Consider a variable point $P$ inside a given triangle $ABC$. Let $D$, $E$, $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$, respectively. Find all points $P$ which minimize the sum \[ {BC\over PD}+{CA\over PE}+{AB\over PF}. \]

Take $r$ such that $1\le r\le n$, and consider all subsets of $r$ elements of the set $\{1,2,\ldots,n\}$. Each subset has a smallest element. Let $F(n,r)$ be the arithmetic mean of these smallest elements. Prove that: \[ F(n,r)={n+1\over r+1}. \]

Determine the maximum value of $m^2+n^2$, where $m$ and $n$ are integers in the range $1,2,\ldots,1981$ satisfying $(n^2-mn-m^2)^2=1$.

a.) For which $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers? b.) For which $n>2$ is there exactly one set having this property?

Click for solution Let $k = \prod p_{i}^{e_{i}}$ be the largest element of the set. Then $k$ divides the least common multiple of the other elements of the set iff the set has cardinality of at least $\max \{ p_{i}^{e_{i}}\}+1$, since for any of the $p_{i}^{e_{i}}$, we must go down at least to $k-p_{i}^{e_{i}}$ to obtain another multiple of $p_{i}^{e_{i}}$. In particular, there is no set of cardinality $3$ satisfying our conditions, because each number larger than or equal to $3$ must be divisible by a number that is larger than two and is a power of a prime. For $n > 3$, we may let $k = \operatorname{lcm}[n-1, n-2] = (n-1)(n-2)$, since all the $p_{i}^{e_{i}}$ must clearly be less than $n$ and this product must also be larger than $ n$ if $n$ is at least $4$. For $n > 4$, we may also let $k = \operatorname{lcm}[n-2, n-3] = (n-2)(n-3)$, for the same reasons. However, for $ n = 4$, this does not work, and indeed no set works other than $\{ 3,4,5,6 \}$. To prove this, we simply note that for any integer not equal to $6$ and larger than $4$ must have some power-of-a-prime factor larger than $3$. Q.E.D.

Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.

Click for solution Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \equiv b \cap c,\ B' \equiv c \cap a,\ C' \equiv a \cap b$, forming a triangle $\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \equiv B'C' \parallel BC,\ b \equiv C'A' \parallel CA,\ c \equiv A'B' \parallel AB$. Thus the triangles $\triangle A'B'C' \sim \triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\angle A', \angle B', \angle C'$ (and also of the angles $\angle A, \angle B, \angle C$), hence, the homothety center is the common incenter I of the triangles $\triangle A'B'C',\ \triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I.

The function $f(x,y)$ satisfies: $f(0,y)=y+1, f(x+1,0) = f(x,1), f(x+1,y+1)=f(x,f(x+1,y))$ for all non-negative integers $x,y$. Find $f(4,1981)$.