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 | + | 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 
, the hypotenuse 
, of length 
, is divided into 
equal parts ( and odd integer). Let 
be the acute angle subtending, from 
, that segment which contains the midpoint of the hypotenuse. Let 
be the length of the altitude to the hypotenuse of the triangle. Prove that:
