The Regional Mathematical Olympiad (RMO), preceded by the all-India pre-RMO (IOQM) or PRMO, is the first step towards representing India at the global platform for 'mathletes', the International Mathematical Olympiad (IMO).
The RMO is a three-hour written test with six or seven problems. On the basis of the performance in the RMO, a certain number of students are selected from each region to participate in the Indian National Mathematical Olympiad (INMO), second step towards the IMO.
In this post we have added the problems and solutions from the RMO 2024.
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| Problem 1 Answer: (b) \(2,1,4,3,6,5,\ldots,n,n-1\) See Solution | Problem 2 Answer: \(n=1,5\) See Solution |
| Problem 3 Answer: See Solution | Problem 4 Answer: See Solution |
| Problem 5 Answer: See Solution | Problem 6 Answer: See Solution |
Let \(n>1\) be a positive integar. Call a rearrangement \(a_1, a_2, \ldots, a_n\) of \(1,2, \ldots, n\) nice.
If for every \(k=2,3, \ldots, n\), we have that \(a_1+a_2+\cdots+a_k\) is not divisible by \(k\).
(a) If \(n>1\) is odd, prove that there is no nice rearrangement of \(1,2, \ldots, n\).
(b) If \(n\) is even, find a nice rearrangement of \(1,2, \ldots, n\).
(a) If $n$ is ODD, then consider any rearrangement $a_1, a_2, \ldots, a_n$.
For $k=n$, we get $a_1+a_2+\cdots+a_n=1+2+\cdots+n=\frac{n(n+1)}{2}=$
$n\left(\frac{n+1}{2}\right)$ which is a multiple of $n$ since $n$ is co-prime to 2 and
thus the condition would be violated rendering the rearrangement 'not nice'.
(b) If $n$ is EVEN, then the following rearrangement is nice.
$$
a_i= \begin{cases}i-1 & , \text { if } i \text { is even } \\ \ i+1 & , \text { if } i \text { is odd }\end{cases}
$$
That is, $\left(a_1, a_2, \ldots, a_n\right)=(2,1,4,3,6,5, \ldots, n, n-1)$. This rearrangement is nice because
$$a_1+a_2+\cdots+a_k= \begin{cases}\frac{k(k+1)}{2}=\left(\frac{k}{2}\right)(k+1) & , \text { if } k \text { is even } \\ \ \frac{k(k+1)}{2}+1= k\left(\frac{k+1}{2}\right)+1 & , \text { if } k \text { is odd }\end{cases}$$
Now it is conspicuous that both possible values of the sum are not divisible by $k$ (except for $k=1$ ) because in the first sum, $\frac{k}{2}$ and $k+1$ are relatively prime numbers and in the second sum, it is the next number of a multiple of $k$ which cannot be a multiple of $k$.
For a positive integer (n), let \(R(n)\) be the sum of the remainders when \(n\) is divided by \(1,2, \ldots, n\). For example, \(R(4)=0+0+1+0=1,\ R(7)=0+1+1+3+2+1+0=8\). Find all positive integers \(n\) such that \(R(n)=n-1\).
Let $n$ be EVEN, then consider the numbers from $\frac{n}{2}+1, \frac{n}{2}+2, \ldots, n-1$. Now, all these numbers bear a quotient of 1 when dividing $n$ because twice of each exceeds $n$. Hence, they leave a remainder of $\frac{n}{2}-1, \frac{n}{2}-2, \ldots, 1$ respectively when dividing $n$ and so the sum of these remainders is no greater than $R(n)$. Thus,
\begin{align*}R(n)&\geq 1+2+\cdots+\left(\frac{n}{2}-1\right)\\&=\frac{\left(\frac{n}{2}-1\right)\left(\frac{n}{2}\right)}{2}\\\Rightarrow n-1&\geq\frac{n^2-2n}{8}\\\Rightarrow n^2-10n+17&\leq 17\\\Rightarrow (n-5)^2&\leq 17\\\Rightarrow n&=2,4,6,8\ (\text{as we assumed $n$ to be EVEN})\end{align*}
Similarly working for $n$ being ODD necessarily yields the remainders $\frac{n-1}{2}, \frac{n-3}{2}, \ldots, 1$ when dividing $n$ and thus
\begin{align*}R(n) &\geq 1+2+\cdots+\frac{n-1}{2}\\\Rightarrow n-1 &\geq \frac{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)}{2}\\\Rightarrow n &\leq 7\end{align*}
Therefore, in total, we just need to check $R(n)$ value for the numbers $n=$ $1,2,3,4,5,6,7,8$. Manually working on them gives $R(1)=0, R(2)=0, R(3)=$ $1, R(4)=1, R(5)=4, R(6)=3, R(7)=8, R(8)=8$. Hence, the positive integers $n$ such that $R(n)=n-1$ are $\boxed{n=1,5}$.
