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Math Help - Help with a constrained extrema problem.

  1. #1
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    Help with a constrained extrema problem.

    Hi there. (This is going to be a long post)

    I want to show that for any triangle that is inscribed in a circle with a radius of R, the largest perimeter of the triangle comes from an equilateral triangle in the circle.


    If i drew a triangle in the circle i noticed that the length of the origin to any vertex of the triangle is R.

    So applying that this is what I did, I can basically split the triangle into 3 parts to find each side of the triangle from the law of cosines:

    c^2 = a^2 + b^2 - 2ab cos C

    For example is c was a side of the triangle I can get its length through R. Here is my bad mspaint interpretation.

    Help with a constrained extrema problem.-7nmne.png

    c^2 = R^2 + R^2 - 2R^2 cos C

    and R = x^2+y^2

    so i can do that for the other sides as well, and form the perimeter equation

    P(a,b,c) = a + b + c

    I can now sub in R as x^2 + y^2 in the perimeter equation to get it in terms of x and y, but after doing that and simplifying, then solving it with the larangian with the constraint as x^y + y^2 = R, I dont get an expression for lamda in terms of x and y. All i get is a constant in terms of the angles cos a, cos b, and cos c.

    I guess it would make sense that the critical point would be in terms of the angles rather than x and y, because if cos a = cos b = cos c then it is an equilateral triangle. But what do i do here?
    Last edited by Kuma; November 19th 2011 at 12:43 PM.
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  2. #2
    Grand Panjandrum
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    Re: Help with a constrained extrema problem.

    Quote Originally Posted by Kuma View Post
    Hi there. (This is going to be a long post)

    I want to show that for any triangle that is inscribed in a circle with a radius of R, the largest perimeter of the triangle comes from an equilateral triangle in the circle.


    If i drew a triangle in the circle i noticed that the length of the origin to any vertex of the triangle is R.

    So applying that this is what I did, I can basically split the triangle into 3 parts to find each side of the triangle from the law of cosines:

    c^2 = a^2 + b^2 - 2ab cos C

    For example is c was a side of the triangle I can get its length through R. Here is my bad mspaint interpretation.

    Click image for larger version. 

Name:	7NMne.png 
Views:	11 
Size:	3.1 KB 
ID:	22812

    c^2 = R^2 + R^2 - 2R^2 cos C

    and R = x^2+y^2

    so i can do that for the other sides as well, and form the perimeter equation

    P(a,b,c) = a + b + c

    I can now sub in R as x^2 + y^2 in the perimeter equation to get it in terms of x and y, but after doing that and simplifying, then solving it with the larangian with the constraint as x^y + y^2 = R, I dont get an expression for lamda in terms of x and y. All i get is a constant in terms of the angles cos a, cos b, and cos c.

    I guess it would make sense that the critical point would be in terms of the angles rather than x and y, because if cos a = cos b = cos c then it is an equilateral triangle. But what do i do here?
    First: You need only consider the unit circle with R=1.

    Second: You can consider one vertex fixed leaving only two free vertices to worry about.

    Third: Each of the two free vertices is characterised by the angle (measured anti-clockwise) between the fixed vertex and the free vertex.

    CB
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