IMAR Mathematical Competition

The idea is to consider a suitable $\mathbb{R}[Y]$-linear combination of $f_0(X)$ and $f_2(1-X)$, namely, $(1-Y)f_0(X)+Y{f}_2(1-X)$, and a suitable $\mathbb{R}[X]$-linear combination of $f_1(Y)$ and $f_3(1-Y)$, namely, $X{f}_1(Y)+(1-X)f_3(1-Y)$, along with a suitable degree $2$ corrective term, $a_{11}XY+a_{10}X+a_{01}Y+a_{00}$, collect them together to form $$f(X,Y)=(1-Y)f_0(X)+X{f}_1(Y)+Y{f}_2(1-X)+(1-X)f_3(1-Y)+a_{11}XY+a_{10}X+a_{01}Y+a_{00}$$ and require the latter to satisfy the conditions in the statement.

Thus, the condition $f(X,0)=f_0(X)$ requires $a_{10}=f_3(1)-f_1(0)$ and $a_{00}=-f_3(1)$. Refer to the hypothesis on the $f_k$ to write $a_{10}=f_0(0)-f_1(0)$ and $a_{00}=-f_0(0)$. Next, the condition $f(0,1-Y)=f_3(Y)$ requires $a_{01}=f_0(0)-f_2(1)$ and $f_2(1)+a_{01}+a_{00}=0$. Notice that the latter holds automatically. With reference again to the hypothesis on the $f_k$, $a_{01}=f_0(0)-f_3(0)$. Similarly, the condition $f(1,Y)=f_1(Y)$ requires $a_{11}=f_0(1)-f_2(0)-a_{01}$ and $f_0(1)+a_{10}+a_{00}=0$. The latter holds again automatically, while the former yields $a_{11}=f_1(0)-f_2(0)-f_0(0)+f_3(0)$. Finally, the condition $f(1-X,1)=f_2(X)$ requires $f_3(0)-f_1(1)-a_{11}-a_{10}=0$ and $f_1(1)+a_{11}+a_{10}+a_{01}+a_{00}=0$, both of which hold by the preceding and the hypothesis on the $f_k$. In terms of the data, the desired polynomial is $$f(X,Y)=(1-Y)f_0(X)+X{f}_1(Y)+Y{f}_2(1-X)+(1-X)f_3(1-Y)-(f_0(0)-f_1(0)+f_2(0)-f_3(0))XY+(f_0(0)-f_1(0))X+(f_0(0)-f_3(0))Y-f_0(0).$$

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Alternative solution.

The polynomial in the previous solution may equally well be obtained as follows: begin by seeking a two-variable polynomial $f$ of the form $$f(X,Y)=g(X,Y)+Xh(Y)+Yk(X)+aXY,$$ where $g$ is a two-variable polynomial satisfying $g(X,0)=f_0(X)$ and $g(0,1-Y)=f_3(Y)$, $h$ and $k$ are one-variable polynomials subject to $h(0)=k(0)=0$ and $h(1)=k(1)$, and $a$ is a real number to be determined from the requirements. The common value $h(1)=k(1)$ will come out of the choice of $h$ and $k$. Notice that such a choice of $g$, $h$ and $k$ yields $f(X,0)=f_0(X)$ and $f(0,1-Y)=f_3(Y)$. Now, the two-variable polynomial $g(X,Y)=f_0(X)+f_3(1-Y)-f_0(0)$ clearly fits the bill; with reference to the hypothesis on the $f_k$, the verification is routine and hence omitted. Next, letting $h(Y)=f_1(Y)-g(1,Y)$ and $k(X)=f_2(1-X)-g(X,1)$, it is again readily checked that $h(0)=k(0)=0$, and $h(1)=f_1(1)-g(1,1)=f_2(0)-g(1,1)=k(1)$. In terms of the data, $h(Y)=f_1(Y)-f_3(1-Y)+f_0(0)-f_1(0)$, $k(X)=-f_0(X)+f_2(1-X)+f_0(0)-f_3(0)$, and $h(1)=k(1)=f_0(0)-f_1(0)+f_2(0)-f_3(0)$. At this stage, requiring $f(1,Y)=f_1(Y)$ yields $a=-f_0(0)+f_1(0)-f_2(0)+f_3(0)$; incidentally, yet not accidentally at all, this is precisely the value of $a_{11}$ in the previous solution. Finally, notice that $h(1)=-a$, to check the remaining requirement: $f(1-X,1)=g(1-X,1)+h(1)(1-X)+k(1-X)+a(1-X)=g(1-X,1)+k(1-X)=f_2(X)$.

Expressing $f$ in terms of the data yields the polynomial in the previous solution, just as mentioned in the beginning.