# Difference between revisions of "1960 IMO Problems/Problem 3"

Line 1: | Line 1: | ||

− | 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: