# Maximum height of intersection of triangles

I'd like some advice regarding the following question, which I have been struggling with for long time.

Let's call the shaded region in the below $S_3$. It is the union of three congruent isosceles triangles with height $1$ and bottom side length $1/3$. Likewise, we call $S_n$ the region of $n$ congruent isosceles triangle with height $1$ and bottom side length $1/n$. Let $\{n_i\}$ be an arbitrary finite subsequence of $\mathbb{N}$ with $k$ elements. I'd like to prove that $$h(\cap_{i=1}^k S_{n_i})\geq \frac{2}{k+1}, \forall k\in\mathbb{N}$$ for an arbitrary $\{n_i\}$, where $h$ maps the set to the height of the set. For example, in the above case $S_2\cap S_4$, the height is $2/3$, so the above inequality obviously holds. In fact, it was proved that the above inequality holds for $k$ from 1 to 7, and I showed numerically that the inequality holds for numerous possible sequences $\{n_i\}$ for some higher $k$. It was also proved that, if the right side of the inequality is replaced with $\frac{2}{2 k-1+1/(2k-3)}$, the inequality holds for an arbitrary positive integer $k\geq 4$. Also, One can assume the following without loss of generality: $$\gcd(n_1,n_2,...n_k)=1\\(k+1)|n_1$$

• It was proved that, if $\{n_i\}$ is an arithmetic progression, the inequality holds for any positive integer $k$.
• Clearly, at least one of $n_i$ is even for non-trivial cases, and therefore the inequality holds for the sequence of any prime numbers.
• What is the argument for $\frac{2}{2k-1+1/(2k-3)}$? May 4, 2015 at 13:30
• It was proven in Y. G. Chen, View-obstruction problems in n-dimensional Euclidean space and a generalization of them, Acta Math. Sinica 37 (1994), no. 4, 551–562., which is written in Chinese, so I could not read. I'm sorry. May 4, 2015 at 14:11
• Just to make sure I got it right, h() returns the supremum ( when such a supremum exists) of the y-coordinate values of its input? May 4, 2015 at 14:35
• Yes. You're right. May 4, 2015 at 15:00

You problem asks to show that, for any set $\{n_i\}$ of $k$ integers, there exists $x\in [0,1]$ such that $$\lVert xn_i\rVert \geq \frac{1}{k+1}\qquad\text{ for all }i=1,\dotsc,k.$$ where $\lVert y\rVert$ is the distance from $y$ to the nearest integer.