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Homework Help: Minimum area of a triangle with fixed incircle

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider an isosceles triangles with height x and a fixed incircle of radius R. Show that the minimum area of such triangle occurs when x = 3R


    2. Relevant equations
    None.


    3. The attempt at a solution
    Well, I know that have to express the area of the triangle in terms of x, then differentiate with respect to x and find where the derivative is zero, and then check whether it's a minimum or maximum. My problem is mostly with the geometric part of the problem. Since I have the height x, I suppose that I need to find the base in terms of x, since area = 1/2 * base * height.
    The only thing I've managed to do is drawing a line from the center of the incircle to one of the points of tangent; this gives a right triangle with a cathetus equal to R and a hyponetuse equal to x - R. I've tried finding similar triangles but it didn't work so well.
    Thanks.
     
  2. jcsd
  3. May 5, 2010 #2
    Denote the 3 vertices of the isosceles triangle by A, B, and C, with sides AB and BC having the same length.
    Drop a perpendicular from B to AC, and denote the point of intersection by D. Note that we have BD = x.
    We shall label the centre of the incircle as O, and shall drop another perpendicular from O to AB. The foot of the perpendicular shall be denoted by E. I believe you're right in saying that OE = R and OB = x - R.
    Here's how I suggest you proceed : Observe that the right-angled triangles ABD and OBE are similar. (Why?) Use this fact to express AD in terms of R and x. Determining the area of the triangle ABC should now be easy.
     
  4. May 6, 2010 #3
    Ah, thanks, I managed to find the area with your "hint". I didn't see that triangle similarity before. From there, differentiating was simple.
     
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