Let \(A B C\) be an acute triangle with \(A B=A C\). Let \(D\) be the point on \(B C\) such that \(A D\) is perpendicular to \(B C\). Let \(O, H, G\) be the circumcentre, orthocentre and centroid of triangle \(A B C\) respectively. Suppose that \(2 \cdot O D=23 \cdot H D\). Prove that \(G\) lies on the incircle of \( \triangle A B C\).
Let $I$ be the incenter, center of incircle, of $\triangle A B C . A B=A C \Rightarrow \triangle A B D \cong$ $\triangle A C D$, so we have $A D$ as the perpendicular bisector of $B C$ and the angle bisector of $\angle A$. Thus, $O, G, H, I$ all lie on $A D$. We know that for any triangle, $O, G, H$ are collinear in that order with $O G: G H=1: 2$.
Let $H D=2 x$, then $O D=\frac{23 \cdot H D}{2}=23 x \Rightarrow O H=21 x$. Now, $O G: G H=$ $1: 2 \Rightarrow O G=7 x, G H=14 x, G D=16 x$ as shown. We also know that centroid divides median in the ratio $2: 1 \Rightarrow A G: G D=2: 1 \Rightarrow A G=$ $32 x \Rightarrow A O=25 x$. Now, $O A=O B=25 x$, so in $\triangle O B D$, we have $B D=$ $\sqrt{O B^2-O D^2}=\sqrt{(25 x)^2-(23 x)^2}=4 x \sqrt{6}$. In $\triangle A B D, A B=\sqrt{A D^2+B D^2}=$ $\sqrt{(48 x)^2+96 x^2}=20 x \sqrt{6}$. Hence, the semi-perimeter $s=A B+B D=20 x \sqrt{6}+$ $4 x \sqrt{6}=24 x \sqrt{6}$ and $[A B C]=\frac{1}{2} \times 2 B D \times A D=192 x^2 \sqrt{6}$.
Thus, inradius $r=\frac{\mid A B C]}{s}=\frac{192 x^2 \sqrt{6}}{24 x \sqrt{6}}=8 x \Rightarrow I D=8 x$. Thus, $G I=G D-I D=$ $16 x-8 x=8 x$ and as we already know that $8 x$ is the radius of incircle, we conclude that $G$ lies on the incircle of $\triangle A B C$.
Let \(a_1, a_2, a_3, a_4\) be real numbers such that \(a_1^2+a_2^2+a_3^2+a_4^2=1\). Show that there exist \(i, j\) with \(1 \leq i<j \leq 4\), such that \(\left(a_i-a_j\right)^2 \leq \frac{1}{5}\).
Without loss of generality, let $a_1\leq a_2\leq a_3\leq a_4$. Let $d_1= a_2-a_1,\ d_2 = a_3-a_2,\ d_3 = a_4-a_3$ and let $d=min{d_1,d_2,d_3}$.
Case (i) - $d=d_1$
So, $d=a_2-a_1$ and let $x=a_2+\frac{d}{2}\Rightarrow a_2=x-\frac{d}{2},\ a_1=x-\frac{3d}{2}$.
So, $a_3=x+\frac{d}{2}+y,\ a_4=x+\frac{3d}{2}+z$, for some $y,z\geq 0$. Then,
\begin{align*}1&=a_1^2+a_2^2+a_3^2+a_4^2\\&=\left(x-\frac{3d}{2}\right)^2+\left(x-\frac{d}{2}\right)^2+\left(x+\frac{d}{2}+y\right)^2+\left(x+\frac{3d}{2}+z\right)^2\\ &=4x^2+5d^2+y^2+2y\left(x+\frac{d}{2}\right)+2z\left(x+\frac{3d}{2}\right)\\ &=\underbrace{(x+y)^2+(x+z)^2}_{non-negative}+\underbrace{2x^2+yd+3zd}_{non-negative}+5d^2\\
\Rightarrow 1&\geq 5d^2\Rightarrow \boxed{d^2\leq\frac{1}{5}}.
\end{align*}
Case (ii) - $d=d_2$
So, $d=a_3-a_2$ and let $x=a_2+\frac{d}{2}\Rightarrow a_2=x-\frac{d}{2},\ a_3=x+\frac{d}{2}$.
