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Difference between revisions of "1990 IMO Problems"

(Created page with "Problems of the 1990 IMO. ==Day I== ===Problem 1=== Chords <math> AB</math> and <math> CD</math> of a circle intersect at a point <math> E</math> inside the circle. Let <...")
 
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<cmath> \frac {2^n + 1}{n^2} </cmath> is an integer.
 
<cmath> \frac {2^n + 1}{n^2} </cmath> is an integer.
  
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[[1990 IMO Problems/Problem 1|Solution]]
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[[1990 IMO Problems/Problem 3|Solution]]
  
 
==Day II==
 
==Day II==
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Given an initial integer <math> n_0 > 1</math>, two players, <math> {\mathcal A}</math> and <math> {\mathcal B}</math>, choose integers <math> n_1</math>, <math> n_2</math>, <math> n_3</math>, <math> \ldots</math> alternately according to the following rules :
 
Given an initial integer <math> n_0 > 1</math>, two players, <math> {\mathcal A}</math> and <math> {\mathcal B}</math>, choose integers <math> n_1</math>, <math> n_2</math>, <math> n_3</math>, <math> \ldots</math> alternately according to the following rules :
  
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I.) Knowing <math> n_{2k}</math>, <math> {\mathcal A}</math> chooses any integer <math> n_{2k + 1}</math> such that
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I.) Knowing <math> n_{2k}</math>, <math>\mathcal{A}</math> chooses any integer <math> n_{2k + 1}</math> such that
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\[ n_{2k} \leq n_{2k + 1} \leq n_{2k}^2.
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<cmath>n_{2k} \leq n_{2k + 1} \leq n_{2k}^2.</cmath>
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\]
+
 
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II.) Knowing <math> n_{2k + 1}</math>, <math> {\mathcal B}</math> chooses any integer <math> n_{2k + 2}</math> such that
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II.) Knowing <math> n_{2k + 1}</math>, <math> {\mathcal B}</math> chooses any integer <math> n_{2k + 2}</math> such that <cmath>\dfrac{n_{2k + 1}}{n_{2k + 2}}</cmath>
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\[ \frac {n_{2k + 1}}{n_{2k + 2}}
 
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\]
 
 
is a prime raised to a positive integer power.
 
is a prime raised to a positive integer power.
  

Latest revision as of 16:44, 6 November 2022

Problems of the 1990 IMO.

Day I

Problem 1

Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $EB$. The tangent line at $E$ to the circle through $D$, $E$, and $M$ intersects the lines $BC$ and $AC$ at $F$ and $G$, respectively. If \[\frac {AM}{AB} = t,\] find $\frac {EG}{EF}$ in terms of $t$.

Solution

Problem 2

Let $n \geq 3$ and consider a set $E$ of $2n - 1$ distinct points on a circle. Suppose that exactly $k$ of these points are to be colored black. Such a coloring is good if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $n$ points from $E$. Find the smallest value of $k$ so that every such coloring of $k$ points of $E$ is good.

Solution

Problem 3

Determine all integers $n > 1$ such that \[\frac {2^n + 1}{n^2}\] is an integer.

Solution

Day II

Problem 4

Let ${\mathbb Q}^ +$ be the set of positive rational numbers. Construct a function $f : {\mathbb Q}^ + \rightarrow {\mathbb Q}^ +$ such that \[f(xf(y)) = \frac {f(x)}{y}\] for all $x$, $y$ in ${\mathbb Q}^ +$.

Solution

Problem 5

Given an initial integer $n_0 > 1$, two players, ${\mathcal A}$ and ${\mathcal B}$, choose integers $n_1$, $n_2$, $n_3$, $\ldots$ alternately according to the following rules :

I.) Knowing $n_{2k}$, $\mathcal{A}$ chooses any integer $n_{2k + 1}$ such that \[n_{2k} \leq n_{2k + 1} \leq n_{2k}^2.\]

II.) Knowing $n_{2k + 1}$, ${\mathcal B}$ chooses any integer $n_{2k + 2}$ such that \[\dfrac{n_{2k + 1}}{n_{2k + 2}}\] is a prime raised to a positive integer power.

Player ${\mathcal A}$ wins the game by choosing the number 1990; player ${\mathcal B}$ wins by choosing the number 1. For which $n_0$ does :


a.) ${\mathcal A}$ have a winning strategy? b.) ${\mathcal B}$ have a winning strategy? c.) Neither player have a winning strategy?

Solution

Problem 6

Prove that there exists a convex 1990-gon with the following two properties :

a.) All angles are equal. b.) The lengths of the 1990 sides are the numbers $1^2$, $2^2$, $3^2$, $\cdots$, $1990^2$ in some order.

Solution

1990 IMO (Problems) • Resources
Preceded by
1989 IMO 		1 • 2 • 3 • 4 • 5 • 6 Followed by
1991 IMO
All IMO Problems and Solutions
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