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Let $X$ be the centroid of $\Delta ABC$, and $AX=5$, $BX=12$, $CX=13$. Find the area of $\Delta ABC$.

Some properties of the centroid that I know:

  • The centroid divides the medians of a triangle in a $2:1$ ratio.
  • The resulting six triangles formed after drawing the three medians of a triangle have equal area.

I don't know how to approach or solve this problem.

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    One straightforward way: use the [median length formula](https://www.artofproblemsolving.com/wiki/index.php/Median_of_a_triangle) to calculate the sides, then use [Heron's formula](https://en.wikipedia.org/wiki/Heron's_formula). That said, there is likely some clever way to take advantage of the fact that $5,12,13$ is a pythagorean triple. – 2017-01-09
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    @dxiv Oh, didn't know about that formula. Thanks – 2017-01-09

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Take $D$ on the line $AX$ such that $X$ is the midpoint of the line segment $AD$.

Then, we see that the quadrilateral $BXCD$ is a parallelogram and that $BX=12,DX=AX=5,BD=XC=13$ with $BD^2=BX^2+DX^2$.

So, $[\triangle{BXC}]=[\triangle{BDC}]=[\triangle{BDX}]=\frac 12\times 12\times 5=30$ from which we have $$[\triangle{ABC}]=3\times[\triangle{BXC}]=\color{red}{90}$$

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    How do you know $\angle XBD$ is $90^\circ$? – 2017-01-09
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    @suomynonA: $\angle{XBD}$ is not equal to $90^\circ$. I got $\angle{BXD}=90^\circ$ from the fact that $BD^2=BX^2+DX^2$. This is the [converse](https://en.wikipedia.org/wiki/Pythagorean_theorem#Converse) of the Pythagorean theorem. – 2017-01-09
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    @suomynonA -- The $90^\circ$ angle is $\angle BXD$, not $\angle XBD$. Check the side lengths of triangle $BXD$. – 2017-01-09
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    Oh sorry I read it wrong, also in the diagram I drew $\angle BXC$ looks about $45^\circ$, lol – 2017-01-09