AHZOLFAGHARI wrote: ... Later, everyone picks up a piece on the adjacent pieces before it is removed ... Does it mean "everyone picks up a piece that is adjacent to some pieces that have been removed before"? If I've correctly understood the problem, it could be solved as follows: 1) Note that Bahram can only take two pieces, so we have to pay attention to those pairs of pieces which account for more than $\frac{1}{2}$. 2) Call those $5$ pieces $A, B, C, D, E$ in clockwise order. For a graph $G$ where $V=\{A, B, C, D, E\}$, an edge is added between $X, Y\in V$if and only if $X, Y$ together accounts for more than $\frac{1}{2}$. 3) Now we cannot have $4$ or more points that has an edge on it except for the case when all these edges share a common vertex. Or else we can find $4$ parts $X, Y, Z, W$ where $X, Y$ is more than $\frac{1}{2}$ and $Z, W$ is more than $\frac{1}{2}$, absurd! If they share a common vertex, then Arash can take that common vertex at first so that Bahram can never take more than a half and thus Arash surely wins. 4) Again if there are $3$ or less vertices having an edge but all edges share a common vertex, Arash can still win by taking the common vertex in the very beginning. Now we only have to deal with the case where all the edges form a triangle. WLOG there only two cases:$(a) \Delta ABC, (b) \Delta ACD$. 4.a) Arash takes $B$ in the beginning. Bahram can only choose $A$ or $C$ in the turn. Arash takes the other one left among $A, C$. Arash wins. 4.b) Arash takes $A$ in the beginning. Then Arash takes $C$ if Bahram takes $B$ beforehand, and takes $D$ if Bahram takes $E$ before hand. $\Box$

I dont remember theire name so i just call those 2 A and B abd i call the pieces QWERT since B only takes 2 pieces we should have 2 pieces such that they are more than half of the cake and this two pieces are not adjant since A can eat the cake in some order such that B has to eat to cakes that are NOT adjant we say that Q+E is higher than half of the cake if A takes Q B shold take T and after the cake is finished the A has eated Q+W+R and B eated E+T and if A eats E in the same way after the cake is finished the pieces that A has eated areE+W+T and B eated Q+R and since (E+W+T)+(Q+W+R)>Q+W+E+R+T we should have that one of them is higher than half of the cake contradiction