Difference between revisions of "1960 IMO Problems/Problem 3"
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== Problem == | == 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> | + | 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> an 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> | ||
\tan{\alpha}=\frac{4nh}{(n^2-1)a}. | \tan{\alpha}=\frac{4nh}{(n^2-1)a}. | ||
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== Solution == | == Solution == | ||
− | Using coordinates, let <math>A=(0,0)</math>, <math>B=(b,0)</math>, and <math>C=(0,c)</math>. Also, let <math>PQ</math> be the segment that | + | Using coordinates, let <math>A=(0,0)</math>, <math>B=(b,0)</math>, and <math>C=(0,c)</math>. Also, let <math>PQ</math> be the segment that contains the midpoint of the hypotenuse with <math>P</math> closer to <math>B</math>. |
<asy> | <asy> | ||
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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. | ||
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+ | ==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== | ||
− | {{ | + | {{IMO7 box|year=1960|num-b=2|num-a=4}} |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
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 |