Four positive integers $m,n,p,q$ with $p\lt m$ and $q\lt n$ are given. Consider $A(0,0)$, $B(p,0)$, $C(m,q)$, $D(m,n)$. Paths $f:A\to D$ and $g:B\to C$ consist of unit steps right or up. Let $S$ be the number of ordered pairs $(f,g)$ with no common points. Prove
$$S=\binom{m+n}{n}\binom{m+q-p}{q}-\binom{m+q}{q}\binom{m+n-p}{n}.$$
Let $A_0,B_0,C_0$ respectively be the midpoints of the altitudes $AH,BK,CL$ of a non-equilateral triangle $ABC$. Let $O$ and $I$ be its circumcenter and incenter. Its incircle touches $BC,CA,AB$ at $D,E,F$, respectively. Prove that $A_0D,B_0E,C_0F,OI$ are concurrent.
A function satisfies $f(0,0)=5^{2003}$, $f(0,n)=0$ for all $n\in\mathbb N$, and, for all integers $m\gt 0$ and $n$,
$$f(m,n)=f(m-1,n)-2\left\lfloor\frac{f(m-1,n)}2\right\rfloor+\left\lfloor\frac{f(m-1,n-1)}2\right\rfloor+\left\lfloor\frac{f(m-1,n+1)}2\right\rfloor.$$
Show that some positive integer $M$ satisfies $f(M,n)=1$ for $|n|\le(5^{2003}-1)/2$ and $f(M,n)=0$ otherwise.
Editorial clarification: the printed condition $n\in\mathbb N$ is preserved above. It omits negative initial indices and, if zero belongs to $\mathbb N$, conflicts with $f(0,0)$. The intended well-posed initial condition used for the proof is $f(0,n)=0$ for every integer $n\ne0$. The bracket notation in the source is read as floor.
Let $M,N,P$ be the midpoints of $BC,CA,AB$ respectively in triangle $ABC$. Let $M_1,N_1,P_1$ be points on its perimeter such that each of $MM_1,NN_1,PP_1$ bisects the perimeter. (a) Prove that these lines have a common point $K$. (b) Show that at least one of $KA/BC$, $KB/CA$, $KC/AB$ is not less than $1/\sqrt3$.
Here “bisects the perimeter” means that the two boundary arcs between a pair of endpoints have equal length.
Let $\mathcal A$ be the set of permutations of $1,2,\ldots,2003$ that fix no proper subset of $\{1,\ldots,2003\}$. For $a=(a_1,\ldots,a_{2003})\in\mathcal A$, let
$$d(a)=\sum_{i=1}^{2003}(a_i-i)^2.$$
(a) Find its minimum $d_0$. (b) Find all permutations attaining $d_0$.
Editorial clarification: “fix no proper subset” means leave no nonempty proper subset invariant as a set under $i\mapsto a_i$. The empty set is always invariant and must be excluded. Thus the permitted permutations are exactly single cycles of length $2003$.