Difference between revisions of "1960 IMO Problems/Problem 3"
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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 subtends the midpoint of the hypotenuse with <math>P</math> closer to <math>B</math>. |
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A,B,C,P,Q; | ||
+ | A=(0,0); | ||
+ | B=(4,0); | ||
+ | C=(0,3); | ||
+ | P=(2.08,1.44); | ||
+ | Q=(1.92,1.56); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | label("A",A,SW); | ||
+ | label("B",B,SE); | ||
+ | label("C",C,NW); | ||
+ | label("P",P,ENE); | ||
+ | label("Q",Q,NNE); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--P); | ||
+ | draw(A--Q); | ||
+ | </asy> | ||
+ | |||
+ | Then, <math>P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)</math>, and <math>Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)</math>. | ||
+ | |||
+ | So, <math>\text{slope}</math><math>(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>. | ||
+ | |||
+ | Thus, <math>\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}</math> | ||
+ | <math>= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}</math>. | ||
+ | |||
+ | 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. | ||
==See Also== | ==See Also== |
Revision as of 14:34, 25 June 2008
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
Using coordinates, let , , and . Also, let be the segment that subtends 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 |