Ana wins. At first turn Ana takes A1, A2, A3. Now pair up the other vertices (A4, A5), (A6, A7)…. (A98, A99). I claim that every time Barbu takes some vertice, Ana can take its pair. Wlog assume Barbu makes a triangle XYZ with Y being the new vertice. Let G be the pair of Y. And say Y and G are on different sides of YX. Since XYZ is an acute triangle, Angle(XZY) must be accute. But since our polygon is cyclic Angle(XGY) = 180 - Angle(XZY) > 90. Hence triangle XYG is obtuse. Hence Ana always has a move.

Ana always wins. Let's say $A$ and $B$ to the Ana and Barbu,respectively. Since $99$ is odd, $A$ will play the last move, so after any move of $B$, there is at least $1$ unchosen vertex of polygon. Let's show that if $B$ chooses triangle $XYZ$ and $L$ is unchosen point, then at least $1$ of triangles $ZLX$ and $ZLY$ is obtuse. Obviously $XYZ$ is acute. If $L$ lies on arc $XZ$ that doesn't contain $Y$, then $\angle XLZ=180-\angle XYZ>90$,so $XLZ$ is obtuse. Similarly if $L$ lies on arc $YZ$ that doesn't contain $X$,then $YLZ$ is obtuse. If $L$ lies on arc $XY$ that doesn't contain $Z$, then since $\angle LXZ+\angle LYZ=180$ at least $1$ of $\angle LXZ$ and $\angle LYZ$ should be obtuse, because they can't be $90 ($ if a regular polygon has odd number of vertices, then center of this polygon doesn't lie on any of its diogonals$)$. So at least one of $XLZ$ and $YLZ$ is obtuse, and $A$ can capture $L$ and write down $ZSL$ such that $ZSL$ is obtuse and $S\in\{X,Y\}$. Note: 300th posts

Jalil_Huseynov wrote: Ana always wins. Let's say $A$ and $B$ to the Ana and Barbu,respectively. Since $99$ is odd, $A$ will play the last move, so after any move of $B$, there is at least $1$ unchosen vertex of polygon. Let's show that if $B$ chooses triangle $XYZ$ and $L$ is unchosen point, then at least $1$ of triangles $ZLX$ and $ZLY$ is obtuse. Obviously $XYZ$ is acute. If $L$ lies on arc $XZ$ that doesn't contain $Y$, then $\angle XLZ=180-\angle XYZ>90$,so $XLZ$ is obtuse. Similarly if $L$ lies on arc $YZ$ that doesn't contain $X$,then $YLZ$ is obtuse. If $L$ lies on arc $XY$ that doesn't contain $Z$, then since $\angle LXZ+\angle LYZ=180$ at least $1$ of $\angle LXZ$ and $\angle LYZ$ should be obtuse, because they can't be $90 ($ if a regular polygon has odd number of vertices, then center of this polygon doesn't lie on any of its diogonals$)$. So at least one of $XLZ$ and $YLZ$ is obtuse, and $A$ can capture $L$ and write down $ZSL$ such that $ZSL$ is obtuse and $S\in\{X,Y\}$. Note: 300th posts You have to take in account that even if XLZ is obtuse, it still doesn’t mean you can always take it. (Because if the last vertex chosen by Barbu was Y, your triangle must have Y as well.

R8kt wrote: Jalil_Huseynov wrote: Ana always wins. Let's say $A$ and $B$ to the Ana and Barbu,respectively. Since $99$ is odd, $A$ will play the last move, so after any move of $B$, there is at least $1$ unchosen vertex of polygon. Let's show that if $B$ chooses triangle $XYZ$ and $L$ is unchosen point, then at least $1$ of triangles $ZLX$ and $ZLY$ is obtuse. Obviously $XYZ$ is acute. If $L$ lies on arc $XZ$ that doesn't contain $Y$, then $\angle XLZ=180-\angle XYZ>90$,so $XLZ$ is obtuse. Similarly if $L$ lies on arc $YZ$ that doesn't contain $X$,then $YLZ$ is obtuse. If $L$ lies on arc $XY$ that doesn't contain $Z$, then since $\angle LXZ+\angle LYZ=180$ at least $1$ of $\angle LXZ$ and $\angle LYZ$ should be obtuse, because they can't be $90 ($ if a regular polygon has odd number of vertices, then center of this polygon doesn't lie on any of its diogonals$)$. So at least one of $XLZ$ and $YLZ$ is obtuse, and $A$ can capture $L$ and write down $ZSL$ such that $ZSL$ is obtuse and $S\in\{X,Y\}$. Note: 300th posts You have to take in account that even if XLZ is obtuse, it still doesn’t mean you can always take it. (Because if the last vertex chosen by Barbu was Y, your triangle must have Y as well. No, since I assumed that $Z$ is the last vertex chosen by $B$ (I wrote that $B$ chooses $XYZ$), a triangle that $A$ will choose should contain point $Z$, not $Y$.

Sorry, somehow I missed that part