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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.

Divide triangle into four equal parts

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 $\le \frac{1}{2}$.

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  • $\begingroup$ 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 $\leq\frac{1}{2}$. $\endgroup$ – Jack D'Aurizio Sep 30 '14 at 8:45
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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).

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