I'm surprised nobody else has posted this. Yes, there are plenty of discussions in Art of Problem Solving, but why not here? And of course there are more problems.

Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA = FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA = DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA = ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD$, $FX$, and $ME$ are concurrent.

Find all positive integers $n$ for which each cell of an $n \times n$ table can be filled with one of the letters I, M, and O in such a way that:

- in each row and each column, one third of the entries are I, one third are M, and one third are O; and
- in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are I, one third are M, and one third are O.

**Note.** The rows and columns of an $n \times n$ table are each labelled $1$ to $n$ in a natural order. Thus each cell corresponds to a pair of positive integer $(i, j)$ with $1 \le i,j \le n$. For $n > 1$, the table has $4n-2$ diagonals of two types. A diagonal of first type consists all cells $(i, j)$ for which $i+j$ is a constant, and the diagonal of this second type consists all cells $(i, j)$ for which $i-j$ is constant.

Let $P = A_1 A_2 \cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, \ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.

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## Comments

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TopNewestI have an extension to Q1. (Almost got a complete solution for it, but i observed this, which follows):

Let this intersection point be labelled $O$. Prove $O$ is the circumcentre of $\Delta XBE$.

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Nice observation. If you complex trig bash it, the solution to your extension becomes very apparent. I'm currently typing up my solution to the original problem and can include your extension at the end.

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Nice, I have my own solution for it as well. Th essence for this extension is to just prove $OE=OX$ since $OX=OB$ is quite apparent.

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Here's my synthetic solution. Let $XF$ intersect $BD$ at $O$ and $EM$ at $O^{'}$. We need to show that $O$ coincides with $O^{'}$. A clear observation shows that $B,C,F,D,X$ are concyclic because $\angle\{DCF\}$=$\angle\{DBF\}$. Moreover $D$ is the circumcenter of triangle $CBA$. A clear angle chase shows that $\angle \{DXF\}$=$\angle\{DBF\}$ showing $O$ as the circumcenter of triangle $OBE$ thus $E{O^{'}}$=$BO$ Thus $O$ coincides with $O{'}$.proved Your extension can be proved easily if you watch $F$ as the incenter of triangle $DAB$ and watching $XFAD$ and $CDEM$ as the parallelogram. Thanks proving $O$ as the circumcenter was really very helpful.

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Oh about it, u were talking to me.. OK I am also trying to solve it with another way...

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Provide a solution using complex numbers.

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Well ur solution was really nice, I am also seeing it and trying to find a mistake..... 😂 😂😂😂😂😂😂😂😇😇😇😇😇😝😝😝😝😝😋

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