I was indeed missing something..

You are missing something The figure above must lie within the square $5\times 5$.

Is this a TST for IMO?

greengrapes, that is not sufficient: a rectangle that's $4\sqrt2 \times 1$ fits easily inside a $5\times5$ square.

JBL wrote: greengrapes, that is not sufficient: a rectangle that's $4\sqrt2 \times 1$ fits easily inside a $5\times5$ square. Oh, yeah.

Peter VDD wrote: Is this a TST for IMO? Yes, it is, and it is the last problem on the TST (so it is problably intended to be the hardest one).

In my opinion they weren't ordered by increasing difficulty. Yes it was a TST for IMO.

Just an idea of the proof:

Just saw this problem and I'll add my solution. I don't think Lithuanian tst's belongs to the preolympiad section, anyway, 1) Notice that if the figure is to fit inside the square then it is optimum to put the center of the figure in the center of the 5x5 square. This is because the figure is symmetric with respect to its center, so if we put the center elsewhere, we're actually trying to fit it in a rectangle which in turn fits in the 5x5 square. 2) Now consider the two diameters of the figure only. They are two segments with length $\sqrt{40}$ which form an angle. It is optimum again to have the big diagonal in the square bisecting the angle among them. This means, that the figure in the first post is actually placed in such a way that all we need to do is try to fit it into a 5x5 square with sides parallel to the sides of the figure. But this clearly can't be done since the figure is 6 squares high and long. Daniel[/url]

Consider the two column 3 and 4, we can easily see that one of them has to be the diagonal of the square since both of them have 5 boxes each. However, the length of the diagonal is $4\sqrt{2}$ which is greater than $5$. So, it is not possible to fit the figure into $5 x 5$ square.

Consider WLOG the smallest rectangle $ABCD$ with $AB$ and $CD$ at angle $\theta\leq\frac{\pi}{4}$ clockwise from horizontal. Also let the center be $(0,0)$. $AB$ and $CD$ go through $\pm(1,3)$. The width of the rectangle is then $2\sqrt{10}\cos\left(\arctan\frac{1}{3}-\theta\right)$. This is increasing for $\theta<\arctan\frac{1}{3}$ and decreasing for $\theta>\arctan\frac{1}{3}$ because $\arctan\frac{1}{3}-\theta\subset(-\pi,\pi)$. At $\theta=0$ the width is $6$. At $\theta=\frac{\pi}{4}$ the width is $4\sqrt{2}<6$. So every rectangle envelope must have one dimension of size at least $4\sqrt{2}$. But $4\sqrt{2}=\sqrt{32}>5$. So a the shape won't fit in a $5\times 5$ square.

no idea———-

Solved with: Kevin Wu, Isaac Zhu, Chris Qiu, Reyna Li, Ryan Yang, Elliot Liu, Alex Zhao, Eric Shen, David Dong, and Alan Bu.