... And I have no idea how to continue on... – 2017-01-02
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    use that the ratio of the intersection of two medians is $2:1$ that means for example $AG:GD=2:1$ – 2017-01-02

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Another approach. Let $ABC$ be a triangle with side lengths $a,b,c$ and lengths of medians given by $m_a,m_b,m_c$. Then $m_a,m_b,m_c$ are the side lengths of a triangle $T$ and the area of $T$ is $\frac{3}{4}$ of the area of $ABC$. Proof without words: enter image description here In our case, $\frac{2}{3}m_a=15,\frac{2}{3}m_b=13,\frac{1}{3}m_c=7$, hence by Heron's formula $$ T = \frac{9}{16}\sqrt{(15+13+14)(-15+13+14)(15-13+14)(15+13-14)}=189 $$ and by the mentioned property: $$ [ABC] = \frac{4}{3}T = \color{red}{252}.$$

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    uh i am a bit confused by this. But thank you too! – 2017-01-02
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    um @Jack D'Aurizio how do that diagram work... can u teach me... i am so confused right now about triangles – 2017-01-03
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    can you add some labels to the diagram, and hopefully i will get it. Thanks a lot! – 2017-01-03
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    @alexzhang: the area of the blue triangle is one fourth of the area of the triangle given by three non-consecutive vertices of the hexagon. On the other hand, the area of the last triangle is exactly half the area of the hexagon (can you see why?), and the area of the hexagon is six times the area of the original triangle. – 2017-01-03
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    oh i see! dividing the hexagon into 6 small triangles and the length of it is 13,14,15. It is 1/3 of the area of the triangle, since the area of the hex equals two times the area of the triangle. Doing the area of 13,14,15 will get us 84. timing it by 3 will give us 252. Thanks for your patience for me! I am really glad that I finally get this. – 2017-01-03
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enter image description here

Since the median is cut into a 2:1 ratio by the centroid, we get: enter image description here

Notice that we can divide this triangle into three smaller triangles, each with one vertex on the centroid. For each of these smaller triangles, we can apply Stewart's Theorem to solve for one side length of the larger triangle (by noting m=n=a/2).

We get the side lengths: $\sqrt{673}$, $4\sqrt{37}$, and $\sqrt{505}$.

Using Heron's formula we can determine that the area of the triangle is 252.

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    That's equal to $(AG-1)(BG-1)(FG-1)/4$. Is it just a coincidence? – 2017-01-02
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    um i actually cant use a calculator.. but that equation is pretty nice! does that come from Stewart's theorem? – 2017-01-02
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    @alexzhang It comes from the parallelogram law. I changed the solution a bit, so the solution involves Stewart's Theorem which you appear to be familiar with. – 2017-01-02
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    it turns out that $a^2+b^2+c^2=1770$ is also correct.. is there anyway i can continue on from this? – 2017-01-02
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    No. Sadly, I cannot find a way to continue from that. – 2017-01-02
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    ok... i try to do heron of those three sides by hand but it looks hard. thank you for the solution though! – 2017-01-02