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What percent of the area of an equilateral triangle is closer to the centroid of the triangle than to any edge of the triangle?

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up vote 5 down vote accepted

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.

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I'll make this the answer as you have more votes. If you have time though, could you confirm the 37/54 ratio mentioned below? Thanks. – C. Williamson Aug 16 '12 at 23:13
I don't think that can be right--it looks to be too big by $50\%$ precisely. Check my expanded answer and see what you think. – Cameron Buie Aug 16 '12 at 23:56

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.

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Looked it over, and it's fine up through the third paragraph. The intersections of the given median and the parabola should be at $x=\frac{-s}6$ and $x=\frac s 2$--given the piece of the triangle we're looking at, the relevant one is $x=\frac{-s}6$. Our integration will find the area bounded by the parabola, the given median, and the $y$-axis--or equivalently, the area under the curve $y=-\frac{x^2\sqrt 3}{s}+\frac{x\sqrt 3}3+\frac{s\sqrt 3}{12}$ from $x=\frac{-s}6$ to $x=0$. (cont'd) – Cameron Buie Jan 26 '13 at 23:06
(cont'd): Hence, the appropriate integration is $$\begin{align}\int_{-s/6}^0\left(\frac{-x^2\sqrt 3}s+\frac{x\sqrt 3}3+\frac{s\sqrt 3}{12}\right)\,dx &= \frac{\sqrt 3}{12s}\int_{-s/6}^0(-12x^2+4sx+s^2)\,dx\\ &= \frac{\sqrt 3}{12s}\bigl[-4x^3+2sx^2+s^2x\bigr]_{x=-s/6}^0\\ &= -\frac{\sqrt 3}{12s}\left[-4\left(\frac{-s}6\right)^3+2s\left(\frac{-s}6\right)^2+s^2\left(\c‌​frac{-s}{6}\right)\right]\\ &= -\frac{\sqrt 3}{12s}\left[\frac{s^3}{54}+\frac{s^3}{18}-\frac{s^3}6\right]\\ &= -\frac{\sqrt 3}{12s}\cdot\frac{-5s^3}{54}\\ &= \frac{5s^2\sqrt 3}{648}\end{align}$$ (cont'd) – Cameron Buie Jan 26 '13 at 23:19
Multiplying by $6$ shows that the total area in the given triangle closer to the centroid than any edge is $\cfrac{5s^2\sqrt 3}{108}$. The total area of the triangle, though, is $\cfrac12\cdot s\cdot\cfrac{s\sqrt 3}2=\cfrac{s^2\sqrt 3}{4}$, so we get a ratio of $\frac5{27}$, not $\frac{37}{54}$. Your method is solid, though (as I thought), so I went ahead and upvoted. – Cameron Buie Jan 26 '13 at 23:27

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