Difference between revisions of "1960 IMO Problems/Problem 3"
m (+ nonexistent template, format, cat) |
(→See also: no header) |
||
Line 9: | Line 9: | ||
{{solution}} | {{solution}} | ||
− | |||
{{IMO box|year=1960|num-b=2|num-a=4}} | {{IMO box|year=1960|num-b=2|num-a=4}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 19:20, 25 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:
![$\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$](http://latex.artofproblemsolving.com/f/2/7/f27044f40ce377a2b7dda5eed1e1c050c9b86bd9.png)
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
1960 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |