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?
On
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:
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:
$$ [ABC] = \frac{4}{3}T = \color{red}{252}.$$
Since the median is cut into a 2:1 ratio by the centroid, we get:
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.