# Dividing a triangle into seventeen equal parts.

I was trying to solve a problem on Pigeonhole principle from Problem Solving Strategies by Arthur Engel.

A target has the form of an equilateral triangle with side 2 units.

1. If it is hit $$5$$ times, then there will be two holes with distance $$\le 1$$.
2. If it is hit $$17$$ times. What is the minimal distance of two holes at most?

I was able to solve $$1$$. Here is what I did.

Divide the triangle into four equal parts, each of them an equilateral triangle of length $$1$$ as shown in the figure. Now using pigeonhole principle, there exists at least $$2$$ points that lie in or on the boundary of the same triangle. It is clear that the distance between then shall be $$\le 1 \qquad \square$$

I think that the same reasoning can be applied to problem $$2$$ and all I have to do is to divide the triangle into $$17$$ equal parts. How can I achieve that?

EDIT: I made a mistake with $$17$$, the triangle needs to be divided into $$16$$ equal parts to solve the problem. For $$16$$ equal parts, we can continue the process by which we divided the larger triangle into $$4$$ to the smaller triangles. That leaves answer $$\le \frac{1}{2}$$.

• Why seventeen? There is a trivial way to divide an equilateral triangle in sixteen ($4\times 4$) equilateral triangles, hence with $17$ hits there are two arrows whose distance is $\leq\frac{1}{2}$. – Jack D'Aurizio Sep 30 '14 at 8:45

Why $17$ equal part? In (i) you divide triangle into $4=5-1$ parts. Do the same with $16=17-1$ parts (it's easy - divide each side into $4$ parts).