International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria


Ivan is Watching You
Ivan's Office

Day 1
    Problem 1
    Problem 2
    Problem 3
    Problem 4
    Problem 5

Day 2
    Problem 6
    Problem 7
    Problem 8
    Problem 9
    Problem 10


    Day 1 questions
    Day 1 solutions
    Day 2 questions
    Day 2 solutions
    Closing Ceremony

Official IMC site

Problem 8

Problem 8. Let \(\displaystyle \Omega=\{(x,y,z)\in \mathbb{Z}^3: y+1\ge x\ge y\ge z\ge 0\}\). A frog moves along the points of \(\displaystyle \Omega\) by jumps of length \(\displaystyle 1\). For every positive integer \(\displaystyle n\), determine the number of paths the frog can take to reach \(\displaystyle (n,n,n)\) starting from \(\displaystyle (0,0,0)\) in exactly \(\displaystyle 3n\) jumps.

(Proposed by Fedor Petrov and Anatoly Vershik, St. Petersburg State University)

Solution. Let \(\displaystyle \Psi=\{(u,v)\in \mathbb{Z}^3: v\ge0, u\ge 2v \}\). Notice that the map \(\displaystyle \pi:\Omega\to\Psi\), \(\displaystyle \pi(x,y,z)=(x+y,z)\) is a bijection between the two sets; moreover \(\displaystyle \pi\) projects all allowed paths of the frogs to paths inside the set \(\displaystyle \Psi\), using only unit jump vectors. Hence, we are interested in the number of paths from \(\displaystyle \pi(0,0,0)=(0,0)\) to \(\displaystyle \pi(n,n,n)=(2n,n)\) in the set \(\displaystyle \Psi\), using only jumps \(\displaystyle {(1,0)}\) and \(\displaystyle {(0,1)}\).

For every lattice point \(\displaystyle (u,v)\in\Psi\), let \(\displaystyle f(u,v)\) be the number of paths from \(\displaystyle (0,0)\) to \(\displaystyle (u,v)\) in \(\displaystyle \Psi\) with \(\displaystyle {u+v}\) jumps. Evidently we have \(\displaystyle f(0,0)=1\). Extend this definition to the points with \(\displaystyle v=-1\) and \(\displaystyle 2v=u+1\) by setting

\(\displaystyle f(u,-1)=0, \quad f(2v-1,v)=0. \tag3 \)

To any point \(\displaystyle (u,v)\) of \(\displaystyle \Psi\) other than the origin, the path can come either from \(\displaystyle (u-1,v)\) or from \(\displaystyle (u,v-1)\), so

\(\displaystyle f(u,v)=f(u-1,v)+f(u,v-1) \quad\text{for \(\displaystyle (u,v)\in\Psi\setminus\{(0,0)\}\).} \tag4 \)

If we ignore the boundary condition (3), there is a wide family of functions that satisfy (4); namely, for every integer \(\displaystyle c\), \(\displaystyle (u,v)\mapsto\binom{u+v}{v+c}\) is such a function, with defining this binomial coefficient to be \(\displaystyle 0\) if \(\displaystyle v+c\) is negative or greater than \(\displaystyle u+v\).

Along the line \(\displaystyle 2v=u+1\) we have \(\displaystyle \binom{u+v}{v} = \binom{3v-1}{v} = 2\binom{3v-1}{v-1} = 2\binom{u+v}{v-1}\). Hence, the function

\(\displaystyle f^*(u,v) = \binom{u+v}{v} -2\binom{u+v}{v-1} \)

satisfies (3), (4) and \(\displaystyle f(0,0)=1\). These properties uniquely define the function \(\displaystyle f\), so \(\displaystyle f=f^*\).

In particular, the number of paths of the frog from \(\displaystyle (0,0,0)\) to \(\displaystyle (n,n,n)\) is

\(\displaystyle f(\pi(n,n,n)) = f(2n,n) = \binom{3n}{n} - 2\binom{3n}{n-1} = \dfrac{\dbinom{3n}{n}}{2n+1}. \)

Remark. There exist direct proofs for the formula \(\displaystyle \binom{3n}{n}/(2n+1)\). For instance, we can replicate the well-known proof of the formula for the Catalan numbers using the Cycle Lemma of Dvoretzky and Motzkin (related to the petrol station replenishment problem). See