So, $a_1=x-\frac{3d}{2}-z,\ a_4=x+\frac{3d}{2}+y$, for some $y,z\geq 0$. Then,
\begin{align*} 1&=a_1^2+a_2^2+a_3^2+a_4^2\\ &=\left(x-\frac{3d}{2}-z\right)^2+\left(x-\frac{d}{2}\right)^2+\left(x+\frac{d}{2}\right)^2+\left(x+\frac{3d}{2}+y\right)^2\\ &=4x^2+5d^2+y^2+2y\left(x+\frac{3d}{2}\right)-2z\left(x-\frac{3d}{2}\right)\\ &=\underbrace{(x+y)^2+(x-z)^2}_{non-negative}+\underbrace{2x^2+3yd+3zd}_{non-negative}+5d^2\\
\Rightarrow 1&\geq 5d^2\Rightarrow \boxed{d^2\leq\frac{1}{5}}.
\end{align*}
Case (iii) - $d=d_3$
So, $d=a_4-a_3$ and let $x=a_3-\frac{d}{2}\Rightarrow a_3=x+\frac{d}{2},\ a_4=x+\frac{3d}{2}$.
So, $a_2=x-\frac{d}{2}-y,\ a_1=x-\frac{3d}{2}-z$, for some $y,z\geq 0$. Then,
\begin{align*} 1&=a_1^2+a_2^2+a_3^2+a_4^2\\ &=\left(x-\frac{3d}{2}-z\right)^2+\left(x-\frac{d}{2}-y\right)^2+\left(x+\frac{d}{2}\right)^2+\left(x+\frac{3d}{2}\right)^2\\ &=4x^2+5d^2+y^2-2y\left(x-\frac{d}{2}\right)-2z\left(x-\frac{3d}{2}\right)\\ &=\underbrace{(x-y)^2+(x-z)^2}_{non-negative}+\underbrace{2x^2+yd+3zd}_{non-negative}+5d^2\\
\Rightarrow 1&\geq 5d^2\Rightarrow \boxed{d^2\leq\frac{1}{5}}.
\end{align*}
Let \(A B C D\) be a cyclic quadrilateral such that \(A B\) is parallel to \(C D\). Let \(O\) be the circumcentre of \(A B C D\), and \(L\) be the point on \(A D\) such that \(O L\) is perpendicular to \(A D\). Prove that
\(O B \cdot(A B+C D)=O L \cdot(A C+B D) \)
Let $\angle ABD=\theta\Rightarrow\angle BDC=\theta$ (since $AB||CD$). By the Inscribed angle theorem, $\angle BAC=\angle BDC=\angle ABD=\angle ACD=\theta$ (thus $ABCD$ is isoceles trapezoid). Hence, $\triangle ABK, \triangle CDK$ are isoceles and so $AB, CD$ have common perpendicular bisector through $K$ which also passes through center of circle $O$, making $E,K,O,F$ collinear as shown.
Now, by the Inscribed angle theorem, $\angle AOD=2\theta\Rightarrow\angle AOL=\theta$ since $\triangle AOD$ is isoceles. Thus, $\cos{\theta}=\frac{OL}{OA}\rightarrow\boxed{eq1}$.
In $\triangle ABK, \cos{\theta}=\frac{AE}{AK}=\frac{BE}{BK}$ and so by the componendo-dividendo rule, $\cos{\theta}=\frac{AE+BE}{AK+BK}=\frac{AB}{AK+BK}$. Similarly, in $\triangle CDK, \cos{\theta}=\frac{CF}{CK}=\frac{DF}{DK}=\frac{CF+DF}{CK+DK}=\frac{CD}{CK+DK}$. Combining these two results and applying componendo-dividendo rule once again, we get, $\cos{\theta}=\frac{AB}{AK+BK}=\frac{CD}{CK+DK}=\frac{AB+CD}{AK+BK+CK+DK}=\frac{AB+CD}{AC+BD}\rightarrow\boxed{eq2}$.
From the equations $eq1$ and $eq2$, we conclude that $\cos{\theta}=\frac{OL}{OA}=\frac{AB+CD}{AC+BD}\Rightarrow \boxed{OB \cdot(A B+C D)=O L \cdot(A C+B D)}$, since $OA=OB=radius$.
Let \(n \geq 2\) be a positive integer. Call a sequence \(a_1, a_2, \cdots, a_k\) of integers an \(n\)-chain
if \(1=a_1<a_2<\cdots<a_k=n\), and \(a_i\) divides \(a_{i+1}\) for all \(i, 1 \leq i \leq k-1\). Let \(f(n)\) be the number of \(n\)-chains where \(n \geq 2\). For example, \(f(4)=2\) corresponding to the 4 -chains $(1, 4)$ and $(1, 2, 4)$.
Prove that \(f\left(2^m \cdot 3\right)=2^{m-1}(m+2)\) for every positive integer \(m\).
