Difference between revisions of "1960 IMO Problems"

 
(Problem 3)
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=== Problem 3 ===
 
=== Problem 3 ===
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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:
 +
<center><math>
 +
\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 +
</math>
 +
</center>
  
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 angel subtending, from <math>A</math>, that segment which contains the mdipoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse fo the triangle. Prove that:
 
\[
 
\tan{\alpha}=\dfrac{4nh}{(n^2-1)a}.
 
\]
 
  
  

Revision as of 00:37, 12 March 2007

Problems of the 2nd IMO 1960 Romania.

Day I

Problem 1

Solution

Problem 2

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:

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



Solution

Day II

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Resources