id stringlengths 10 15 | question stringlengths 63 2.3k | solutions stringlengths 20 28.5k |
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IMOSL-2024-C3 | Let \(n\) be a positive integer. There are \(2n\) knights sitting at a round table. They consist of \(n\) pairs of partners, each pair of which wishes to shake hands. A pair can shake hands only when next to each other. Every minute, one pair of adjacent knights swaps places.
Find the minimum number of exchanges of ... | Answer: The minimum number of exchanges is \(\frac{n(n - 1)}{2}\) .
Common remarks. The solution is divided into three lemmas. We provide multiple proofs of each lemma.
Solution. Join each pair of knights with a chord across the table. We'll refer to these chords as chains.
First we show that \(n(n - 1) / 2\) ... |
IMOSL-2024-C4 | On a board with 2024 rows and 2023 columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are 202... | Comment. One of the main difficulties of solving this question is in determining the correct expression for \(n\) . Students may spend a long time attempting to prove bounds for the wrong value for \(n\) before finding better strategies.
Students may incorrectly assume that Turbo is not allowed to backtrack to squar... |
IMOSL-2024-C5 | Let \(N\) be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers 1, 2, ..., \(N\) on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers \((k, n)\) , where \(k \geq 0\) and \(n\) is one of the integers on the board, a... | Answer: The answer is that Geoff wins when \(N\) is of the form \(2^{n}\) for \(n\) odd or of the form \(t2^{n}\) for \(n\) even and \(t > 1\) odd.
Common remarks. We will say that a set \(S\) wins if the current player wins given \(S\) as the current set of integers on the board. Otherwise, we will say that \(S\) l... |
IMOSL-2024-C6 | Let \(n\) and \(T\) be positive integers. James has \(4n\) marbles with weights 1, 2, ..., \(4n\) . He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most \(T\)... | Answer: The minimum value of \(T\) is \(4n\) .
Solution 1. We must have \(T \geqslant 4n\) , as otherwise we can never move the marble of weight \(4n\) . We will show that \(T = 4n\) by showing that, for any initial configuration, there is a sequence of moves, never increasing the absolute value of the difference ab... |
IMOSL-2024-C7 | Let \(N\) be a positive integer and let \(a_{1}\) , \(a_{2}\) , ... be an infinite sequence of positive integers. Suppose that, for each \(n > N\) , \(a_{n}\) is equal to the number of times \(a_{n - 1}\) appears in the list \(a_{1}\) , \(a_{2}\) , ..., \(a_{n - 1}\) .
Prove that at least one of the sequences \(a_{1... | Solution 1. Let \(M > \max (a_{1}, \ldots , a_{N})\) . We first prove that some integer appears infinitely many times. If not, then the sequence contains arbitrarily large integers. The first time each integer larger than \(M\) appears, it is followed by a 1. So 1 appears infinitely many times, which is a contradiction... |
IMOSL-2024-C8 | Let \(n\) be a positive integer. Given an \(n \times n\) board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a \(2 \times 2\) square with exactly one cell coloured black... | Answer: The answer is \(n = 2^{k}\) where \(k\) is a nonnegative integer.
Solution 1. We first prove by induction that it is possible the colour the whole board black for \(n = 2^{k}\) . The base case of \(k = 1\) is trivial. Assume the result holds for \(k = m\) and consider the case of \(k = m + 1\) . Divide the \... |
IMOSL-2024-G1 | Let \(ABCD\) be a cyclic quadrilateral such that \(AC< BD< AD\) and \(\angle DBA< 90^{\circ}\) . Point \(E\) lies on the line through \(D\) parallel to \(AB\) such that \(E\) and \(C\) lie on opposite sides of line \(AD\) , and \(AC = DE\) . Point \(F\) lies on the line through \(A\) parallel to \(CD\) such that \(F\) ... | Solution 1. Let \(T\) be the midpoint of arc \(\overline{BAC}\) and let lines \(BA\) and \(CD\) intersect \(EF\) at \(K\) and \(L\) , respectively. Note that \(T\) lies on the perpendicular bisector of segment \(BC\) .
