Maths Olympiad Prep

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Stage 10 · Mixed

10 problems · Hardest shortlist tier · mathsolympiadprep.com

The answer key prints on its own page at the end.

  1. Given an integer n>1n > 1 and an integer aa that is coprime with nn. There is a country consisting of nn islands D1,D2,,DnD_1, D_2, \dots, D_n. For any two different islands DiD_i and DjD_j, there is a one-way ferry from DiD_i to DjD_j if and only if ijia(modn)ij \equiv ia \pmod{n}. A tourist hopes to visit as many islands as possible. He can first fly to any island he chooses to start the tour, and afterwards can only use the one-way ferry to tour freely between islands in this country. Find the maximum possible number of different islands that the tourist can visit.

    Number theory Solution and answer checking →

  2. Let V\mathcal{V} be a finite set of points in the plane. We say that V\mathcal{V} is balanced if for any two distinct points A,BVA, B \in \mathcal{V}, there exists a point CVC \in \mathcal{V} such that AC=BCAC = BC. We say that V\mathcal{V} is center-free if for any distinct points A,B,CVA, B, C \in \mathcal{V}, there does not exist a point PVP \in \mathcal{V} such that PA=PB=PCPA = PB = PC.

    a. Show that for all n3n \geqslant 3, there exists a balanced set consisting of nn points.

    b. For which n3n \geqslant 3 does there exist a balanced, center-free set consisting of nn points?

    Geometry Solution and answer checking →

  3. Find all positive integers n2n \ge 2 for which there exist nn real numbers
    a1<a2<<an a_1 < a_2 < \dots < a_n
    and a real number r>0r > 0 such that the n(n1)2\frac{n(n-1)}{2} differences ajaia_j - a_i for 1i<jn1 \le i < j \le n are equal, in some order, to the numbers
    r1,r2,,rn(n1)2. r^1, r^2, \dots, r^{\frac{n(n-1)}{2}}.

    Algebra Solution and answer checking →

  4. Let A0=(a1,,an)A_{0} = (a_{1}, \ldots, a_{n}) be a finite sequence of real numbers. For each k0k \geq 0, from the sequence Ak=(x1,,xn)A_{k} = (x_{1}, \ldots, x_{n}) we construct a new sequence Ak+1A_{k+1} in the following way.
    1. We choose a partition {1,,n}=IJ\{1, \ldots, n\} = I \cup J, where II and JJ are two disjoint sets, such that the expression
    iIxijJxj \left|\sum_{i \in I} x_{i} - \sum_{j \in J} x_{j}\right|
    attains the smallest possible value. (We allow the sets II or JJ to be empty; in this case the corresponding sum is 00.) If there are several such partitions, one is chosen arbitrarily.
    2. We set Ak+1=(y1,,yn)A_{k+1} = (y_{1}, \ldots, y_{n}), where yi=xi+1y_{i} = x_{i} + 1 if iIi \in I, and yi=xi1y_{i} = x_{i} - 1 if iJi \in J.
    Prove that for some kk, the sequence AkA_{k} contains an element xx such that xn/2|x| \geq n / 2.

    Combinatorics Solution and answer checking →

  5. Let nn be a positive integer. We say that a polynomial PP with integer coefficients is nn-good if there exists a polynomial QQ of degree 2 with integer coefficients such that Q(k)(P(k)+Q(k))Q(k)(P(k)+Q(k)) is never divisible by nn for any integer kk.
    Determine all integers nn such that every polynomial with integer coefficients is an nn-good polynomial.

    Number theory Solution and answer checking →

  6. Let ABCDEA B C D E be a convex pentagon with CD=DEC D = D E and EDC2ADB\angle E D C \neq 2 \cdot \angle A D B. Suppose that a point PP is located in the interior of the pentagon such that AP=AEA P = A E and BP=BCB P = B C. Prove that PP lies on the diagonal CEC E if and only if area(BCD)+area(ADE)=area(ABD)+area(ABP)\operatorname{area}(B C D) + \operatorname{area}(A D E) = \operatorname{area}(A B D) + \operatorname{area}(A B P).

    Geometry Solution and answer checking →

  7. Let Z0\mathbb{Z}_{\geqslant 0} be the set of all nonnegative integers. Find all the functions f:Z0Z0f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0} satisfying the relation
    f(f(f(n)))=f(n+1)+1 f(f(f(n)))=f(n+1)+1
    for all nZ0n \in \mathbb{Z}_{\geqslant 0}.

    Algebra Solution and answer checking →

  8. Suppose there are 101 persons sitting around a round table in an arbitrary order. The kkth person possesses kk pieces of cards, k=1,,101k = 1, \dots, 101. We call it a transition if one transits one of his cards to one of his adjacent persons. Find the minimum positive number kk, such that whatever the order of the seating, there is a way of no more than kk transitions so that each person possesses 51 cards.

    Combinatorics Solution and answer checking →

  9. Given a positive integer nn, let DD be the set of positive divisors of nn, and let f:DZf: D \to \mathbb{Z} be a function. Prove that the following are equivalent:
    (A) for any positive divisor mm of nn,
    ndmf(d)(n/dm/d); n \mid \sum_{d|m} f(d) \binom{n/d}{m/d};
    (B) for any positive divisor kk of nn,
    kdkf(d). k \mid \sum_{d|k} f(d).

    Number theory Solution and answer checking →

  10. Suppose there are beetles on a chessboard consisting of 2012×20122012 \times 2012 unit squares. Each unit square can accommodate at most one beetle. At a moment, all beetles fly and land on the chessboard again. For a beetle, we call the vector from its flying unit to its landing unit the beetle's "displacement vector". We call the sum of all beetle's "displacement vectors" the "total displacement vectors".
    Find the maximum length of "total displacement vector" considering the number of beetles and all possible positions of flying and landing. (posed by Qu Zhenhua)

    Geometry Solution and answer checking →

Answer key — Stage 10 · Mixed

Worked solutions for every problem are on the site, one page per problem.

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