The idea of hosting a Junior Olympiad of Mathematics (JOM) for the junior students in the IMO camp ignited my thoughts of creating problems for the junior students.
In particular, I proposed my first problem to the Olympiad after being inspired by IMO 2012, Problem 4:
a problem that caused turmoils both among the students during the contests, and among the leaders and coordinators during the coordination sessions.
Nevertheless, some beauty of the problem still exists in this IMO problem, which I extracted it and turned it into a problem that made itself into N6 on the JOM shortlist.
Problem 5 of the same Olympiad also inspired me to create another geometry problem,
also for the first edition of the Junior Olympiad.
Unwilling to succumb to the fact that I didn't have enough time to attempt this problem on the contest (blame problem 4),
I later attempted it again and found an elegant solution using poles and polars.
That led me to the unexpected discovery of a geometric identity that eventually became problem G2 on the JOM shortlist.
Below are the 10 problems I proposed for the first three editions of JOM. Have fun solving!
JOM 2015, A6.
Let $(a_{n})_{n\ge 0}$ and $(b_{n})_{n\ge 0}$ be two sequences with arbitrary real values $a_0, a_1, b_0, b_1$. For $n\ge 1$, let $a_{n+1}, b_{n+1}$ be defined in this way:
$$a_{n+1}=\dfrac{b_{n-1}+b_{n}}{2}, b_{n+1}=\dfrac{a_{n-1}+a_{n}}{2}$$
Prove that for any constant $c>0$ there exists a positive integer $N$ s.t. for all $n>N$, $|a_{n}-b_{n}|< c$.
Let $(c_n)_{n\ge 0}$ be a sequence with $c_n=a_n-b_n$, then
$c_{n+2}=-\frac{c_n+c_{n+1}}{2}$. In the same way
we obtain $c_{n+3} = -\frac{c_{n+2}+c_{n+1}}{4}=\frac{c_n-c_{n+1}}{4}$.
Now, by triangle inequality $|c_{n+2}|\le \frac 12 (|c_n|+|c_{n+1}|)$
and $|c_{n+3}|\le\frac 14 (|c_{n+1}+|c_n|). $
Hence, $|c_{n+2}|+|c_{n+3}|\le \frac 34 (|c_n|+|c_{n+1}|)$.
Repeating the step gives $|c_{2n}|+|c_{2n+1}|\le \left(\frac 34\right)^n (|c_0|+|c_1|)$.
We can take $N=2\log _{\frac{3}{4}}\left(\frac{c}{|c_0|+|c_1|}\right)+2$
so that for all $n\ge N$,
$|c_n|$
$\le |c_n|+|c_{n+1}|$
$\le \left(\frac 34\right)^{\frac 12 n} (|c_0|+|c_1|)$
$< \left(\frac 34\right)^{\log _{\frac{3}{4}}\left(\frac{c}{|c_0|+|c_1|}\right)} (|c_0|+|c_1|)$
$=c$,
as desired.
JOM 2014, C6. Let $n$ be a positive integer.
At the beginning, two frogs, $A$ and $B$ are at point 0 of a number line.
At each second, they jump to the right and stop after arriving at $n$, following the constraints below:
(i) Each of them must jump either 1 or 2 step(s) every second.
(ii) Both of them must arrive at $n$ at the same time.
(iii) Frog $A$ must never be behind $B$ during the jump.
Let $f(n)$ be the number of arrangements of the sequence of jumps for the two frogs. Prove that
$f(n)\ge 2^{n-1}$.
Let $g(n)$ be the number of arrangements of the sequence of jumps that follows the constraint of $f(n)$ except frog $A$ arrives at $n+1$ and $B$ arrives at $n$ simultaneously.
The strategy is to induct on both $f(n)$ and $g(n)$ to prove that $f(n), g(n)\ge 2^{n-1}$.
Denote $(a_1+a_2+\cdots +a_k, b_1+b_2+\cdots +b_k)$ to be the sequence of jumps by frogs $A$ and $B$ where $a_i, b_i\in\{1, 2\}$.
Base case. $f(1)=1$ because $(1,1)$ is the only sequene. $f(2)=2$ because we have $(2,2), (1+1, 1+1)$.
$g(1)=1$ as we have $(2,1)$ while $g(2)=2$ as we have $(2+1, 1+1)$ and $(1+2, 1+1)$.
Inductive step.
Now assume that there is an $n\ge 2$ such that $f(k), g(k)\ge 2^{k-1}$ for all $k=1,2,\cdots ,n$.
