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.
- If it is hit $5$ times, then there will be two holes with distance $\le 1$.
- 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 $\le \frac{1}{2}$.
Best Answer
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).