Junior Olympiad of Mathematics, 2013

Introduction.

Inspired by the Experimental Lincoln Mathematical Olympiad, Justin and I decided to initiate a similar competition for Malaysian students who were selected into the IMO training camps. The organizers were the senior students (students who have already entered the camp more than once), and the participants were the junior students. The senior students then collectively come out with 29 insightful problems to be considered as the contest problems, selected 5 problems for the contest, and graded the scripts of the contestants.

Problems.

Day 1. (January 17, 2013, 2:30pm to 4:30pm, UTC+8)

Problem 1.

What is the minimum value of $\dfrac{m^m}{1\cdot 3\cdot 5\cdot \cdots \cdot (2m-1)}$ for positive integers $m$?

Zi Song Yeoh

Let the above number be $a_m$, for all $m$. Note that $a_1=1$. It suffices to prove that for all $m$, $a_m\ge 1$. Since $1+3+\dots+2m-1=m^2$, it follows that $\displaystyle m=\dfrac{1+3+\dots+2m-1}{m}\ge \sqrt[m]{1\cdot 3\cdot 5\dots 2m-1}$ by the $AM-GM$ inequality, which means that $a_m\ge 1$.
Alternative: Notice that by pairing up $2m>a>0$ with $2m-a>0$ by $AM-GM$ we have $2m=a+2m-a\ge 2\sqrt{a(2m-a)}\Rightarrow m^2\ge a(2m-a)$. So $1\cdot 3\dots (2m-1)=1(2m-1)3(2m-3)\dots (m-1)(m+1)\le m^m$ for $m$ even and $1(2m-1)3(2m-3)\dots (m-2)(m+2)(m)\le m^m$ for $m$ odd.

Problem 2.

Find all positive integers $a\in \{1,2,3,4\}$ such that if $b=2a$, then there exist infinitely many positive integers $n$ such that $$\underbrace{aa\dots aa}_\textrm{$2n$}-\underbrace{bb\dots bb}_\textrm{$n$}$$is a perfect square.

Zhi Yee Hoo

Let $x=111...1$ ($n$ ones). Notice that $aa...a-bb...b=x(10^{n-1}a-a)=9ax^{2}.$ So we know that the answer is yes for all $n$ if $a=1, 4$, and no for all $n$ if $a=2, 3.$ Hence the answer is $a=1, 4.$

Day 2. (January 18, 2013, 7:30am to 11:30am, UTC+8)

Problem 3.

The cells of an $n \times n$ table are filled with the numbers $1,2,\dots,n$ for the first row, $n+1,n+2,\dots,2n$ for the second, and so on until we have $n^2-n,n^2-n+1,\dots,n^2$ for the $n$-th row. Peter picks $n$ numbers from this table such that no two of them lie on the same row or column. Peter then calculates the sum $S$ of the numbers he has chosen. Prove that Peter always gets the same number for $S$, no matter how he chooses his $n$ numbers.

(Collectively constructed by organizers)

Let Peter choose $a_i$ from row $i$. Notice that the number in the $i$-th row and $j$-th column is $(k-1)n+j$. Let $a_i=(i-1)n+b_i$. Since no two numbers lie on the same column, the $b_i$ are all distinct. It follows that $\{b_1,b_2,\dots,b_n\}=\{1,2,\dots,n\}$. In particular, we have $$S=\sum_{i=1}^na_i=\sum_{i=1}^n(i-1)n+b_i=\sum_{i=1}^n(i-1)n+\frac{n(n+1)}{2}$$ which is clearly independent of the choice of the $a_i$.

Problem 4.

Let $n$ be a positive integer. A pseudo-Gangnam Style is a dance competition between players $A$ and $B$. At time $0$, both players face to the north. For every $k\ge 1$, at time $2k-1$, player $A$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $B$ is forced to follow him; at time $2k$, player $B$ can either choose to stay stationary, or turn $90^{\circ}$ clockwise, and player $A$ is forced to follow him.
After time $n$, the music stops and the competition is over. If the final position of both players is north or east, $A$ wins. If the final position of both players is south or west, $B$ wins. Determine who has a winning strategy when:
(a) $n=2013^{2012}$;
(b) $n=2013^{2013}$.

Anzo Teh

(a) At time $1$, let $A$ choose to be stationary. For the subsequent steps (i.e. $k\ge 1$), if $B$ stays stationary at time $2k$ then $A$ moves $90^{\circ}$ at time $2k+1$. On the other hand, if $B$ turns $90^{\circ}$ at time $2k$ then $A$ stays stationary at time $2k+1.$ So their positions form a cycle of period 8. In particular, at time $8k+1$ both players will come to original position, i.e. north. Since $2013^{2012}\equiv 1\pmod {8},$ their final position will be north and $A$ will win.
(b) For $k\ge 1$, if $A$ stays stationary at time $2k-1$ then $B$ moves $90^{\circ}$ at time $2k$. On the other hand, if $A$ turns $90^{\circ}$ at time $2k-1$ then $B$ stays stationary at time $2k.$ Again, it is a cycle of period 8, and at time $8k+4$ both players are facing south. Since $2013^{2013}\equiv 5\pmod 8$, we know that at time $2013^{2013}-1$ both players are at south, and for the final step $A$ can only choose to stay facing south or turn $90^{\circ}$ clockwise to face west. So $B$ will definitely win.

Problem 5.

Consider a triangle $ABC$ with height $AH$ and $H$ on $BC$. Let $\gamma_1$ and $\gamma_2$ be the circles with diameter $BH,CH$ respectively, and let their centers be $O_1$ and $O_2$. Points $X,Y$ lie on $\gamma_1,\gamma_2$ respectively such that $AX,AY$ are tangent to each circle and $X,Y,H$ are all distinct. $P$ is a point such that $PO_1$ is perpendicular to $BX$ and $PO_2$ is perpendicular to $CY$.

Prove that the circumcircles of $PXY$ and $AO_1O_2$ are tangent to each other.

Justin Lim

Notice that $AH=AX=AY$ by the power of a point theorem, since $AH$ is clearly tangent to both $\gamma_1$ and $\gamma_2$. Now note that $$\angle XBC=\angle AXH=90-\frac{\angle XAH}{2}=90-\angle GHD=180-(90+\angle GHD)=180-\angle GHC$$. Thus $B,C,X,Y$ are concyclic. But since $O_1X=O_1B$, we have $PB=PX$ and similarly $PY=PC$, thus $P$ is the circumcenter of $B,C,X,Y$. Now note that $$\angle PO_1O_2=180-\angle BO_1P=180-\angle XO_1P=\frac{\angle XO_1B}{2}=90-\angle XBH=90-\angle XHA=90-\angle AO_1H$$ thus $\angle AO_1P=\angle AO_2P=90$. It follows that $AP$ is a diameter in circle $AO_1O_2$, thus $AP$ contains its center. But we know that $AX=AY$ and $PX=PY$, thus $AP$ is the perpendicular bisector of $XY$. So the center of circle $XPY$ lies on $AP$ as well. And since $P$ lies on both circles, they are tangent at $P$, as desired.