In counting $f(n+1)$, we know that the second last position for $A$ and $B$ can be
$(n,n)$, $(n-1, n-1)$ or $(n, n-1)$. This means
$f(n+1)=f(n)+f(n-1)+g(n-1)\ge 2^{n-1}+2^{n-2}+2^{n-2}=2^n.$
In counting $g(n+1)$, the last step can be
$(1,1), (2,1), (1,2), (2,2)$ whereby the second last position is
$(n+1, n)$, $(n,n)$, $(n+1, n-1)$, $(n, n-1)$.
Thus $g(n+1)\ge g(n)+f(n)+g(n-1)\ge 2^{n-1}+2^{n-1}+2^{n-2}\ge 2^n$. Q.E.D.
Proposer's note: equality for $f(n)$ holds iff $n\le 4.$
JOM 2014, Problem 5 (G5).
Given $\triangle ABC$ with circumcircle $\Gamma$ and circumcentre $O$.
Let $X$ to be a point on $\Gamma$.
Let $XC_1$, $XB_1$ to be feet of perpendiculars from $X$ to lines $AB$ and $AC$.
Let $\omega_C$ to be circle with centre the midpoint of $AB$ and passing through $C_1$ .
Define $\omega_B$ similarly.
Prove that $\omega_B$ and $\omega_C$ has a common point on $XO$.
Let $P$ be the projection from $A$ to $XO$ and we claim that $P$ lies on $\omega_B$ and $\omega_C$.
Let the midpoint of $AC$ be $M_B$.
Since $\angle XB_1A=XPA=90^{\circ}$, we have $X, B_1, P, A$ concyclic and
$\angle (PB_1, B_1A)=\angle (PX, XA)$ and
$\angle (B_1P, PX)=\angle (B_1A, AX)$.
Also, since $O, M_B, A, P$ are concyclic (becuse $\angle OM_BA=\angle OPA=90^{\circ}$),
we have $\angle (OP, PM_B)=\angle (OA, AM_B)$.
So $\angle (PB_1, B_1A)=\angle (PB_1, B_1M_B)$
$=\angle (PX, XA)$
$=\angle (OX, XA)$
$=\angle (XA, AO)$
$=\angle (XA, AB_1)+\angle (AB_1, AO)$
$=\angle (XP, PB_1)+\angle (AM_B, AO)$
$=\angle (OP, PB_1)+\angle (PM_B, PO)$
$=\angle (PM_B, PB_1)$.
This means $M_BB_1=M_BP$ and $P$ lies on $\omega_B$.
Likewise, it lies on $\omega_C$.
JOM 2014, N11.
Prove that for all nonzero integers $c$ there exists composite positive integers $a, b$ such that:
(i) $a - b = c$.
(ii) $\gcd(a, b) = \gcd(a, c) = \gcd(b, c) = 1$.
Solution 1.
We may assume that $c>0$ since if $a-b=c$ then $b-a=-c$.
Let $p, q$ be distinct primes not divisible by $c$.
Consider the set $P=\{ap^2 | a\in\mathbb{N}\}$.
Since $p^2\perp q^2$,
$P$ contains the complete residues mod $q^2$,
and there exist infinitely many $a$ such that $ap^2\equiv c\pmod{q^2}$.
So there exists $a, b$ with $ap^2-bq^2=c$, and $b$ positive for $a$ sufficiently large.
Clearly, $ap^2$ and $bq^2$ are composite.
Notice that $a$ may not be relatively prime to $c$.
However, for any $x\in\mathbb{N}$ we have
$(a+xq^2)p^2-(b+xp^2)q^2=c$.
Moreover, $c, p, q$ are pairwise relatively prime.
Thus by Dirichlet's theorem, there are infinitely many $x$ such that
$a+xq^2$ and $c$ are relatively prime.
The rest of the conclusion follows from that
$\gcd(a, b) = \gcd(a, c) = \gcd(b, c)$.
Solution 2.
W.L.O.G. let $c>0$. Let $a=(2c+1)!+2c+1$ and $b=(2c+1)!+c+1$.
$\gcd (a,c)=\gcd (2c+1, c)=1$ since $c|(2c+1)!$.
Moreover, $2c+1|(2c+1)!+2c+1$ and $c+1|(2c+1)!+c+1$ and both $2c+1, c+1>1$.
JOM 2013, A5.
Find all polynomials $P\in \mathbb{R}[x]$ such that
$P(x)+P(y)+P(z)=0\Rightarrow x+y+z=0$.