Since \(ABCD\) is cyclic, \(\frac{BD}{\sin\angle BAD} = \frac{AC}{\sin\an... |
IMOSL-2024-G2 | Let \(ABC\) be a triangle with \(AB < AC < BC\) , incentre \(I\) and incircle \(\omega\) . Let \(X\) be the point in the interior of side \(BC\) such that the line through \(X\) parallel to \(AC\) is tangent to \(\omega\) . Similarly, let \(Y\) be the point in the interior of side \(BC\) such that the line through \(Y\... | Solution 1. Let \(A'\) be the reflection of \(A\) in \(I\) , then \(A'\) lies on the angle bisector \(AP\) . Lines \(A'X\) and \(A'Y\) are the reflections of \(AC\) and \(AB\) in \(I\) , respectively, and so they are the tangents to \(\omega\) from \(X\) and \(Y\) . As is well- known, \(PB = PC = PI\) , and since \(\an... |
IMOSL-2024-G3 | Let \(A B C D E\) be a convex pentagon and let \(M\) be the midpoint of \(A B\) . Suppose that segment \(A B\) is tangent to the circumcircle of triangle \(C M E\) at \(M\) and that \(D\) lies on the circumcircles of triangles \(A M E\) and \(B M C\) . Lines \(A D\) and \(M E\) intersect at \(K\) , and lines \(B D\) an... | Common remarks. Each of solutions we present consists of three separate parts:
(a) proving \(K P \parallel M C\) and \(L Q \parallel M E\) ;
(b) proving \(K L \parallel A B\) and, optionally, showing that points \(C\) , \(E\) , \(K\) , and \(L\) are concyclic;
(c) completing the solution either using homotheties... |
IMOSL-2024-G4 | Let \(A B C D\) be a quadrilateral with \(A B\) parallel to \(C D\) and \(A B< C D\) . Lines \(A D\) and \(B C\) intersect at a point \(P\) . Point \(X\neq C\) on the circumcircle of triangle \(A B C\) is such that \(P C = P X\) . Point \(Y\neq D\) on the circumcircle of triangle \(A B D\) is such that \(P D = P Y\) . ... | Solution 1. Let \(M\) and \(N\) be the midpoints of \(A D\) and \(B C\) , respectively and let the perpendicular bisector of \(A B\) intersect the line through \(P\) parallel to \(A B\) at \(R\) .
Lemma. Triangles \(Q A B\) and \(R N M\) are similar.
Proof. Let \(O\) be the circumcentre of triangle \(A B C\) , an... |
IMOSL-2024-G5 | Let \(A B C\) be a triangle with incentre \(I\) , and let \(\Omega\) be the circumcircle of triangle \(B I C\) Let \(K\) be a point in the interior of segment \(B C\) such that \(\angle B A K < \angle K A C\) . The angle bisector of \(\angle B K A\) intersects \(\Omega\) at points \(W\) and \(X\) such that \(A\) and \(... | Common remarks. The key step in each solution is to prove that \(\angle Z A K = \angle I A Y\) and \(\angle W A K = \angle I A X\) . The problem is implied by these equalities, as we then have that
\[\angle W A Y = \angle W A K + \angle K A I + \angle I A Y = \angle I A X + \angle K A I + \angle Z A K = \angle Z A X... |
IMOSL-2024-G6 | Let \(ABC\) be an acute triangle with \(AB < AC\) , and let \(\Gamma\) be the circumcircle of \(ABC\) . Points \(X\) and \(Y\) lie on \(\Gamma\) so that \(XY\) and \(BC\) intersect on the external angle bisector of \(\angle BAC\) . Suppose that the tangents to \(\Gamma\) at \(X\) and \(Y\) intersect at a point \(T\) on... | Solution 1. Let \(N\) be the midpoint of \(\overline{BAC}\) on \(\Gamma\) , and let \(NX\) and \(NY\) intersect \(BC\) at \(W\) and \(Z\) , respectively.
Claim. Quadrilateral \(WXYZ\) is cyclic, and its circumcentre is \(J\) .
Proof. As \(N\) is the midpoint of \(\overline{BAC}\) , \(W\) and \(Z\) lie on \(BC\) ,... |
IMOSL-2024-G7 | Let \(ABC\) be a triangle with incentre \(I\) such that \(AB < AC < BC\) . The second intersections of \(AI\) , \(BI\) , and \(CI\) with the circumcircle of triangle \(ABC\) are \(M_A\) , \(M_B\) , and \(M_C\) , respectively. Lines \(AI\) and \(BC\) intersect at \(D\) and lines \(BM_C\) and \(CM_B\) intersect at \(X\) ... | Solution 1.
Let \(O\) be the circumcentre of \(\triangle ABC\) . First we note from standard properties of the Miquel point \(S\) we have:
\(\triangle SMC_MB \sim \triangle SBC \sim \triangle SPQ\) ; \((*)\)
- \(I\) and \(S\) are inverses with respect to circle \(ABC\) ;
\(\angle OS... |
IMOSL-2024-G8 | Let \(A B C\) be a triangle with \(A B< A C< B C\) , and let \(D\) be a point in the interior of segment \(B C\) . Let \(E\) be a point on the circumcircle of triangle \(A B C\) such that \(A\) and \(E\) lie on opposite sides of line \(B C\) and \(\angle B A D = \angle E A C\) . Let \(I\) , \(I_{B}\) , \(I_{C}\) , \(J_... | Solution 1. Let \(X\) be the intersection of \(I_{B}J_{C}\) and \(J_{B}I_{C}\) . We will prove that, provided that \(A B< A C< B C\) , the following two conditions are equivalent:
(1) \(A X\) bisects \(\angle B A C\) ;
(2) \(I_{B}\) , \(I_{C}\) , \(J_{B}\) , and \(J_{C}\) are concyclic.
