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 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> | ||
− | + | \tan{\alpha}=\frac{4nh}{(n^2-1)a}. | |
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+ | ==See Also== | ||
{{IMO box|year=1960|num-b=2|num-a=4}} | {{IMO box|year=1960|num-b=2|num-a=4}} | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 10:45, 28 October 2007
Problem
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:
Solution
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See Also
1960 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |