1959 IMO Problems

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

See Also

1959 IMO (Problems) • Resources)
Preceded by
First IMO
1 2 3 4 5 6 Followed by
1960 IMO
All IMO Problems and Solutions
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