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Question: 4 points are given inside or on the boundary of a unit square. I have a conjecture that there must be 2 points at a distance $\leq 1$.

Progress: I’ve found that this question is a corollary of the circle packing problem, which reads that the minimal separation of putting 4 unit circles in a square large enough is $1$. However, when I search for the proof, all I found was intuitive proofs based on packing circles or a single “easy to obtain”.

I really have no clue how to prove this theorem without using intuitions on the circle packing problem. Hoping for a clearer proof that only makes use of elementary geometry.

Really appreciate it if some ideas are given!

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$ Commented 2 days ago

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Among the four points,the four triangles whose vertices are the given points must contain a right or an obtuse triangle. Let the three edges of this triangle be $a,b,c$, where WLOG $c$ is the longest,then by the cosine law, $a^2 + b^2 \leq c^2 \leq 2$, so one of $a$ and $b$ must $\leq 1$.

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    $\begingroup$ Connected : the case of 9 points. $\endgroup$ Commented 2 days ago
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    $\begingroup$ Here is a proof of that the four triangles cannot all be acute. $\endgroup$ Commented 2 days ago
  • $\begingroup$ Thanks @Dan for the supplementaries! I just take this as granted my bad :( $\endgroup$ Commented 2 days ago
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Can't you just... unwrap the boundary into a 4-unit-length "cyclic" line segment?

put one point at one end (its "reflection" will be on the other), let's say left-most

all points get associated with part of the segment to the right of them up until the next point (or end of the line, which also as a point)

now you're dividing 4-unit-length line into 4 pieces - if all 4 are more than 1 they won't fit

wrapping it into a square would only make some distances shorter

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  • $\begingroup$ Kinda get it… It’d be great if you could add a picture to enunciate it:) $\endgroup$ Commented yesterday
  • $\begingroup$ This only seems to handle the case where all 4 points are on the boundary. $\endgroup$ Commented yesterday
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    $\begingroup$ Actually this can deal with all convex case because we can draw a smaller bounding rectangle for an smaller convex quadrangle, with a easier argument simply using triangular inequality. $\endgroup$ Commented yesterday

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