Difference between revisions of "1960 IMO Problems"

(Day I)
(Day II)
Line 31: Line 31:
  
 
=== Problem 4 ===
 
=== Problem 4 ===
 
+
Construct triangle <math>ABC</math>, given <math>h_a</math>, <math>h_b</math> (the altitudes from <math>A</math> and <math>B</math>), and <math>m_a</math>, the median from vertex <math>A</math>.
  
 
[[1960 IMO Problems/Problem 4 | Solution]]
 
[[1960 IMO Problems/Problem 4 | Solution]]
  
 
=== Problem 5 ===
 
=== Problem 5 ===
 +
Consider the cube <math>ABCDA'B'C'D'</math> (with face <math>ABCD</math> directly above face <math>A'B'C'D'</math>).
 +
 +
a) Find the locus of the midpoints of the segments <math>XY</math>, where <math>X</math> is any point of <math>AC</math> and <math>Y</math> is any piont of <math>B'D'</math>;
  
 +
b) Find the locus of points <math>Z</math> which lie on the segment <math>XY</math> of part a) with <math>ZY = 2XZ</math>.
  
 
[[1960 IMO Problems/Problem 5 | Solution]]
 
[[1960 IMO Problems/Problem 5 | Solution]]
  
 
=== Problem 6 ===
 
=== Problem 6 ===
 +
Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. let <math>V_1</math> be the volume of the cone and <math>V_2</math> be the volume of the cylinder.
 +
 +
a) Prove that <math>V_1 \neq V_2</math>;
 +
 +
b) Find the smallest number <math>k</math> for which <math>V_1 = kV_2</math>; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
  
 
[[1960 IMO Problems/Problem 6 | Solution]]
 
[[1960 IMO Problems/Problem 6 | Solution]]

Revision as of 10:48, 28 October 2007

Problems of the 2nd IMO 1960 Romania.

Day I

Problem 1

Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$.

Solution

Problem 2

For what values of the variable $x$ does the following inequality hold:

\[\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?\]


Solution

Problem 3

In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ and odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$



Solution

Day II

Problem 4

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Solution

Problem 5

Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).

a) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any piont of $B'D'$;

b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.

Solution

Problem 6

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Solution

Resources