Suppose a square of sidelengh $l$ is inside an unit square and does not contain its centre. Show that $l\le 1/2.$ Marius Cavachi
Problem
Source:
Tags: rotation, geometry proposed, geometry
mavropnevma
31.05.2011 22:24
Again, maybe the OP should mention he is posting problems from the 2011 Romanian Selection Tests, so that they can be gathered into the Resources section. Here, a celebrated Erdös result kills the problem: if two squares of sides $a$ and $b$ are packed without overlap within a third square, then its side must be at least $a+b$. The problem at hand allows a simpler solution, and the Erdös result would probably have been disallowed as an argument, had any contestant invoked it.
abconjecture
01.06.2011 09:05
Thanks. Do you have the other problems of the TST?
jgnr
10.06.2011 09:09
mavropnevma wrote: if two squares of sides $a$ and $b$ are packed without overlap within a third square, then its side must be at least $a+b$. Do you have a proof for this theorem? I've seen it a few times before, but never found a solution.
Binomial-theorem
15.04.2012 01:30
Assume on the contrary that $l> \frac12$ works. However, even if one point on the square was on the corner of the unit square, then the horizontal and vertical distance will both go past the center of the unit square, and we have an obvious contradiction. All other points that aren't on the corner of the square will be closer to the origin, and therefore using the same logic, it will go past the center of the square. This is better shown with a picture:
[asy][asy]
dot((0,0));
label("C", (0,0), N);
draw((-1, 1)--(1,1)--(1, -1)--(-1, -1)--(-1, 1));
dot((-1, -0.5));
dot((-1, 1));
dot((0.5, 1));
dot((0.5, -0.5));
draw((-1, 1)--(0.5, 1)--(0.5, -0.5)--(-1, -0.5)); [/asy][/asy]
Now I will prove that all $l\le \frac12$ work. If you put one vertice of the square with side length $l$ on the unit square, then the maximum distance of the opposite vertice to this point will be on the center of the unit square, and all other distances will be on a line in-between these two points. Therefore, all $l\le \frac12$ work and our proof is complete $\blacksquare$.
math_explorer
15.04.2012 04:34
Binomial-theorem wrote: All other points that aren't on the corner of the square will be closer to the origin, and therefore using the same logic, it will go past the center of the square. I'm unconvinced that this is obvious. A square not including a corner of the larger square is closer to the origin, but can rotate so its diagonal isn't pointed directly at the origin.