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Difference between revisions of "1959 IMO Problems"

 
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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.
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[[1959 IMO Problems/Problem 4 | Solution]]
  
 
=== 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>.
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 +
(a) Prove that the points <math>\displaystyle N </math> and <math>\displaystyle 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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 +
(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>.
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[[1959 IMO Problems/Problem 5 | Solution]]
  
 
=== 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.
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[[1959 IMO Problems/Problem 6 | Solution]]
  
 
== Resources ==
 
== Resources ==
 
* [[1959 IMO]]
 
* [[1959 IMO]]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1959 IMO 1959 Problems on the Resources page]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1959 IMO 1959 Problems on the Resources page]

Revision as of 21:20, 26 July 2006

Problems of the 1st IMO 1959 Romania.

Day I

Problem 1

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

Solution

Problem 2

For what real values of $\displaystyle x$ is

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

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

Solution

Problem 3

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

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

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

Solution

Day II

Problem 4

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

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

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

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

Solution

Problem 6

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

Solution

Resources

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