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Difference between revisions of "1959 IMO Problems"
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Problems of the 1st [[IMO]] 1959 Romania.
+
Problems of the 1st [[IMO]] 1959 in Romania.
  
 
== Day I ==
 
== Day I ==
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=== Problem 1 ===
 
=== Problem 1 ===
  
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Prove that <math>\displaystyle\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>\displaystyle n</math>.
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Prove that <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>.
  
 
[[1959 IMO Problems/Problem 1 | Solution]]
 
[[1959 IMO Problems/Problem 1 | Solution]]
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=== Problem 2 ===
 
=== Problem 2 ===
  
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For what real values of <math>\displaystyle x</math> is
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For what real values of <math>x</math> is
  
 
<center>
 
<center>
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</center>
 
</center>
  
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given (a) <math>A = \sqrt{2}</math>, (b) <math>\displaystyle A=1</math>, (c) <math>\displaystyle A=2</math>, where only non-negative real numbers are admitted for square roots?
+
given (a) <math>A = \sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots?
  
 
[[1959 IMO Problems/Problem 2 | Solution]]
 
[[1959 IMO Problems/Problem 2 | Solution]]
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=== Problem 3 ===
 
=== Problem 3 ===
  
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Let <math>\displaystyle a,b,c</math> be real numbers.  Consider the quadratic equation in <math>\displaystyle \cos{x}</math> :
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Let <math>a,b,c</math> be real numbers.  Consider the quadratic equation in <math>\cos{x}</math> :
  
 
<center>
 
<center>
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<math>\displaystyle a\cos ^{2}x + b\cos{x} + c = 0.</math>
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<math>a\cos ^{2}x + b\cos{x} + c = 0.</math>
 
</center>
 
</center>
  
−
Using the numbers <math>\displaystyle a,b,c</math>, form a quadratic equation in <math>\displaystyle \cos{2x}</math>, whose roots are the same as those of the original equation.  Compare the equations in <math>\displaystyle \cos{x}</math> and <math>\displaystyle \cos{2x}</math> for <math>\displaystyle a=4, b=2, c=-1</math>.
+
Using the numbers <math>a,b,c</math>, form a quadratic equation in <math>\cos{2x}</math>, whose roots are the same as those of the original equation.  Compare the equations in <math>\cos{x}</math> and <math>\cos{2x}</math> for <math>a=4, b=2, c=-1</math>.
  
 
[[1959 IMO Problems/Problem 3 | Solution]]
 
[[1959 IMO Problems/Problem 3 | Solution]]
  
 
== Day II ==
 
== Day II ==
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=== Problem 4 ===
 
=== Problem 4 ===
  
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Construct a right triangle with a given hypotenuse <math>\displaystyle c</math> such that the median drawn to the hypotenuse is the [[geometric mean]] of the two legs of the triangle.
+
Construct a right triangle with a given hypotenuse <math>c</math> such that the median drawn to the hypotenuse is the [[geometric mean]] of the two legs of the triangle.
  
 
[[1959 IMO Problems/Problem 4 | Solution]]
 
[[1959 IMO Problems/Problem 4 | Solution]]
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=== Problem 5 ===
 
=== Problem 5 ===
  
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An arbitrary point <math>\displaystyle M </math> is selected in the interior of the segment <math>\displaystyle AB </math>.  The squares <math>\displaystyle AMCD </math> and <math>\displaystyle MBEF </math> are constructed on the same side of <math>\displaystyle AB </math>, with the segments <math>\displaystyle AM </math> and <math>\displaystyle MB </math> as their respective bases.  The circles about these squares, with respective centers <math>\displaystyle P </math> and <math>\displaystyle Q </math>, intersect at <math>\displaystyle M </math> and also at another point <math>\displaystyle N </math>.  Let <math>\displaystyle N' </math> denote the point of intersection of the straight lines <math>\displaystyle AF </math> and <math>\displaystyle BC </math>.
+
An arbitrary point <math>M </math> is selected in the interior of the segment <math>AB </math>.  The squares <math>AMCD </math> and <math>MBEF </math> are constructed on the same side of <math>AB </math>, with the segments <math>AM </math> and <math>MB </math> as their respective bases.  The circles about these squares, with respective centers <math>P </math> and <math>Q </math>, intersect at <math>M </math> and also at another point <math>N </math>.  Let <math>N' </math> denote the point of intersection of the straight lines <math>AF </math> and <math>BC </math>.
  
