Argentina_2018

The sides of each of the 80 angles pass through the endpoints of a side of the 10-gon $P$. We say that such an angle and such a side are adjacent; each side is adjacent to exactly 8 angles. Call black the 59 angles that are known to be equal and $\alpha$ the measure of a black angle. Let $a$ be a side of $P$. The locus of points $X$ such that $a$ subtends angle $\alpha$ at $X$ is the union of two circular arcs. Denote by $y_a$ the one of them which lies on the same side of $a$ as the polygon $P$. We also name $y_a$ the entire circle containing $y_a$. All black angles adjacent to $a$ have their vertices on $y_a$.

For a side $a$ let $m_a$ be the number of black angles adjacent to $a$. The hypothesis can be stated as $\sum m_a\ge 59$. We show that the inequality implies that $P$ is cyclic. Consider two cases.

Let there be a side $a$ with $m_a\ge 7$. Then $y_a$ contains the vertices of at least 7 black angles; these vertices are different from the endpoints of $a$. Hence at least $2+7=9$ vertices of $P$ lie on $y_a$.

Suppose that there is a vertex $A\in y_a$, and let $AB=b$, $AC=c$ be the sides with common vertex $A$. Then $y_b\ne y_a$, $y_c\ne y_a$ by $A\in y_a$. So arcs $y_a$ and $y_b$ has (at most) two common points; one such point is $B$. Because all vertices except $A$ are on $y_a$, we see that $y_b$ contains at most one vertex different from $A$ and $B$, implying $m_b\le 1$. On the other hand it is immediate that $y_a=y_a$ for every $s\ne b,c$, hence $A\in y_a$ for $s\ne b,c$. Thus $m_s\le 7$ for each of the 8 sides $s$ different from $b$ and $c$. In conclusion $\sum m_a\le 1+1+8\cdot 7=58$, contradicting the hypothesis.

Suppose now that $m_a\le 6$ for each side $a$. Then a direct computation using $\sum m_a\ge 59$ shows that $m_a=6$ holds for at least 9 sides $a$: the last side satisfies $m_a=5$ or $m_a=6$. We show that $y_b=y_c$ for every two consecutive sides $b=AB,c=AC$; this is enough to imply that $P$ is cyclic. Indeed $y_b$ contains at least $m_b-1$ vertices different from $A,B$ and $C$. Likewise $y_c$ contains at least $m_c-1$ vertices different from $A,B$ and $C$. Both arcs combined contain at least $m_b+m_c-2\ge 6+5-2=9$ vertices $D\ne A,B,C$. It follows that there are two vertices $D,E\ne A,B,C$ that are common for $y_b$ and $y_c$. One more such vertex is $A$, so $y_b=y_c$, as stated.

Now that $P$ is cyclic, each side $BC=a$ there corresponds an angle ${\alpha }_a$ such that $\overrightarrow{BC}={\alpha }_a$ for every vertex $V\ne B,C$. Also ${\alpha }_a$ occurs among the 80 angles of division exactly $8k$ times where $k$ is the number of sides with length $a$. Because at least 59 angles are known to be equal, $P$ has at least 8 equal sides. The angle ${\alpha }_a$ corresponding to them occurs 64, 72 or 80 times. It is straightforward

now that the 80 angles of division can assume at most 3 different values. If these are exactly 3 then one of them occurs 64 times, and each of the other two occurs 8 times.