MOPSS, 27th September 2025
(SMMC 2017 A1, AoPS) The five sides and five diagonals of a regular pentagon are drawn on a piece of paper. Two people play a game, in which they take turns to colour one of these ten line segments. The first player colours line segments blue, while the second player colours line segments red. A player cannot colour a line segment that has already been coloured. A player wins if they are the first to create a triangle in their own colour, whose three vertices are also vertices of the regular pentagon. The game is declared a draw if all ten line segments have been coloured without a player winning. Determine whether the first player, the second player, or neither player can force a win.
(Tournament of Towns Fall 2013, Senior A Level P4, AoPS) Integers \(1, 2, \dots, 100\) are written on a circle, not necessarily in that order. Can it be that the absolute value of the difference between any two adjacent integers is at least \(30\) and at most \(50\)?
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- Assume that such an arrangement is possible.
- Consider the integers \(1, 2, \dots, 25, 76, 77, \dots, 100\).
- Show that no two of these integers are adjacent.
- Conclude that the integers \(1, 2, \dots, 25, 76, 77, \dots, 100\) and \(26, 27, \dots, 75\) are placed alternately along the circle.
- Show that \(26\) is adjacent to \(76\) only.
- Derive a contradiction.
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Let us assume that the integers \(1, 2, \dots, 100\) can be arranged along the circumference of a circle in some order such that the absolute value of the difference between any two adjacent integers is at least \(30\) and at most \(50\). Since the difference of any two of the integers \(1, 2, \dots, 25, 76, 77, \dots, 100\) is less than \(30\) or greater than \(50\) it follows that no two of these integers are adjacent. Consequently, the elements of the sets \[ \{1, 2, \dots, 25, 76, 77, \dots, 100\}, \{26, 27, \dots, 75\} \] are placed alternately along the circle. However, \(26\) is adjacent to \(26\) only, which is impossible. This shows that there is no arrangement of the integers \(1, 2, \dots, 100\) along the circumference of a circle satisfying the given condition.
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Note that a \(2 \times 3\) rectangle can be tiled by two non-overlapping trominos. Since a \(12 \times 15\) chessboard can be covered by non-overlapping \(2 \times 3\) rectangles, it follows that a \(12 \times 15\) chessboard can be tiled by non-overlapping trominos.
For any integer \(n \geq 6\), show that a square can be cut into \(n\) squares, not necessarily of the same size.
Show that for any integer \(n \geq 1\), a \(2^n \times 2^n\) chessboard with one square removed can be tiled by non-overlapping trominos, that is, \(L\)-shaped pieces consisting of three squares.
(Tournament of Towns Spring 2025, Senior O Level P1, AoPS) Find the minimum positive integer such that some four of its natural divisors sum up to \(2025\).
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If \(n\) is a positive integer and \(a\), \(b\), \(c\), and \(d\) are four distinct positive divisors of \(n\) satisfying \(a < b < c < d\), then observe that the inequalities \[ d \leq n, c \leq \frac{n}{2}, b \leq \frac{n}{3}, a \leq \frac{n}{4} \] hold.
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Suppose there exists a positive integer \(n\) such that some four of its natural divisors sum up to \(2025\). Let these four divisors be \(a\), \(b\), \(c\), and \(d\). Without loss of generality1, we can assume that \(a < b < c < d\). Note that \[ d \leq n, c \leq \frac{n}{2}, b \leq \frac{n}{3}, a \leq \frac{n}{4}, \] which shows that \[ 2025 = a + b + c + d \leq n + \frac{n}{2} + \frac{n}{3} + \frac{n}{4} = \frac{25n}{12}. \] This implies that \[ n \geq \frac{12 \cdot 2025}{25} = 972. \] Note that the positive integers \(972, 486, 324, 243\) divide \(972\) and sum up to \(2025\). This proves that \(972\) is the smallest positive integer such that some four of its natural divisors sum up to \(2025\).
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Why there is no loss of generality in assuming \(a < b < c < d\)? ↩
(SMMC 2017 B2, AoPS) Determine all pairs \((p, q)\) of positive integers such that \(p\) and \(q\) are primes, and \(p^{q - 1} + q^{p - 1}\) is the square of an integer.
