Documents

YearFilenameLanguageSource
1993IMO-1993-problems-eng.pdfen
Problem 1

Let f(x)=xn+5xn1+3f(x) = x^n + 5x^{n-1} + 3, where n>1n > 1 is an integer. Prove that f(x)f(x) cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Problem 2

Let DD be a point inside acute triangle ABCABC such that ADB=ACB+π/2\angle ADB = \angle ACB + \pi/2 and ACBD=ADBCAC \cdot BD = AD \cdot BC.

(a) Calculate the ratio (ABCD)/(ACBD)(AB \cdot CD)/(AC \cdot BD).

(b) Prove that the tangents at CC to the circumcircles of ACD\triangle ACD and BCD\triangle BCD are perpendicular.

Problem 3

On an infinite chessboard, a game is played as follows. At the start, n2n^2 pieces are arranged on the chessboard in an nn by nn block of adjoining squares, one piece in each square. A move in the game is a jump in a horizontal or vertical direction over an adjacent occupied square to an unoccupied square immediately beyond. The piece which has been jumped over is removed.

Find those values of nn for which the game can end with only one piece remaining on the board.

Problem 4

For three points P,Q,RP, Q, R in the plane, we define m(PQR)m(PQR) as the minimum length of the three altitudes of PQR\triangle PQR. (If the points are collinear, we set m(PQR)=0m(PQR) = 0.)

Prove that for points A,B,C,XA, B, C, X in the plane, m(ABC)m(ABX)+m(AXC)+m(XBC).m(ABC) \leq m(ABX) + m(AXC) + m(XBC).

Problem 5

Does there exist a function f:NNf: \mathbf{N} \to \mathbf{N} such that f(1)=2f(1) = 2, f(f(n))=f(n)+nf(f(n)) = f(n) + n for all nNn \in \mathbf{N}, and f(n)<f(n+1)f(n) < f(n + 1) for all nNn \in \mathbf{N}?

Problem 6

There are nn lamps L0,,Ln1L_0, \ldots, L_{n-1} in a circle (n>1n > 1), where we denote Ln+k=LkL_{n+k} = L_k. (A lamp at all times is either on or off.) Perform steps s0,s1,s_0, s_1, \ldots as follows: at step sis_i, if Li1L_{i-1} is lit, switch LiL_i from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:

(a) There is a positive integer M(n)M(n) such that after M(n)M(n) steps all the lamps are on again;

(b) If n=2kn = 2^k, we can take M(n)=n21M(n) = n^2 - 1;

(c) If n=2k+1n = 2^k + 1, we can take M(n)=n2n+1M(n) = n^2 - n + 1.