# Proving by induction that an equilateral triangle will always be divided into (n+1)^2 small triangles?

I'm working on a proof that looks like this:

Let $n$ be a positive integer. Given an equilateral triangle, place $n$ points on each side, dividing the side into $n+1$ equal segments.

Use the points to draw $n$ line segments parallel to each side of the triangle ($3n$ line segments in all).

Prove by induction that this will always divide the triangle into exactly $(n+1)^2$ little equilateral triangles.

But I am unsure of how to proceed past the base case.

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From a triangle with $(n+1)$-fold subdivision you can remove $n+1$ small triangles that have an edge on the base line and the $n$ triangles between them that have a vertex on the base line. The result is a triangle with $n$-fold subdivision. Thus if $f(n)$ denotes the number of small triangles, we have $$f(n+1)=f(n)+(n+1)+n.$$

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Hint:

If you split sides of $\triangle ABC$ in $n$ parts, then $\triangle AB'C'$ is a triangle whose edges were splitted in $(n-1)$ parts.

$\hspace{150pt}$

Cheers!

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thanks you for this, although i am not sure how to continue forward with a proof using this. any other suggustions? – Anders Bijou Dec 9 '12 at 22:42
@AndersBijou Well, you have a base for $n = 1$, you assume that your theorem is true for all $n < k$, and then you need to prove that it is true for $n = k$. To prove the step combine the hint and the assumption, then count the missing triangles from the last row. Finally, add everything and check if left- and right-hand sides are equal. – dtldarek Dec 9 '12 at 22:50
Sorry, I posted something which made no sense, much like this proof for me. – Anders Bijou Dec 9 '12 at 23:23
@AndersBijou I would say your solution is close to what I would have written. What doesn't make sense for you? – dtldarek Dec 9 '12 at 23:28
The solution was incorrect as the last line was incorrect. And now i've accidentally totally deleted it. – Anders Bijou Dec 9 '12 at 23:29

HINT: Your induction hypothesis is that when you divide the triangle with $n$ points on a side, and have $(n+1)^2$ small triangles. You want to show that when you divide it with $n+1$ points on a side, you get $(n+2)^2$ small triangles.

Imagine that you’ve divided it with $n+1$ points on a side. Show that if you ignore the bottom row of small triangles, the rest is an equilateral triangle that has been divided with $n$ points on a side. (You just have to check that its bottom edge has been divided with $n$ points into $n+1$ pieces.) By the induction hypothesis there are then $(n+1)^2$ small triangles above the bottom row of small triangles. Now show that the bottom row contains just enough small triangles to make up the difference between $(n+1)^2$ and $(n+2)^2$, i.e., $(n+2)^2-(n+1)^2$.

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