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- Check that \((p, q) = (2, 2)\) works.
- If \(p, q\) are odd, show that \(p^{q - 1} + q^{p - 1} \equiv 2 \pmod{4}\), so it cannot be a perfect square.
- Without loss of generality, let \(p\) be an odd prime, and \(q = 2\). Then \(p^{q - 1} + q^{p - 1} = 2^{p - 1} + p\).
- Write \(2^{p - 1} + p = m^2\).
- Factorize to obtain \((m - 2^{\frac{p - 1}{2}})(m + 2^{\frac{p - 1}{2}}) = p\), which gives \[ m + 2^{\frac{p - 1}{2}} = p. \]
- Note that \(p < 2^{\frac{p - 1}{2}}\) holds1 for \(p \geq 7\).
- Conclude that \(p\) can only be \(3\) or \(5\).
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Try to show that \(2^{\frac{n - 1}{2}} > n\) holds for all integers \(n \geq 7\). ↩
(SMMC 2020 A1, AoPS) There are \(1001\) points in the plane such that no three are collinear. The points are joined by \(1001\) line segments such that each point is an endpoint of exactly two of the line segments. Prove that there does not exist a straight line in the plane that intersects each of the 1001 line segments in an interior point. An interior point of a line segment is a point of the line segment that is not one of the two endpoints.
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- On the contrary, let us assume that there exists a straight line \(L\) in the plane that intersects each of the \(1001\) line segments in an interior point.
- Show that \(L\) cannot pass through any of the \(1001\) points.
- Color the points on one side of \(L\) red, and the points on the other side of \(L\) blue.
- Show that the number of the line segments is equal to twice the number of red points, to obtain a contradiction.
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On the contrary, let us assume that there exists a straight line \(L\) in the plane that intersects each of the \(1001\) line segments in an interior point. Note that \(L\) cannot pass through any of the \(1001\) points. Indeed, if \(L\) passes through one of the \(1001\) points, then the two line segments that have this point as an endpoint, cannot intersect \(L\) at their interior points, since no three of the given points are collinear. Let us color the points on one side of \(L\) red, and the points on the other side of \(L\) blue. Since \(L\) intersects each of the \(1001\) line segments in an interior point, each of the \(1001\) line segments has one red endpoint and one blue endpoint. Moreover, each of the \(1001\) points is an endpoint of exactly two of the line segments. Therefore, the number of line segments is equal to twice the number of red points, which is imposssible since \(1001\) is odd. This completes the proof.
Let \(a_1, a_2, \dots, a_n\) be positive integers. Let \(b_k\) denote the number of those \(a_i\) for which \(a_i\) is greater than or equal to \(k\). Prove that \[ a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_{a_n}. \]
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- Try to interpret both sides of the equation in terms of counting objects.
- Can you think of a way to represent the integers \(a_1, a_2, \ldots, a_n\) using dots?
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Consider the set \[ S = \{(i, j) : 1 \leq i \leq n, 1 \leq j \leq a_i\}. \] The size of this set is equal to \(a_1 + a_2 + \cdots + a_n\). On the other hand, for any positive integer \(j\), the number of \(i\) such that \((i, j) \in S\) is exactly \(b_j\). This shows that the size of \(S\) is also equal to \(b_1 + b_2 + \cdots + b_{a_n}\). This yields the desired result.
Alternatively, note that
\(\begin{align*} a_1 + a_2 + \cdots + a_n & = \sum_{i=1}^n \sum_{j=1}^{a_i} 1 \\ & = \sum_{i=1}^n \sum_{j=1}^{a_n} 1_{a_i \geq j}(i,j) \\ & = \sum_{j=1}^{a_n} \sum_{i=1}^n 1_{a_i \geq j}(i,j) \\ & = \sum_{j=1}^{a_n} b_j, \end{align*} \\)
where \(1_{a_i \geq j}(i,j)\) is \(1\) if \(a_i \geq j\), and \(0\) otherwise.
Lecture notes for Math Olympiad (IOQM, RMO, INMO)
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