Let the n-gons be A1 A2 ... An and B1 B2 ... Bn, such that clockwise we encounter them in this order : A1, B1, A2, B2, ... {C1} = B1Bn \cap A1A2 {C2} = B1B2 \cap A1A2 {C3} = B1B2 \cap A2A3 ................................... Let a = length of perimeter of C1 C2 .. C2n which belongs to A1 A2 .. An, and b the similar for the second n-gon. Let p be the common value of the perimeters of the n-gons. Triangles Bi C(2i-1) C2i and Bj C(2j-1) C2j are similar, for all i,j. Therefore : (C(2i-1)Bi+BiC2i)/C(2i-1)C2i = (C(2j-1)Bj+BjC2j)/C(2j-1)C2j = (sum C(2i-1)Bi+sum BiC2i)/(sum C(2i-1)C2i) = (p-b)/a Also, notice that in a similar manner for the other triangles, we obtain that each ratio like above is equal to (p-a)/b. But by the sine law the two ratios are equal, and therefore : (p-b)/a = (p-a)/b ===> pb - b^2 = pa - a^2 ===> (a-b)(a+b) - p(a-b) = 0 ===> (a-b)(a+b-p)=0 But by triangle inequality we have C(2i-1)C2i<C(2i-1)Bi + BiC(2i+1). Summing up yields b<p-a, therefore a+b<p. So a=b, and we are done.

I have seen a rather nicer solution: Let O1 and O2 be their centers. Shift the first one with the vector O1O2 such that the problem becomes isomorphic. Note that through this transfer the share of each n-gon in the total perimeter doesn’t change and that in the latter situation the problem is trivial. This problem for n=2 has been used in USSR math Olympiad.

What you mean when you say that the problem becomes isomorphic? Lich: Chamos literally traduced is a way to refer to young people here im venezuela. So kinda everybody that participates on the IMO are chamos :D:D

I meant by isomorphic that: If you look the figure in every direction its the same thing so the problem becomes trivial.

I don't know why it lies in Combinatorics. I will move it Geometry. Though, I think problem was solved.

Here is the solution given in our yearly collection of training problems. It was probably copied from the official solution. It is easy to see that the intersection $S\cap T$ has $2n$ sides if and only if the sides of $S\cap T$ alternate: blue, red, blue, red, etc. Denote the vertices of $S\cap T$ as $C_1D_1C_2D_2\ldots,C_nD_n$ so that the sides $C_1D_1$, $C_2D_2$, $\ldots$, $C_nD_n$ are blue. Denote the vertices of $S$ by $A_1, A_2,\ldots, A_n$ and the vertices of $T$ by $B_1,B_2,\ldots,B_n$ so that $C_1D_1\subset B_1B_2,\ldots, C_nD_n\subset B_nB_1$ and $D_nC_1\subset A_1A_2,\ldots, D_{n-1}C_n\subset A_nA_1$. One can check that all the triangles $D_nB_1C_1, D_1B_2C_2,\ldots, D_{n-1}B_nC_n$ and $C_1A_2D_1, C_2A_3D_2,\ldots, C_nA_1D_n$ are similar. Therefore $$ \frac{D_nC_1}{D_nB_1+B_1C_1}=\frac{D_1C_2}{D_1B_2+B_2C_2}=\cdots=\frac{D_{n-1}C_n}{D_{n-1}B_n+B_nC_n}$$ $$=\frac{C_1D_1}{C_1A_2+A_2D_1}=\frac{C_2D_2}{C_2A_3+A_3D_2}=\cdots=\frac{C_nD_n}{C_nA_1+A_1D_n}. $$ Hence $$\frac{D_nC_1+\cdots+D_{n-1}C_n}{D_nB_1+B_1C_1+\cdots+D_{n-1}B_n+B_nC_n}=\frac{C_1D_1+\cdots+C_nD_n}{C_1A_2+A_2D_1+\cdots+C_nA_1+A_1D_n}.$$ Let $x=D_nC_1+D_1C_2+\cdots+D_{n-1}C_n$ and $y=C_1D_1+C_2D_2+\cdots+C_nD_n$, so that $x$ is the sum of the red sides of $S\cap T$ and $y$ is the sum of the blue sides. If $a$ is the length of a side of $S$ (or $T$), then the equality above can be written in the following form $$\frac{x}{na-y}=\frac{y}{na-x}$$ or equivalently $$(na-x-y)(x-y)=0.$$ Since the perimeter of $S\cap T$ is strictly less than the perimeter $na$ of $S$ or $T$, we obtain $x=y$.