1960 IMO Problems/Problem 3
Contents
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:
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.
Solution 2
Let be points on side such that segment contains midpoint , with closer to and (without loss of generality) . Then if is an altitude, then is between and . Combined with the obvious fact that is the midpoint of (for is odd), we have
See Also
1960 IMO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 4 |