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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: Incomplete | Comment 5 of 11 |
(In reply to Incomplete by hoodat)

Consider, for example, a case where the angle at V is 60° so that an isosceles triangle is also equilateral. Consider a side length of 10 for this equilateral triangle and point W on the side opposite V situated such that x1 = 2.5 and x2 = 7.5.

First of all, the area of the isosceles (equilateral) triangle is 5*sqr(3)*10/2 ~= 43.30.  Now take the midpoint, M, of the side of the equilateral triangle that contains V and is on the same side as x2. A triangle is formed with sides 5 and 7.5 with a 60° angle between them. Call the endpoint of x2 other than W, B. Side MW can be found by the law of cosines from triangle MBW:

MW^2 = 5^2 + 7.5^2 - 2*5*7.5*cos(60°) = 43.75

MW = sqrt(43.75) ~= 6.614

Further work shows that at this point bisects the line segement defined by MW extended cut off by the two sides of angle V, and the new triangle formed has area of about 32.5, which is less than that of the equilateral triangle.

(I had intended to carry the calculations out exactly, but Geometer's Sketchpad shows the results well enough that the minimum occurs when x1 = x2, rather than when the triangle is isosceles.)

 Posted by Charlie on 2010-03-29 17:44:09

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