Let's solve RMO 2023
Let $\mathbb{N}$ be the set of all positive integers and $S=\left\{(a, b, c, d) \in \mathbb{N}^4: a^2+b^2+c^2=d^2\right\}$. Find the largest positive integer $m$ such that $m$ divides $a b c d$ for all $(a, b, c, d) \in S$.
Solution
Notice that $(2, 2, 1, 3)\in S\Rightarrow m$ is a divisor of $12 = 2\times 2\times 1\times 3$. We can verify easily by taking $0^2, 1^2, 2^2, 3^2$ that any perfect square leaves a remainder of either $0$ (or) $1$ when divided by $3$ and $4$. Suppose, if none of $a, b, c$ is divisible by $3$, then $a^2+b^2+c^2$ will leave a remainder of $0$ when divided by $3\Rightarrow 3$ divides $d$. Hence, one of $a,b,c,d$ will be divisible by $3$. Similarly, suppose let atmost one of the numbers $a, b, c$ be even, then $a^2+b^2+c^2$ will leave a remainder of $2$ (or) $3$ when divided by $4$ which cannot be a perfect square. So, atleast two of the numbers $a, b, c, d$ will be even $\Rightarrow 4$ divides $abcd$. Hence, $abcd$ is divisible by $12$ for all $(a, b, c, d) \in S$ and we know $m$ divides $12$. Hence, the largest possible value of $m$ is $12$.
Let $\omega$ be a semicircle with $A B$ as the bounding diameter and let $C D$ be a variable chord of the semicircle of constant length such that $C, D$ lie in the interior of the arc $A B$. Let $E$ be a point on the diameter $A B$ such that $C E$ and $D E$ are equally inclined to the line $A B$. Prove that
(a) the measure of $\angle C E D$ is a constant;
(b) the circumcircle of triangle $C E D$ passes through a fixed point.
Solution
First, let us prove that for given points $C$ and $D$, the point $E$ is unique. $\angle DEA = \angle CEB=\alpha$ because we cannot have $\angle DEA=\angle CEA$ as points $C, D, E$ cannot be collinear as otherwise point $E$ would lie outside of the diameter $AB$. Now, reflect point $C$ about the diameter $AB$ to get $C'$ and notice that $\angle C'EB=\alpha=\angle DEA$, so $D, E, C'$ are collinear because the vertically opposite angles are equal. Since $C, D$ are given, the distinct lines $DC'$ and $AB$ are given and so its intersection $E$ will always be unique as shown below.
Let $O$ be the center of $\omega$. Now let us construct point $E$ in a different way. Let the circumcircle of $\triangle OCD$ intersect $AB$ at $E'$, other than $O$. Notice that $\angle CE'B = \angle ODC = \angle OCD = \angle OE'D= \angle DE'A$ as $\triangle OCD$ is isoceles and the points $O, D, C, E'$ are concyclic. So, we have $\angle CE'B = \angle DE'A$, which means $CE'$ and $DE'$ are equally inclined to $AB$ and hence $E' = E$ because $E$ is an unique point.
If $R$ is the radius of $\omega$, we know that
$CD=2R\sin{\angle COD}\Rightarrow \angle COD=\sin^{-1}{\left(\frac{CD}{2R}\right)}\Rightarrow \angle COD = \angle CED$ is constant (note that $\angle COD<90^{\circ}$ and hence $sin^{-1}$ of it is unique). Also the circumcircle of $\triangle CED$ always passes through the fixed point $O$.
For any natural number $n$, expressed in base 10, let $s(n)$ denote the sum of all its digits. Find all natural numbers $m$ and $n$ such that $m<n$ and
$$(s(n))^2=m \quad \text { and } \quad(s(m))^2=n.$$
Solution
$m<n\Rightarrow m<\left(s(m)\right)^2$. So, let us find the set of natural numbers that satisfy this condition.
CLAIM : $p>\left(s(p)\right)^2, \forall p\geq 1000$
Proof:
Let us prove the claim using induction on the number of digits of $p$. Denote the number digits of $p$ by $N(p)$. For base case,
$\underline{N(p)=4:}$
Maximum value of $s(p)=4\times 9=36$ and $36^2=1296$. Hence, $p<1296$. But now if $p<1296\Rightarrow s(p)<1+9+9+9=28\Rightarrow \left(s(p)\right)^2 < 28^2=784<p$, hence not possible.
Let the claim be true for $\underline{N(p)=k}$, for some integer $\underline{k\geq4:}$
Take any number $p$ with $N(p)=k+1$ and let its first $k$ digits be $t$, then by induction hypothesis, $t>\left(s(t)\right)^2$ and $s(p)\leq s(t)+9$ and $p\geq 10t$.
