HMMT-Nov 2010 team, Problem 8

A function $f(x, y)$ is linear in $x$ and in $y . f(x, y)=\frac{1}{x y}$ for $x, y \in\{3,4\}$. What is $f(5,5)$ ?

Solution: The main fact that we will use in solving this problem is that $f(x+2, y)-f(x+1, y)=$ $f(x+1, y)-f(x, y)$ whenever $f$ is linear in $x$ and $y$. Suppose that $f(x, y)=a x y+b y+c x+d=$ $x(a y+c)+(b y+d)$ for some constants $a, b, c$, and $d$. Then it is easy to see that \[ \begin{aligned} f(x+2, y)-f(x+1, y) & =(x+2)(a y+c)+(b y+d)-(x+1)(a y+c)-(b y+d)=a y+c \\ f(x+1, y)-f(x, y) & =(x+1)(a y+c)+(b y+d)-x(a y+c)-(b y+d)=a y+c \end{aligned} \] which implies that $f(x+2, y)-f(x+1, y)=f(x+1, y)-f(x, y)$. In particular, $f(5, y)-f(4, y)=$ $f(4, y)-f(3, y)$, so $f(5, y)=2 f(4, y)-f(3, y)$. Similarly, $f(x, 5)=2 f(x, 4)-f(x, 3)$. Now we see that: \[ \begin{aligned} f(5,5) & =2 f(5,4)-f(5,3) \\ & =2[2 f(4,4)-f(3,4)]-[2 f(4,3)-f(3,3)] \\ & =4 f(4,4)-2 f(3,4)-2 f(4,3)+f(3,3) \\ & =\frac{4}{16}-\frac{4}{12}+\frac{1}{9} \\ & =\frac{1}{4}-\frac{1}{3}+\frac{1}{9} \\ & =\frac{1}{9}-\frac{1}{12} \\ & =\frac{1}{36} \end{aligned} \] so the answer is $\frac{1}{36}$. $\fbox{\frac{1}{36}}$.

Answer: \frac{1}{36}