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 Minimal area. (Posted on 2010-03-29)
Take some point V and draw two rays from it. Choose some other point W in between those two rays. Then, construct a line that touches both rays and passes through W.

Now, this line forms a closed triangle together with the two rays. The point W divides this line into two segments (x1, x2). What is the ratio of these two segments such that the area of the enclosed triangle is minimal?

Does this minimal area even exist?

 See The Solution Submitted by Vee-Liem Veefessional Rating: 4.6667 (3 votes)

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 re(3): Thoughts | Comment 10 of 11 |
(In reply to re(2): Thoughts by Bractals)

My own theory is that it's worth D2 as the long skinny triangles of my solution--the increments of area as one swivels the line about W--can be seen to be balancing in their addition and subtraction when their heights measured from W are equal. In the infinitesimal case it does not matter that the base is not at right angles to the height, as the thickness of this dA is the same on both slivers when the heights are equal. It's not necessary to go into further calculus than that setup, and the realization that that's a minimum rather than a maximum, as, continuing toward being parallel to the sides of the V angle makes the area larger and larger.

... or looked at another way: The two slivery incremental/decremental triangles have the same apex angle as the angles are vertical to each other. The shorter one can be overlayed on the longer and obviously has less area. So the line has to be rotated so the longer one gets taken in, and its area subtracted from the target triangle and the shorter one gets added to the target triangle, until this is no longer possible, when the infinitesimally thin triangles become the same length.

It's easier to think about than to write it down, but it's basically solved in the head. The actual integration need not be done.

Edited on March 30, 2010, 10:18 pm
 Posted by Charlie on 2010-03-30 21:58:44

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