Let circles \(A I B\) a... |
IMOSL-2024-N1 | Find all positive integers \(n\) with the following property: for all positive divisors \(d\) of \(n\) , we have that \(d + 1 \mid n\) or \(d + 1\) is prime. | Answer: \(n \in \{1, 2, 4, 12\}\) .
Solution 1. It is easy to verify that \(n = 1, 2, 4, 12\) all work. We must show they are the only possibilities. We write \(n = 2^{k} m\) , where \(k\) is a nonnegative integer and \(m\) is odd. Since \(m \mid n\) , either \(m + 1\) is prime or \(m + 1 \mid n\) .
In the former... |
IMOSL-2024-N2 | Determine all finite, nonempty sets \(\mathcal{S}\) of positive integers such that for every \(a\) , \(b \in \mathcal{S}\) there exists \(c \in \mathcal{S}\) with \(a \mid b + 2c\) . | Answer: The possible sets are \(\mathcal{S} = \{t\}\) and \(\mathcal{S} = \{t, 3t\}\) for any positive integer \(t\) .
Solution 1. Without loss of generality, we may divide all elements of \(\mathcal{S}\) by any common factor, after which they cannot all be even. As \(a \nmid b + 2c\) for \(a\) even and \(b\) odd, t... |
IMOSL-2024-N3 | Determine all sequences \(a_{1}\) , \(a_{2}\) , ... of positive integers such that, for any pair of positive integers \(m \leqslant n\) , the arithmetic and geometric means
\[\frac{a_{m} + a_{m + 1} + \cdots + a_{n}}{n - m + 1} \quad \text{and} \quad (a_{m}a_{m + 1}\cdot \cdot \cdot a_{n})^{\frac{1}{n - m + 1}}\]
... | Answer: The only such sequences are the constant sequences (which clearly work).
Solution 1. We say that an integer sequence \(b_{1}\) , \(b_{2}\) , ... is good if for any pair of positive integers \(m \leqslant n\) , the arithmetic mean \(\frac{b_{m} + b_{m + 1} + \cdots + b_{n}}{n - m + 1}\) is an integer. Then th... |
IMOSL-2024-N4 | Determine all positive integers \(a\) and \(b\) such that there exists a positive integer \(g\) such that \(\gcd (a^{n} + b, b^{n} + a) = g\) for all sufficiently large \(n\) . | Answer: The only solution is \((a, b) = (1, 1)\) .
Solution 1. It is clear that we may take \(g = 2\) for \((a, b) = (1, 1)\) . Supposing that \((a, b)\) satisfies the conditions in the problem, let \(N\) be a positive integer such that \(\gcd (a^{n} + b, b^{n} + a) = g\) for all \(n \geq N\) .
Lemma. We have tha... |
IMOSL-2024-N5 | Let \(\mathcal{S}\) be a finite nonempty set of prime numbers. Let \(1 = b_{1}< b_{2}< \dots\) be the sequence of all positive integers whose prime divisors all belong to \(\mathcal{S}\) . Prove that, for all but finitely many positive integers \(n\) , there exist positive integers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\... | Solution 1. If \(\mathcal{S}\) has only one element \(p\) , then \(b_{i} = p^{i - 1}\) and we can easily find \(a_{1}\) , ..., \(a_{n}\) with \(2 = \left[\sum_{i = 0}^{n - 1}\frac{1}{p^{i}}\right] = \sum_{i = 0}^{n - 1}\frac{a_{i}}{p^{i - 1}}\) by taking \(a_{1} = a_{2} = \dots = a_{n - 1} = 1\) and choosing \(a_{n} = ... |
IMOSL-2024-N6 | Let \(n\) be a positive integer. We say that a polynomial \(P\) with integer coefficients is \(n\) - good if there exists a polynomial \(Q\) of degree 2 with integer coefficients such that \(Q(k)(P(k) + Q(k))\) is never divisible by \(n\) for any integer \(k\) .
Determine all integers \(n\) such that every polynomia... | Answer: The set of such \(n\) is any \(n > 2\) .
Solution 1. First, observe that no polynomial is 1- good (because \(Q(X)(P(X) + Q(X))\) always has roots modulo 1) and the polynomial \(P(X) = 1\) is not 2- good (because \(Q(X)(Q(X) + 1)\) is always divisible by 2).
Now, if \(P\) is \(d\) - good with some \(Q\) , ... |
IMOSL-2024-N7 | Let \(\mathbb{Z}_{>0}\) denote the set of positive integers. Let \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) be a function satisfying the following property: for \(m\) , \(n\in \mathbb{Z}_{>0}\) , the equation
\[f(mn)^{2} = f(m^{2})f(f(n))f(mf(n))\]
holds if and only if \(m\) and \(n\) are coprime.
For each... | Answer: All numbers with the same set of prime factors as \(n\) .
Common remarks. We refer to the given property as \(P(m,n)\) . We use the notation \(\operatorname {rad}(n)\) for the radical of \(n\) : the product of the distinct primes dividing \(n\) .
Solution 1. We start with a series of straightforward deduc... |
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