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Someone could help me with this problem. It is a problem that made me a friend and the truth is not how to solve.

If in the next figure the segments of length $h_i$ are perpendiculars to the base BC of the right triangle ABC and the segments of length $d_i$ are perpendiculars to AC.

Which is the value of $\,\,\,\displaystyle{\sum_{i=1}^{\infty}h_i}$?

enter image description here

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Please have a look at… where there is a discussion about accepting answers to ones questions. – Gerry Myerson Aug 15 '12 at 1:32
up vote 1 down vote accepted

Here, $d_1=h_1\sin A$ and $h_2=d_1\sin A\implies h_2=h_1\sin^2 A$.

Following in similar fashion, $h_3=h_2\sin^2A=h_1\sin^4A$.

Thus $$\sum_{n=1}^{\infty}h_n=h_1(1+\sin^2A+\sin^4A+\cdots)=h_1\cdot \frac{1}{1-\sin^2 A}=h_1\cdot\frac{1}{\cos^2 A}=h_1 \cdot \sec^2 A$$ This is a geometric series with common ratio=$\sin^2 A<1$.

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I still wonder if there's a geometric interpretation of the result. (I also wonder why I didn't catch your original error. :) – Blue Aug 15 '12 at 3:01

Hint: evaluate $h_2$ by similarity of triangles. $h_3/h_2$ will be the same as $h_2/h_1$, so build a geometric series.

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if $\theta$ is the angle at $C$, then it's also the angle between $AB$ and $d_1$:

$$d_1=h_1\cos\theta$$ Similarly, since it's also the angle between $d_1$ and $h_2$ we have $$h_2=d_1\cos\theta$$ So to get from one line to the next you multiply by $\cos\theta$. To go from $h_{n}$ to $h_{n+1}$ we multiply by $\cos^2\theta$. Set up the geometric series and solve.

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