Difference between revisions of "1960 IMO Problems/Problem 3"
m |
(→Solution 2) |
||
(One intermediate revision by the same user not shown) | |||
Line 40: | Line 40: | ||
Since <math>[ABC]=\frac{1}{2}bc=\frac{1}{2}ah</math>, <math>bc=ah</math> and <math>\tan{\alpha}=\frac{4nh}{(n^2-1)a}</math> as desired. | Since <math>[ABC]=\frac{1}{2}bc=\frac{1}{2}ah</math>, <math>bc=ah</math> and <math>\tan{\alpha}=\frac{4nh}{(n^2-1)a}</math> as desired. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>P, Q, R</math> be points on side <math>BC</math> such that segment <math>PR</math> contains midpoint <math>Q</math>, with <math>P</math> closer to <math>C</math> and (without loss of generality) <math>AC \le AB</math>. Then if <math>AD</math> is an altitude, then <math>D</math> is between <math>P</math> and <math>C</math>. Combined with the obvious fact that <math>Q</math> is the midpoint of <math>PR</math> (for <math>n</math> is odd), we have | ||
+ | <cmath>\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.</cmath> | ||
==See Also== | ==See Also== |
Latest revision as of 00:14, 17 May 2015
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 |