Difference between revisions of "1960 IMO Problems/Problem 3"
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==See Also== | ==See Also== | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 19:05, 15 September 2012
Problem
In a given right triangle , the hypotenuse
, of length
, is divided into
equal parts (
an 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:
![$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$](http://latex.artofproblemsolving.com/d/7/6/d7667ee6a4058553b8290f72ffe77f099ab473bf.png)
Solution
Using coordinates, let ,
, and
. Also, let
be the segment that contains the midpoint of the hypotenuse with
closer to
.
Then, , and
.
So, , and
.
Thus,
.
Since ,
and
as desired.
See Also
1960 IMO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 4 |