IMO 1993
IMO 1993 — 0/6 solved.
IMO 1993
Official IMO 1993 problems · 0/6 solved.
| # | Status | Time |
|---|---|---|
| 1 | — | — |
| 2 | — | — |
| 3 | — | — |
| 4 | — | — |
| 5 | — | — |
| 6 | — | — |
Problem 1
Let $f(x)=x^n+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients.
Problem 2
Let $D$ be a point inside acute triangle $ABC$ such that $\angle ADB=\angle ACB+\pi/2$ and $AC\cdot BD=AD\cdot BC$.
(a) Compute the ratio $(AB\cdot CD)/(AC\cdot BD).$
(b) Prove that the tangents at $C$ to the circumcircles of $\triangle ACD$ and $\triangle BCD$ are perpendicular.
Problem 3
On an infinite chessboard, a game is played as follows. At the start, $n^2$ pieces are arranged on the chessboard in an $n$ by $n$ 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 $n$ for which the game can end with only one piece remaining on the board.
Problem 4
For three points $P,Q,R$ in the plane, we define $m(PQR)$ as the minimum length of the three altitudes of $\triangle PQR$. (If the points are collinear, we set $m(PQR)=0$.)
Prove that for points $A,B,C,X$ in the plane, $$ m(ABC)\le m(ABX)+m(AXC)+m(XBC). $$
Problem 5
Does there exist a function $f:\textbf{N}\rightarrow\textbf{N}$ such that $f(1)=2,f(f(n))=f(n)+n$ for all $n\in\textbf{N}$ and $f(n)<f(n+1)$ for all $n\in\textbf{N}$?
Problem 6
There are $n$ lamps $L_0, \ldots , L_{n-1}$ in a circle ($n > 1$), where we denote $L_{n+k} = L_k$. (A lamp at all times is either on or off.) Perform steps $s_0, s_1, \ldots$ as follows: at step $s_i$, if $L_{i-1}$ is lit, switch $L_i$ from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:
(a) There is a positive integer $M(n)$ such that after $M(n)$ steps all the lamps are on again;
(b) If $n = 2^k$, we can take $M(n) = n^2 - 1$;
(c) If $n = 2^k + 1$, we can take $M(n) = n^2 - n + 1.$