Difference between revisions of "1960 IMO Problems/Problem 3"
m (+ nonexistent template, format, cat) |
|||
| Line 1: | Line 1: | ||
| + | == Problem == | ||
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> | ||
| Line 4: | Line 5: | ||
</math> | </math> | ||
</center> | </center> | ||
| + | |||
| + | == Solution == | ||
| + | {{solution}} | ||
| + | |||
| + | == See also == | ||
| + | {{IMO box|year=1960|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 15:20, 12 March 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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 | ||
