2006 South East Mathematical Olympiad

Day 1

1

Suppose $a>b>0$, $f(x)=\dfrac{2(a+b)x+2ab}{4x+a+b}$. Show that there exists an unique positive number $x$, such that $f(x)=\left(\dfrac{a^{\frac{1}{3}}+b^{\frac{1}{3}}}{2} \right)^3$.

2

In $\triangle ABC$, $\angle ABC=90^{\circ}$. Points $D,G$ lie on side $AC$. Points $E, F$ lie on segment $BD$, such that $AE \perp BD $ and $GF \perp BD$. Show that if $BE=EF$, then $\angle ABG=\angle DFC$.

3

There is a standard deck of $52$ cards without jokers. The deck consists of four suits(diamond, club, heart, spade) which include thirteen cards in each. For each suit, all thirteen cards are ranked from “$2$” to “$A$” (i.e. $2, 3,\ldots , Q, K, A$). A pair of cards is called a “straight flush” if these two cards belong to the same suit and their ranks are adjacent. Additionally, "$A$" and "$2$" are considered to be adjacent (i.e. "A" is also considered as "$1$"). For example, spade $A$ and spade $2$ form a “straight flush”; diamond $10$ and diamond $Q$ are not a “straight flush” pair. Determine how many ways of picking thirteen cards out of the deck such that all ranks are included but no “straight flush” exists in them.

4

Given any positive integer $n$, let $a_n$ be the real root of equation $x^3+\dfrac{x}{n}=1$. Prove that (1) $a_{n+1}>a_n$; (2) $\sum_{i=1}^{n}\frac{1}{(i+1)^2a_i} <a_n$.

Day 2

1

In $\triangle ABC$, $\angle A=60^\circ$. $\odot I$ is the incircle of $\triangle ABC$. $\odot I$ is tangent to sides $AB$, $AC$ at $D$, $E$, respectively. Line $DE$ intersects line $BI$ and $CI$ at $F$, $G$ respectively. Prove that $FG=\frac{BC}{2}$.

2

Find the minimum value of real number $m$, such that inequality \[m(a^3+b^3+c^3) \ge 6(a^2+b^2+c^2)+1\] holds for all positive real numbers $a,b,c$ where $a+b+c=1$.

3

(1) Find the number of positive integer solutions $(m,n,r)$ of the indeterminate equation $mn+nr+mr=2(m+n+r)$. (2) Given an integer $k (k>1)$, prove that indeterminate equation $mn+nr+mr=k(m+n+r)$ has at least $3k+1$ positive integer solutions $(m,n,r)$.

4

Given a circle with its perimeter equal to $n$( $n \in N^*$), the least positive integer $P_n$ which satisfies the following condition is called the “number of the partitioned circle”: there are $P_n$ points ($A_1,A_2, \ldots ,A_{P_n}$) on the circle; For any integer $m$ ($1\le m\le n-1$), there always exist two points $A_i,A_j$ ($1\le i,j\le P_n$), such that the length of arc $A_iA_j$ is equal to $m$. Furthermore, all arcs between every two adjacent points $A_i,A_{i+1}$ ($1\le i\le P_n$, $A_{p_n+1}=A_1$) form a sequence $T_n=(a_1,a_2,,,a_{p_n})$ called the “sequence of the partitioned circle”. For example when $n=13$, the number of the partitioned circle $P_{13}$=4, the sequence of the partitioned circle $T_{13}=(1,3,2,7)$ or $(1,2,6,4)$. Determine the values of $P_{21}$ and $P_{31}$, and find a possible solution of $T_{21}$ and $T_{31}$ respectively.