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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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 solution | Comment 3 of 11 |

At any given position of the line with segments x1 and x2, pivoting the line about W so that the longer segment of the two gets closer to V will result in a net decrease in the triangle's area: Consider the thin triangles added and subtracted. In the present diagram x2 is longer than x1, and so x2 should approach V. In incremental triangle taken away from the triangle in question would then have an area dA = dTheta * (x2)^2 / 2 while the incremental triangle added to it would have area dA = dTheta * (x1)^2 / 2, where dTheta is the small angle (approaching zero in calculus terms, thus the differential notation) through which the line is pivoted.

This continues until x1 and x2 are equal, when the positive and negative differentials of the area become equal and the area has reached a minimum.

The ratio is 1.

Edited to use Theta rather than attempt to get a real theta and wind up with accented E.

Edited on March 30, 2010, 9:20 am
 Posted by Charlie on 2010-03-29 16:19:33

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