Answer. All $P\in\mathbb{R}[x]$ such that $P(x) = ax$ or $P(x) = ax^{2n}Q(x)$ for some $a\neq 0$, $n\ge 0$, and $Q(x)$
with no real root. Solution 1 (Justin Lim). If $P(x) = 0$ for some $x$, then $3P(x) = 0 \Rightarrow x = 0$.
Suppose $P(0) \neq 0$, then $P$ has no real root, so we're done.
Suppose $P(0) = 0$. If $P$ takes both positive and negative
values, then $P$ is surjective on $R$.
Take infinitely many pairs $x,y$ such that $P(x)+P(y) = 0 \Rightarrow x+y =
0 \Rightarrow P(x) + P(-x) = 0.$
It follows that $P(x) + P(-x)\equiv 0$ for all real $x$. For fixed x, choose $z$ such
that $P(x) + P(x) + P(z) = 0 \Rightarrow z = -2x \Rightarrow 2P(x) = P(2x)$, which gives $\deg(P) = 1$.
Suppose $P$ takes only positive values or only negative values. Then clearly $P$ has even multiplicity
at 0, so $P(x) = x^{2n}Q(x)$ for some polynomial $Q(x)$ without real roots.
Conversely, it is easy to
check that the above polynomials work, since for $P(x) = ax$, $P(x) + P(y) + P(z) = a(x + y + z)$ and
$a(x + y + z) = 0 \Leftrightarrow x + y + z=0$.
For $P(x) = x^{2n}Q(x)$, clearly $P(x) + P(y) + P(z) \neq 0$ for $x, y, z$
with not all of them equal to 0.
JOM 2013, Problem 4 (C3).
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) $n=2013^{2013}$.
(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.
JOM 2013, G2.
Let $\omega_1$ and $\omega_2$ be two circles, with centres $O_1$ and $O_2$ respectively,
intersecting at $X$ and $Y$. Let a line tangent to both $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively.
Let $E,F$ be points on $O_1O_2$ such that $XE$ is tangent to $\omega_1$ and $YF$ is tangent to $\omega_2$. Let
$AE\cap \omega_1=A, C$ and $BF\cap\omega_2=B,D$.
Show that line $BO_2$ is tangent to the circumcircle of $\triangle ACD$ and line $AO_2$ is tangent to $\triangle BCD$.
We show that both $C$ and $D$ lie on the circle with $AB$ as diameter.
Denote $T$ as the midpoint of $AB$ (by power of point theorem $T, X, Y$ are collinear).
From $TA\perp AO_1$ and $TB\perp BO_2$ this will entail that $AO_1$ and $BO_2$ tangent to the circle with $AB$ as diameter.
One way to see it is that by symmetry, we have $YE$ tangent to $\omega_1$, too.
Thus $E$ is the pole of $XY$ w.r.t. $\omega_1$.
Also notice that the polar of $A$ w.r.t. $\omega_1$ is $AB$ itself.
Thus by La Hire's theorem, the pole of $AE$ is the intersection of the polar of $A$ (i.e. $AB$) and the polar of $E$ (i.e. $XY$),
which is $T$.
With $AE$ intersecting $\omega_1$ at $A$ and $C$, we know that $TA$ and $TC$ are both tangent to $\omega_1$.
Similarly, $TB$ and $TD$ are both tangent to $\omega_2$.
This entails $TA=TB=TC=TD$, Q.E.D.
JOM 2013, G5.
Let $ABCD$ be a convex quadrilateral, with $AD,BC$ intersecting at $F$. Choose lines
$\ell_1$, $\ell_2$ such that $\ell_1$ passes through $F$ and is tangent to circle $FAB$, and $\ell_2$ passes through $F$ and is tangent to circle $FCD$.
Let $\ell_1, ell_2$ intersect $AC, BD$ at $W, X, Y, Z$.
Prove that $A,B,C,D$ are concyclic if and only if $W,X, Y, Z$ are concyclic.
Suppose that $ABCD$ is cyclic.
Now $\angle WFA$
$=\angle FBA$
$=\angle FDC$
$=\angle ZFB$,
$\angle WAF$
$=\angle DAC$
$=\angle DBC$
$=\angle ZBF$.
Considering triangles $WAF$ and $ZBF$,
we know that $\angle WFA$
$=\angle ZFB$,
$\angle WAF$
$=\angle ZBF$,
$\angle AWF=\angle BZF\Rightarrow $
$\angle XWY=\angle XZY$.
So $WXYZ$ is cyclic.
