Minimum area of a triangle with fixed incircle

[Referos]

Homework Statement

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

Homework Equations

None.

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.

 

[pizzasky]

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.

 

[Referos]

Ah, thanks, I managed to find the area with your "hint". I didn't see that triangle similarity before. From there, differentiating was simple.