If $s(t)\leq 18$, then $s(p)\leq 18+9=27\Rightarrow \left(s(p)\right)^2<1000<p$ and the claim follows. So, let $s(t) >18$, then
$\left(s(p)\right)^2\leq \left(s(t)+9\right)^2=\left(s(t)\right)^2+81+18\times \left(s(t)\right)<\left(s(t)\right)^2+81+\left(s(t)\right)^2<t+t+t=3t<10t\leq p\Rightarrow \left(s(p)\right)^2\leq p$
which proves our induction claim. $\blacksquare$
So, $m<1000$ by the CLAIM $\Rightarrow s(m)\leq 27\Rightarrow n\leq 27^2\Rightarrow m<27^2$. Now, we know that $s(n)$ and $n$ leave the same remainder when divided by $9$ due to the divisibility rule and since $n$ is a perfect square, $s(n)$ leaves a remainder from one of $0,1,4,7$ when divided by $9$. So, the possible values of $s(n)$ are $\{1,4,7,9,10,13,16,18,19,22,25\}$. Out of these, only for $m=\{4^2,7^2,13^2\}$, we get $m<\left(s(m)\right)^2$. Now, in these, only $m=13^2$ satisfies $m=\left(s(\left(s(m)\right)^2)\right)^2$. Hence, $m=169,\ n=256$ is the the only solution.
Let $\Omega_1, \Omega_2$ be two intersecting circles with centres $O_1, O_2$ respectively. Let $l$ be a line that intersects $\Omega_1$ at points $A, C$ and $\Omega_2$ at points $B, D$ such that $A, B, C, D$ are collinear in that order. Let the perpendicular bisector of segment $A B$ intersect $\Omega_1$ at points $P, Q$; and the perpendicular bisector of segment $C D$ intersect $\Omega_2$ at points $R, S$ such that $P, R$ are on the same side of $l$. Prove that the midpoints of $P R, Q S$ and $O_1 O_2$ are collinear.
Solution
Let $M, N$ denote the foot of perpendicular from $O_1, O_2$ to $AC, BD$ respectively. So, $M, N$ are the midpoints of $AC, BD$ respectively $\Rightarrow GM=GC-MC=\frac{AB}{2}+BC-\left(\frac{AB}{2}+\frac{BC}{2}\right)=\frac{BC}{2}$. Similarly, $NI=\frac{BC}{2}\Rightarrow GM=NI$.
CLAIM : In any trapezium, the line joining midpoints of the lateral sides will be parallel to its base.
This statement can be proved by extending the lateral sides to meet at a point and use similar triangles. Let $T$ be the midpoint of $GI$. Since $GM=NI\Rightarrow T$ will also be the midpoint of $MN$. Notice that $GIRP, GISQ, MNO_2O_1$ are trapeziums. Now consider a line $t$ through $T$ parallel to $NO_2$ and by the CLAIM above, $t$ will pass through the midpoint of $O_1O_2$ and since $NO_2\ ||\ IS\ ||\ IR$, the line $t$ will also pass through the midpoints of $QS, PR$. Hence, the midpoints of $P R, Q S$ and $O_1 O_2$ are collinear on the line $t$.
Let $n>k>1$ be positive integers. Determine all positive real numbers $a_1, a_2, \ldots, a_n$ which satisfy
$$
\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}=\sum_{i=1}^n a_i=n .
$$
Solution
Since $a_i's$ are positive and $k\in\mathbb{N}$, by AM-GM,
$$\frac{\underbrace{1+1+\cdots+1}_{k-1\ times}+\frac{1}{a_i^{k}}}{k}\geq \left(1\cdot 1\cdots 1\cdot\frac{1}{a_i^{k}}\right)^{\frac{1}{k}}=\frac{1}{a_i}$$
$$\Rightarrow \sqrt{a_i}\geq \sqrt{\frac{k}{k-1+\frac{1}{a_i^{k}}}}=\sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}$$
and the equality holds if and only if $\frac{1}{a_i^{k}}=1\Rightarrow a_i=1,\ \forall i\in\{1,2,\ldots,n\}$.
By the QM-AM inequality,
$$\sqrt{\frac{a_1+a_2+\cdots+a_n}{n}}\geq \frac{\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}}{n}\geq \frac{\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}}{n}$$
We know that $a_1+a_2+\cdots+a_n=n$. So we get,
$$1\geq \frac{\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}}{n}\Rightarrow \sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}\leq n$$
and the inequality holds true if and only if $a_i=1,\ \forall i\in\{1,2,\ldots,n\}$. Hence to satisfy the equality given in the question $a_i=1,\ \forall i\in\{1,2,\ldots,n\}$.
Consider a set of 16 points arranged in a $4 \times 4$ square grid formation. Prove that if any 7 of these points are coloured blue, then there exists an isosceles right-angled triangle whose vertices are all blue.
Solution