Art of Problem Solving

1960 IMO Problems/Problem 3: Difference between revisions

Temperal (talk | contribs)
See also: no header
1=2 (talk | contribs)
Line 10: Line 10:


{{IMO box|year=1960|num-b=2|num-a=4}}
{{IMO box|year=1960|num-b=2|num-a=4}}
==See Also==


[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Geometry Problems]]

Revision as of 09:45, 28 October 2007

Problem

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:

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

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

See Also