Area in an equilateral triangle that is closer to the centroid than to any edge What percent of the area of an equilateral triangle is closer to the centroid of the triangle than to any edge of the triangle?
 A: This reduces to the following problem: What percent of the area of a $30$-$60$-$90$ triangle is closer to the vertex at the $60^\circ$ angle than it is to the side opposite the $60^\circ$ angle? (Why?)
Hint: Set any such triangle with the right-angle vertex at the origin and the side opposite the $60^\circ$ angle on the positive $x$-axis. The distance of any given point (of the triangle) from the side opposite the $60^\circ$ angle will be simply the $y$ coordinate. Use the distance formula to find the distance from the vertex at the $60^\circ$ angle.

Expansion: Let's in particular look at the triangle with vertices $(0,0)$, $(0,1)$, and $(\sqrt{3},0)$--that is, bounded by the $x$- and $y$-axes and the line $y=1-\cfrac x{\sqrt{3}}$. (It doesn't actually matter that it's this triangle and not a more general one, because of similarity relations.) This triangle has a total area of $\cfrac{\sqrt{3}}2$.
The distance from any point $(x,y)$ to the vertex at the $60^\circ$ angle is $\sqrt{x^2+(y-1)^2}$, so a point $(x,y)$ will be equidistant from that vertex as from the side opposite the $60^\circ$ angle if and only if $y=\sqrt{x^2+(y-1)^2}$.
Observing that $y\geq 0$ for all $(x,y)$ in the given triangle, so a point $(x,y)$ in the triangle will have such equidistance if and only if $$y^2=x^2+(y-1)^2$$ if and only if $$2y=x^2+1$$ if and only if $$y=\frac{x^2+1}{2}.$$ The points of the triangle that are closer to the vertex at the $60^\circ$ angle than to the side opposite it will be the points of the region bounded below by this parabola, on the left by the $y$-axis, and above by the line $y=1-\cfrac{x}{\sqrt{3}}$. The line segment and parabola intersect at the point $\left(\cfrac{1}{\sqrt{3}},\cfrac23\right)$, so the area of this region is $$\int_0^{1/\sqrt{3}}\left(1-\frac{x}{\sqrt{3}}-\frac{x^2+1}{2}\right)\,dx=\left[\frac{x}{2}-\frac{x^2}{2\sqrt{3}}-\frac{x^3}{6}\right]_{x=0}^{1/\sqrt{3}}=\frac{5\sqrt{3}}{54}.$$ As the whole triangle has area $\cfrac{\sqrt{3}}2$, then this gives us a ratio of $\frac{5}{27}$, or about $18.5\%$.

From what I can tell, your method should also work just fine, so I suspect you simply made some arithmetic errors, or goofed on your integrand or limits of integration. If there's anything about my answer about which you're uncertain, please don't hesitate to let me know.
A: First, consider an equilateral triangle so that one of the edges is centered on the x-axis. If the side length is $s$, then some simple calculations show that the centroid is at $(0,\frac{s\sqrt{3}}{6})$.
Drawing the medians of the triangle divide it into 6 congruent pieces; we can calculate the area closer to the centroid in one of these and multiply by 6. 
One can use the distance formula a bit to see that an equation that gives all the points equidistant between the centroid and the x-axis (this only applies to the lower third of the equilateral triangle) is $\frac{x^2\sqrt{3}}{s}+\frac{s\sqrt{3}}{12}$. Then, find the equation for one of the medians (one of them is: $\frac{x\sqrt{3}}{3}+\frac{s\sqrt{3}}{6}$). 
Finding the points of intersection between the median and above quadratic equation, and performing the simple integration yields the area in one of the six sub-triangles that lies closer to the centroid. Then, calculating the whole area and finding the ratio (or percent) is easy: I got the answer that 37/54 of the area is closer to the centroid, or roughly 68.5%. Someone may want to check this calculation in case of a small mistake, but I think the method is definitely correct.
