Denote $ B_{i}\in A_{i + 1}A_{i + 3}\cap A_{i + 2}A_{i + 4}$, where the indices are taken modulo $ 5$. The basic thing is that the equality of the areas imply the following parallelism: $ A_{i}A_{i + 1}\| A_{i + 2}A_{i + 4}$. It is straightforward to see that $ M \equiv X \in A_{1}B_{1}\cap A_{2}B_{2}\cap A_{3}B_{3}\cap A_{4}B_{4}\cap A_{5}B_{5}$, based on the above parallelism. Instead, another solution, more nicer I think, is by observing that there exists an affine transformations that maps the initial pentagon in a regular one, and the problem is now more like trivial. Remark: A very similar problem, related to this one, is the following I posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=149421

I did not expect to see it here. This is was a problem that I gave to my team leader to propose somewhere and he did that for Romanian TST. So just to say that this is my problem, even so it may appear too easy.

pohoatza wrote: Now a straightforward solution is by proving that $ M \equiv X \in A_{1}B_{1}\cap A_{2}B_{2}\cap A_{3}B_{3}\cap A_{4}B_{4}\cap A_{5}B_{5}$, based on the above parallelism. Well, why they intersect at one point, and why the point $ M$ satisfies the condition?

Can anybody elaborate more on this?

Since $[A_1A_2A_3 ] = [A_2A_3A_4]$, we have $A_1A_4||A_2A_3$. Similarly are proven the relations $A_{i-1}A_{i+2}||A_iA_{i+1}$' where all indices are taken modulo $5$. Denote by $U$ the intersection of $A_1A_3$ and $A_2A_4$. Notice that $UA_4||A_1A_5$ and $U A_1||A_4A_5$' Thus $A_1A_5A_4U$ is a parallelogram, and $A_5 U$ passes through the midpoint $H$ of $A_1A_4$. Denote by $V$ the midpoint of $A_2A_3$' Since $A_1A_2A_3A_4$ is a trapezoid, it is well-known that $HU$ (i.e. $A_5 U$) passes through $V$. Let $G$ be the centroid of the pentagon. Notice that $G$ lies on $A_5 V$, since the centroids of $A_2A_3, A_1A_4$, and $A 5$ , all lie on $A_5 V$. Similarly, the segments joining the vertices $A_i$ with the midpoints of $A_{i+2}A_{i+3}$ will all pass through $G$. We shall prove that $G$ satisfies the requirement. Notice that $[A_5 GA_2 ] = [A_5GA_3]$ (from $A_2 V = A_3 V$). Similarly, it is proven that $[A_2 GA_5 ]=[A_2 GA_4 ] = [A_1 GA_4 ] = [A_1 GA_3 ] = [A_3 GA_5 ]$. Denote by $S_2$ this common value, by $S_1$ the common value of $[A_1 A_2 A_3 ] = ...,$ and by $S_T$ the area of the pentagon. It follows that $[A_2 GA_3 ] = S_T -[A_5 A_1 A_2 ]-[A_5A_4A_3] - [A_5 GA_2 ] - [A_5GA_3] = S_T - 2S_1 - 2S_2 $, therefore all $[A_iGA_{i+1}]$ are equal, and thus $G$ satisfies the desired requirement. So done.