For the other direction we neeed the following lemma:
Given a non-cyclic convex quadrilateral $ABCD$, we have
$\angle CDA+\angle CBA>180^{\circ}$
$\Leftrightarrow\angle CBD > \angle CAD$
$\Leftrightarrow\angle ADB > \angle ACB$
$\Leftrightarrow\angle BDC > \angle BAC$
$\Leftrightarrow\angle ABD > \angle ACD$.
Indeed, if we let $C'$ be on $AC$ with $ABC'D$ cyclic,
then $C$ lies outside the circle circumscribing $ABC'D$, and
$C'$ is on angle domain $DBC$ and angle domain $ADC$.
Therefore,
$\angle CBD >\angle C'BD= \angle CAD$,
$\angle ADB =\angle AC'B > \angle ACB$,
$\angle BDC > \angle BDC' = \angle BAC$,
$\angle ABD > \angle AC'D = \angle ACD$.
Now assume that $WXYZ$ is cyclic, but $ABCD$ is not.
W.l.o.g. let $\angle CDA+\angle CBA>180^{\circ}$.
Now $\angle ZFB=\angle FDC > \angle FBA=\angle WFA$.
Also that $\angle FWA=\angle FZB$, sine $\angle XWY=\angle XZY$.
So $\angle CD=\angle WAF=180^{\circ}-\angle FWA$
$=\angle WFA$
$>180^{\circ}-\angle FZB-\angle BFZ=\angle FBZ=\angle CBD$.
But from the lemma above we have $\angle CBD>\angle CAD$, contradiction.
The case where $\angle CDA+\angle CBA< 180^{\circ}$ is completely analogous.
JOM 2013, N5. Let $p$ be an odd prime such that $2^p+1|p^p+1$. Prove that any prime divisor of $2^p+1$ other than $3$ is greater than $6p$.
Let $q\neq 3$ be a prime dividing $2p + 1$ and it is clear that $q$ divides $p^p + 1$.
It is also clear that $3|2^p + 1$ (since $p$ is odd) and hence $p^p\equiv -1\pmod{3}$.
So $p\equiv -1\pmod{3}\Rightarrow p^p\equiv -1\pmod{6}$.
Also that $2^p\equiv -1\pmod{q}\Rightarrow 2^{2p}\equiv 1\pmod{q}$.
Let $r$ to be the minimal positive integer such that $2^r\equiv 1\pmod{q}$.
It is clear that $r|2p$, so $r\in\{1, 2, p, 2p\}$.
But $r=1,2$ implies that $q|1$ or $q|3$,
which is impossible, and by (1), $r\neq p$, so $r = 2p$.
But since $\phi(q)=q-1$ by Fermat's little theorem,
$2p|q-1$, so $\exists k\in\mathbb{N}$ s.t. $q=2kp+1$.
We will now show that $k\neq 1, 2$.
We have $q|p^p+1\Rightarrow q|\left(\frac{q-1}{2k}\right)^p+1$
$\Rightarrow q|\frac{(q-1)^p+(2k)^p}{(2k)^p}$
$\Rightarrow q|-1+(2k)^p$, since $q$ is relatively prime to $2k$.
If $k=1$, from $q|1+2^p$ follows $q|2$, which is impossible.
If $k=2$, $q=4p+1$, and so $q=4p+1\equiv 4(2)+1\equiv 0\pmod{3}$,
contradicting the fact that $q$ is prime.
Hence $k\ge 3$ and $q\ge 6p+1 > 6p$, as desired.
JOM 2013, N6. Find all functions
$f:\mathbb{Z}\to\mathbb{Z}$ such that for all prime $p$ the following condition holds:
$$p|ab+bc+ca\Leftrightarrow p|f(a)f(b)+f(b)f(c)+f(c)f(a).$$
Answer: $f(a)\equiv a$ or $f(a)\equiv -a$.
Denote $ab+bc+ca$ as $L$ and $f(a)f(b)+f(b)f(c)+f(c)f(a)$ as $R$.
It is not hard to see that $L=0\Leftrightarrow R=0$,
since 0 is the only integer divisible by all primes.
Hence plugging $a=b=c=0$ gives $f(0)=0$.
Also, $|L|=1$ iff $|R|=1$, since 1 is the only integer divisible by none of the primes.
Hence plugging $a=0, b, c\in\{1, -1\}$ gives
$f(1), f(-1)\in\{1, -1\}$.
Letting $a=0, b=1$ gives $L=c$, $R=f(1)f(c)$,
but since $|f(1)|=1$ we have $|R|=|f(c)|$ so
$p|c$ iff $p|f(c)$ for each prime $p$...(1).
Now suppose that $f(m)=f(n)$ for some integers $m$ and $n$.
