Difference between revisions of "1960 IMO Problems/Problem 3"

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==See Also==
==See Also==
{{IMO box|year=1960|num-b=2|num-a=4}}
{{IMO7 box|year=1960|num-b=2|num-a=4}}
[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Geometry Problems]]

Revision as of 19:05, 15 September 2012


In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:



Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.

[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, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)$.

So, $\text{slope}$$(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}$, and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.

Thus, $\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}$ $= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}$.

Since $[ABC]=\frac{1}{2}bc=\frac{1}{2}ah$, $bc=ah$ and $\tan{\alpha}=\frac{4nh}{(n^2-1)a}$ as desired.

See Also

1960 IMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 7 Followed by
Problem 4
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