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I have got an assignment having olympiad problems. One entry of these questions went this way:

There are n line segments in the plane With the sum of lengths equal to one. Prove that there exist a straight line such that sum of length of projection of the segments on the line equal to $\frac{2}{\pi}$.

I was trying to use a bit of geometric construction and trigonometry but I don't think the data is sufficient for the method I want to use. I shall be highly thankful if you guys can suggest me how to approach for the problem. Thanks in advance.

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    $\begingroup$ Show that $2/\pi$ is the average of the sum of projected lengths over all possible orientations of a projected-onto line. (By continuity, at least one orientation achieves the average.) To do this, express the "average of the sum of projected lengths" into "the sum of the average of projected lengths"; the average of projected lengths for any one segment happens to be a straightforward integral. $\endgroup$ – Blue Dec 3 '16 at 17:07
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Let $l_i$ be the lengths of segments, and $\theta_i$ be the angles they make with +X-axis. $$\sum_{i=1}^{n}l_i=1$$ Sum of lengths of projections across line at angle $\theta$:-

$$f(\theta)=\sum_{i=1}^{n}l_i|\cos(\theta_i-\theta)|$$

Average value of $|\cos(\theta)|=\frac{2}{\pi}$. Hence average of $f(\theta)$:- $$avg(f(\theta))=\sum_{i=1}^{n}l_i\cdot avg(|\cos(\theta_i-\theta)|)=\frac{2}{\pi}\sum_{i=1}^{n}l_i=\frac{2}{\pi}$$

Since $f(\theta)$ is continuos, it must achieve its average value.

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