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Difference between revisions of "1993 IMO Problems"

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==Problem 1==
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Problems of the 1993 [[IMO]].
 +
 
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==Day I==
 +
===Problem 1===
 
Let <math>f(x)=x^n+5x^{n-1}+3</math>, where <math>n>1</math> is an integer. Prove that <math>f(x)</math> cannot be expressed as the product of two nonconstant polynomials with integer coefficients.
 
Let <math>f(x)=x^n+5x^{n-1}+3</math>, where <math>n>1</math> is an integer. Prove that <math>f(x)</math> cannot be expressed as the product of two nonconstant polynomials with integer coefficients.
  
 
[[1993 IMO Problems/Problem 1|Solution]]
 
[[1993 IMO Problems/Problem 1|Solution]]
  
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==Problem 2==
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===Problem 2===
 
Let <math>D</math> be a point inside acute triangle <math>ABC</math> such that <math>\angle ADB=\angle ACB+\pi/2</math> and <math>AC\cdot BD=AD\cdot BC</math>.
 
Let <math>D</math> be a point inside acute triangle <math>ABC</math> such that <math>\angle ADB=\angle ACB+\pi/2</math> and <math>AC\cdot BD=AD\cdot BC</math>.
  
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[[1993 IMO Problems/Problem 2|Solution]]
 
[[1993 IMO Problems/Problem 2|Solution]]
  
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==Problem 3==
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===Problem 3===
 
On an infinite chessboard, a game is played as follows. At the start, <math>n^2</math> pieces are arranged on the chessboard in an <math>n</math> by <math>n</math> 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 <math>n</math> for which the game can end with only one piece remaining on the board.
 
On an infinite chessboard, a game is played as follows. At the start, <math>n^2</math> pieces are arranged on the chessboard in an <math>n</math> by <math>n</math> 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 <math>n</math> for which the game can end with only one piece remaining on the board.
  
 
[[1993 IMO Problems/Problem 3|Solution]]
 
[[1993 IMO Problems/Problem 3|Solution]]
  
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==Problem 4==
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==Day II==
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===Problem 4===
 
For three points <math>P,Q,R</math> in the plane, we define <math>m(PQR)</math> as the minimum length of the three altitudes of <math>\triangle PQR</math>. (If the points are collinear, we set <math>m(PQR)=0</math>.)
 
For three points <math>P,Q,R</math> in the plane, we define <math>m(PQR)</math> as the minimum length of the three altitudes of <math>\triangle PQR</math>. (If the points are collinear, we set <math>m(PQR)=0</math>.)
  
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[[1993 IMO Problems/Problem 4|Solution]]
 
[[1993 IMO Problems/Problem 4|Solution]]
  
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==Problem 5==
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===Problem 5===
 
Does there exist a function <math>f:\textbf{N}\rightarrow\textbf{N}</math> such that <math>f(1)=2,f(f(n))=f(n)+n</math> for all <math>n\in\textbf{N}</math> and <math>f(n)<f(n+1)</math> for all <math>n\in\textbf{N}</math>?
 
Does there exist a function <math>f:\textbf{N}\rightarrow\textbf{N}</math> such that <math>f(1)=2,f(f(n))=f(n)+n</math> for all <math>n\in\textbf{N}</math> and <math>f(n)<f(n+1)</math> for all <math>n\in\textbf{N}</math>?
  
 
[[1993 IMO Problems/Problem 5|Solution]]
 
[[1993 IMO Problems/Problem 5|Solution]]
  
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==Problem 6==
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===Problem 6===
 
There are <math>n</math> lamps <math>L_0, \ldots , L_{n-1}</math> in a circle (<math>n > 1</math>), where we denote <math>L_{n+k} = L_k</math>. (A lamp at all times is either on or off.) Perform steps <math>s_0, s_1, \ldots</math> as follows: at step <math>s_i</math>, if <math>L_{i-1}</math> is lit, switch <math>L_i</math> from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:
 
There are <math>n</math> lamps <math>L_0, \ldots , L_{n-1}</math> in a circle (<math>n > 1</math>), where we denote <math>L_{n+k} = L_k</math>. (A lamp at all times is either on or off.) Perform steps <math>s_0, s_1, \ldots</math> as follows: at step <math>s_i</math>, if <math>L_{i-1}</math> is lit, switch <math>L_i</math> from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:
  

Latest revision as of 20:34, 4 July 2024

Problems of the 1993 IMO.

Day I

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.

Solution

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.

Solution

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.

Solution

Day II

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, \[m(ABC)\le m(ABX)+m(AXC)+m(XBC).\]

Solution

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 $n\in\textbf{N}$?

Solution

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 $M(n) = n^2 - n + 1.$

Solution

See Also

1993 IMO (Problems) • Resources
Preceded by
1992 IMO 		1 • 2 • 3 • 4 • 5 • 6 Followed by
1994 IMO
All IMO Problems and Solutions
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