Difference between revisions of "1960 IMO Problems/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 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:
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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>
 
<center><math>
 
\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 
\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 
</math>
 
</math>
 
</center>
 
</center>

Revision as of 01:25, 12 March 2007

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:

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