Plugging $(a,b,c)=(m,-2m, -2m)$ gives
$L=0$, so $2f(m)f(-2m)+f(-2m)f(-2m)=R=0$.
Plugging $(a,b,c)=(n, -2m, -2m)$ gives
$R=2f(n)f(-2m)+f(-2m)f(-2m)$
$=2f(m)f(-2m)+f(-2m)f(-2m)$
$=0$.
This forces $L=0$, and therefore
$4m^2-4mn=0$, or $4m(m-n)=0$.
This forces $m=n$, or $m=0$.
The latter case means that $f(n)=f(0)=0$ for some integer $n$,
but from (1) we have $p|n$ for all primes $n$.
Hence $n=m=0$, in this case.
In either case we have $m=n$, so $f$ is injective...(2).
Now we prove that $f$ is odd, i.e. $f(-a)=-f(a)$ for all integers $a$.
Assume that $f(a)+f(-a)\neq 0$ for some $a$.
Let $b=-a$, then $L=-a^2$ for all integers $c$.
This means $p|a$ iff $p|L$ iff $p|R$ iff $p|f(a)f(-a)+f(c)(f(a)+f(-a))$.
Now, denote $u=\gcd(f(a)+f(-a), f(a)f(-a))$,
and for each integer $x$, denote $p(x)$ as the set of primes that divide $x$
(so $p(0)$ is the set of all primes, and $p(\pm 1)$ is empty).
We established the equivalence $p(c)=p(f(c))$ for all integers $c$.
Denote $f(a)+f(-a)=xu$ and $f(a)f(-a)=yu$, so $\gcd(x,y)=1$ and
$p(x)\subseteq p(a)$.
Let's choose some $c$ such that $p(c)=p(a)\backslash p(x)$,
then since $p(xy)=p(x)\cup p(y)$, we have
$p(x)\cap p(yf(c))$
$=p(x)\cap (p(y)\cup p(c))$
$=(p(x)\cap p(y))\cup(p(x)\cap p(c))$
$=\phi\cup\phi$
$=\phi$
so $x$ and $yf(c)$ are relatively prime.
Recall that $R=u(x+yf(c))$.
If $p(a)\backslash p(x)\neq\phi$, then there exists infinitely $c$ with
$p(c)=p(a)\backslash p(x)\neq\phi$, so
we can choose such $c$ satisfying $x+yf(c)\neq\pm 1$
(the fact that $f$ is injective implies that there are infinitely many possible values of $f(c)$).
Otherwise, $c=\pm 1$ and with $x\neq 0$ we know that either $x+y$ or $x-y$ is not in the set $\{1, -1\}$
(injectivity of $f$ shows that $f(1), f(-1)$ are precisely $1, -1$).
In either case there is a prime $q$ dividing $x+yf(c)$,
but with $\gcd (x, yf(c))=1$, $q$ divides neither of them.
This gives $q\not\in p(x), p(y), p(c)$ and with $p(x)\cup p(c)=p(a)$ we have
$q\not\in p(a)$, or $q\nmid -a^2=L$, contradiction.
Finally, we establish the equality $f(kx)=kf(x)$ for all integers $k, x$.
We induct on $k$ and the base case $k=1$ is obvious.
Now let $f(kx)=kf(x)$ for some $k> 0$ and for all $x$.
Letting $(a,b,c)=kx, -k(k+1)x, -(k+1)x$ gives $L=0$, so
$0=R=f(kx)f(-k(k+1)x)+f(kx)f(-(k+1)x)+f(-k(k+1)x)f(-(k+1)x)$
$=f(k(k+1)x)f((k+1)x)-f(kx)f((k+1)x)-f(kx)f(k(k+1)x)$
(notice the implicit use of the identity $f(-a)=-f(a))$.
The inductive hypothesis $f(kx)=kf(x)$ allows us to factor $k$ out form the equation, so
$0=kf((k+1)x)f((k+1)x)-kf(x)f((k+1)x)-kf(x)kf((k+1)x)$
$=kf((x+1)x)(f((k+1)x)-f(x)-kf(x))$,
since $k > 0$ anf $f((x+1)x)\neq 0$ for $x > 0$,
$f((k+1)x)=(k+1)f(x)$, as desired.
In particular, $f(c)=cf(1)$.
If $f(1)=1$, then $f(c)=c$ for all integers $c$ and if $f(1)=-1$,
then $f(c)=-c$ for all integers $c$.
Both functions satisfy the problem condition since $L=R$ in both cases.
Q.E.D.