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Difference between revisions of "1960 IMO Problems"

(→‎Problem 3)
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=== Problem 1 ===
 
=== Problem 1 ===
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Determine all three-digit numbers <math>N</math> having the property that <math>N</math> is divisible by 11, and <math>\dfrac{N}{11}</math> is equal to the sum of the squares of the digits of <math>N</math>.
  
 
[[1960 IMO Problems/Problem 1 | Solution]]
 
[[1960 IMO Problems/Problem 1 | Solution]]
  
 
=== Problem 2 ===
 
=== Problem 2 ===
 +
For what values of the variable <math>x</math> does the following inequality hold:
  
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<cmath>\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?</cmath>
  
  
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In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:
 
In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:
 
<center><math>
 
<center><math>
−
\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
+
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 
</math>
 
</math>
 
</center>
 
</center>

Revision as of 08:37, 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

Solution

Problem 5

Solution

Problem 6

Solution

Resources

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