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(a) Prove that the points <math>\displaystyle N </math> and <math>\displaystyle N' </math> coincide.
+
(a) Prove that the points <math>N </math> and <math>N' </math> coincide.
  
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(b) Prove that the straight lines <math>\displaystyle MN </math> pass through a fixed point <math>\displaystyle S </math> independent of the choice of <math>\displaystyle M </math>.
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(b) Prove that the straight lines <math>MN </math> pass through a fixed point <math>S </math> independent of the choice of <math>M </math>.
  
−
(c) Find the locus of the midpoints of the segments <math>\displaystyle PQ </math> as <math>\displaystyle M </math> varies between <math>\displaystyle A </math> and <math>\displaystyle B </math>.
+
(c) Find the locus of the midpoints of the segments <math>PQ </math> as <math>M </math> varies between <math>A </math> and <math>B </math>.
  
 
[[1959 IMO Problems/Problem 5 | Solution]]
 
[[1959 IMO Problems/Problem 5 | Solution]]
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=== Problem 6 ===
 
=== Problem 6 ===
  
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Two planes, <math>\displaystyle P </math> and <math>\displaystyle Q </math>, intersect along the line <math>\displaystyle p </math>.  The point <math>\displaystyle A </math> is in the plane <math>\displaystyle P </math>, and the point <math>\displaystyle {C} </math> is in the plane <math>\displaystyle Q </math>; neither of these points lies on the straight line <math>\displaystyle p </math>.  Construct an isosceles trapezoid <math>\displaystyle ABCD </math> (with <math>\displaystyle AB </math> parallel to <math>\displaystyle DC </math>) in which a circle can be constructed, and with vertices <math>\displaystyle B </math> and <math>\displaystyle D </math> lying in the planes <math>\displaystyle P </math> and <math>\displaystyle Q </math>, respectively.
+
Two planes, <math>P </math> and <math>Q </math>, intersect along the line <math>p </math>.  The point <math>A </math> is in the plane <math>P </math>, and the point <math>{C} </math> is in the plane <math>Q </math>; neither of these points lies on the straight line <math>p </math>.  Construct an isosceles trapezoid <math>ABCD </math> (with <math>AB </math> parallel to <math>DC </math>) in which a circle can be constructed, and with vertices <math>B </math> and <math>D </math> lying in the planes <math>P </math> and <math>Q </math>, respectively.
  
 
[[1959 IMO Problems/Problem 6 | Solution]]
 
[[1959 IMO Problems/Problem 6 | Solution]]

Revision as of 09:59, 30 May 2012

Problems of the 1st IMO 1959 in Romania.

Day I

Problem 1

Prove that $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.

Solution

Problem 2

For what real values of $x$ is

$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A,$

given (a) $A = \sqrt{2}$, (b) $A=1$, (c) $A=2$, where only non-negative real numbers are admitted for square roots?

Solution

Problem 3

Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ :

$a\cos ^{2}x + b\cos{x} + c = 0.$

Using the numbers $a,b,c$, form a quadratic equation in $\cos{2x}$, whose roots are the same as those of the original equation. Compare the equations in $\cos{x}$ and $\cos{2x}$ for $a=4, b=2, c=-1$.

Solution

Day II

Problem 4

Construct a right triangle with a given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

Solution

Problem 5

An arbitrary point $M$ is selected in the interior of the segment $AB$. The squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with the segments $AM$ and $MB$ as their respective bases. The circles about these squares, with respective centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.

(a) Prove that the points $N$ and $N'$ coincide.

(b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$.

(c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

Solution

Problem 6

Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is in the plane $P$, and the point ${C}$ is in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB$ parallel to $DC$) in which a circle can be constructed, and with vertices $B$ and $D$ lying in the planes $P$ and $Q$, respectively.

Solution

Resources

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