I don't know if this question is a famous one. One of my fellows asked me these questions to tease me, but I was able to find a solution for only one of these:
There is one needle of length $2$ placed on the $x$-axis from $(0, 0)$ to $(2, 0)$. It had been dipped into ink, so when the needle moves, the ink colors the plane as it goes. Ignore the width of the needle(i.e., the needle is a segmented line).
- There are $n$ number of points in the plane. What is the least area for the needle to touch each point?
- There is a line of length $n$. We say a needle passes through the line when the needle literally passes through every points in the line. What is the least area for the given needle to pass through the length $n$ line?
- What if the line is a length $n$ curve?
According to him, I've only answered him right the first question. My solution was this: just move the needle backwards a lot, then the angle between the needle and one point ($\theta _1$) becomes smaller as the needle keeps going backward. So we can make the angle smaller as we want, which implies the colored area caused by moving the needle by $\theta_1$ also becomes smaller as we want. Following this argument, we can say that we can make the total colored area smaller as we want(i.e., the sum of limit of the total area caused by changing $\theta_1$ to $\theta_n$ goes to 0).
What makes me crazy is the last two questions. I answered him the following way: as before, move the needle backwards infinitely. Then the angle between the needle and the two endpoints of the curve becomes smaller as it keep moves backward.
Now think it as shooting a shot from the needle to one of a dot of the line. So first shoot at the one endpoint of the line, move back, shoot at the next dot, move back, and repeating this procedure. Then it requires in total $\theta$ angle, the angle between the needle and the two endpoints of the given line. But, since we can make the angle smaller as we want, again the total area caused by changing $\theta$ goes to 0.
But he said this is a wrong proof. And I think he mentioned that the answer of questions #2 and #3 are different. But according to my argument, there's no reason for having different answers for the two last questions.
Progress There was a decent discussion on the topic with Bill Trok, and I want to show you the whole progress, regardless of their validity.
Consider a bijection between [0, $\theta$] to [c, d], where the last interval represents the line segment. Then using the bijection $f$, I think we can argue that we need - by using the shooting technique mentioned above - $\theta$ amount of change in angle in total to shoot every element in [c, d]. But since we can make that $\theta$ as small as we want, the area would go to 0 in the limit. Hence, the answer seems to be 0 in the limit. The techinical problem suggested by Bill Trok was that in this way, maybe some 'unmeasurable' problem would occur. Sadly, I yet haven't taken any measure theory class.
First, let the needle lie on the y-axis and the segmented line on the positive x-axis right next to the origin. Then first, move the needle down then rotate the needle by $\theta$ to color a part of the line. Now as we move by $\theta$, the colored area is as close as tan$\theta$, which is bigger than $\theta$ itself. Then move the needle backward again then make the tip of the needle pointing at the uncolored segment, doing this procedure again. The conclusion was... a little bit ambiguous, though.
Bill Tork has linked to contents related to Kekeya set. According to this, we can just rotate the needle as far as we want with an arbitrary small amounts of space, so the answers seem to be 0 in the limit.
Help me! I cannot come up with any more clever ideas.
P.S. I couldn't classify this question by myself, so I just put it in geometry. Any suitable modifications are welcomed.