I was looking at some equilateral triangles and started drawing up some equations and I came across the following (it may take some time for my actual question to come):

enter image description here

Suppose that we put a point $D$ anywhere within the triangle. Then we draw lines to each vertex of the triangle, we can now see that we form $3$ triangles, namely $ABD, ACD$ and $BCD$. Further draw lines perpendicular from $D$ to each side of the triangle, the lines $DE, DG$ and $DF$.

We can also note that $AH$ is the height of the triangle (we'll call the height $h$). For simplicity we will also call the sides $AC = AB = BC = S$. We will also rename the perpendicular lines from $D$ to each side; $L_1, L_2$ and $L_3$ respectively (for our purposes it doesn't matter which line is which $L_n$).

We can easily see that everything done so far is valid no matter where we put the point $D$ because we can always the draw the lines.

Now the definitions are done, so we can start looking at the area of $ABC$. This is just $\frac{bh}{2} = \frac{Sh}{2}$. The area can also be derived from adding together the area of triangles $ABD, ACD$ and $BCD$. The area for each small triangle is just $\frac{SL_n}{2}$.

Now set the two equations for the areas to equal each other: $$\frac{Sh}{2} = \frac{SL_1}{2} + \frac{SL_2}{2} + \frac{SL_3}{2}$$ $$h =L_1 + L_2 + L_3$$

Which is quite interesting. No matter where we put a point $D$, the perpendicular lengths from $D$ to the sides always sum up to the height of the triangle.

My next step was to extend this to all regular polygons, which led me to derive the following:

$$\frac{2A_n}{S} = \sum_{i=1}^{n} L_i$$

Where $A_n$ is the area of some regular polygon and the subscript $n$ denotes how many sides it has.

I was wondering if anyone has some material about this? Where it's e.g. extended to other shapes/higher dimensions etc. Thank you for your help.


Essentially the same proof shows a corresponding theorem for a regular tetrahedron -- the altitude is the sum of the length of the perpendicular dropped to each of the four sides (from any interior point). This is, in fact, one starting point for the notion of "barycentric coordinates", which might be a starting place for you to look to see what's already been done with ideas like these.

(BTW, the same proof and corresponding theorem holds in all dimensions. In dimension 1, it's not very interesting, though, and in dimensions higher than 3 it's hard to visualize.)

  • $\begingroup$ Sounds interesting, I'll look into barycentric coordinates! $\endgroup$ Oct 4 '17 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.