RMO 2023 Questions, Solutions, Discussions
| RMO 2023 | |||
|---|---|---|---|
| Question | Solution | Discussion | AoPS |
(RMO 2023a P2, AoPS) Given a prime number \(p\) such that \(2p\) is equal to the sum of the squares of some four consecutive positive integers. Prove that \(p-7\) is divisible by \(36\).
Click here for the spoiler!
Show that the sum of four consecutive squares is congruent to \(6\) modulo \(8\), and conclude that \(p \equiv 3 \bmod 4\). Considering congruence conditions modulo \(3\), prove that the smallest of the four consecutive numbers is a multiple of \(3\). Deduce that the sum of the four consecutive squares is \(5\) modulo \(9\).
(RMO 2023b P1, AoPS) 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 \(abcd\) for all \((a, b, c, d) \in S\).
Click here for the spoiler!
- Show that \((1, 2, 2, 3)\) lies in \(S\), and deduce that \(m\) divides \(12\).
- Let \((a, b, c, d)\) be an element of \(S\). Show that at least one of \(a, b, c, d\) is divisible by \(3\), and at least one of them is even.
- Prove that if \(d\) is even, then at least one of \(a, b, c\) is even, and that if \(d\) is odd, then at least two of \(a, b, c\) are even.
- Conclude that \(m\) is divisible by \(2 \cdot 2 \cdot 3\).
(RMO 2023b P6, AoPS) 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.
The following is from AoPS, and is due to Rohan (Goyal?) as mentioned here by L567.
Click here for the spoiler!
- Show that if the small square (as in Fig. 1) does not contain a blue point, then we are done.
- It remains to consider the case when the small square (as in Fig. 1) contains at least one blue point.
- Rotating the configuration about the center of the small square (if necessary), assume that the top-left vertex of the small square (as in Fig. 1) is blue.
-
Prove that the gray square contains at most three blue points.
-
Consider the case when at least two blue points lie on the bigger dashed circle. Show that the smaller dashed circle does not contain any blue point, in this case. Hence, the gray square contains at most three blue points.
-
Similarly, if the smaller dashed circle contains at least two blue points, then the gray square (as in Fig. 2) contains at most three blue points.
-
- It suffices to consider that each one of the red, purple, and green \(L\)-shapes, has at most one end-point which is blue (otherwise, we are done).
- Use the above to show that the bottom-right point (as in Fig. 2) is blue.
- Consider the point at the bottom-right corner, and the center of the gray square, and a blue end-point of an \(L\)-shape, to show that these three points form the vertices of an isosceles triangle having the required properties.
Click here for the spoiler!
Note that the \(16\) points are the vertices of the four squares as in Fig. 1. If no vertex of the small square is blue, then by the pigeonhole principle, at least one of the remaining three squares has at least three blue vertices, and hence there exists an isosceles right-angled triangle with blue vertices. Let us assume that the small square (as in Fig. 1) has a blue vertex. Rotating the configuration about the center of the small square (if necessary), we may and do assume that the top-left vertex of the small square is blue. Henceforth, on the contrary, let us assume that there are no isosceles right-angled triangles with blue vertices.
Claim. The gray square (as in Fig. 1) contains at most three blue points.
Proof of the Claim. Note that the points within the gray square lies on the two dashed circles (as in Fig. 1). Therefore, to prove the Claim, it suffices to show that if one of the dashed circles (as in Fig. 1) contains at least two blue points, then the other dashed circle does not contain any blue point.
If at least two blue points lie on the bigger dashed circle, then using the assumption, it follows that no more blue points lies on it, and hence these two blue points lie along a diameter. It follows that no blue point lies on the other dashed circle.
If at least two blue points lie on the smaller dashed circle, then using a similar argument, it follows that no blue point lies on the bigger dashed circle.
The Claim follows.
Note that if both the end-points of one of the red, purple, and green \(L\)-shapes (as in Fig. 2) are blue, then these points together with the center of gray square form the vertices of an isosceles right-angled triangle, contradicting the assumption. Hence, each one of these three \(L\)-shapes has at most one end-point which is blue. Since there are \(7\) blue points, using the Claim, it follows that the bottom-right point is blue. Note that the center of the gray square, the bottom-right point, and a blue end-point of the purple \(L\)-shape, form the vertices of a triangle having the required properties.
Lecture notes for Math Olympiad (IOQM, RMO, INMO)
Click on the icons below to download.
Topics Links Algebra Combinatorics Geometry Number Theory INMO Training Camp 2025, MP IMO Training Camp MOPSS Simon Marais Mathematics Competition Resources