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> |
Revision as of 13:32, 22 August 2008
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.
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 |