Optimizing Triangle Dimensions for Circumscribing a Circle

[John O' Meara]

Homework Statement

Find the dimensions of an isosceles triangle of least area that can be circumscribed about a circle of radius R.

Homework Equations

Let a,b, and c denote the lengths of the sides of this triangle and c=b. Then the angles opposite these sides are A, B and C respectively, (B=C). Let B' be the angle between the radius R and the side a. See attached figure.

The Attempt at a Solution

The quantity to be minimumized is the area of the triangle. I am not too sure how to go about this but the area of the triangle S=\frac{a}{2}(R+R \sin(B')). Now 2B'+2A=pi, therefore B' =\frac{\pi}{2}-A \mbox{ therefore} \sin (B') = \sin(\frac{\pi}{2}-A) = \sin(\frac{\pi}{2}+A). Therefore S=\frac{a}{2}R(1+\sin(\frac{\pi}{2}+A)).
If I differentiate this implicitly I get \frac{a}{2}+\frac{a}{2}R\frac{d}{dR}R \sin(\frac{\pi}{2}+A) + \frac{a}{2} \sin(\frac{\pi}{2}+A)=0.
I need to find an expression for the \sin(\frac{\pi}{2}+A). I cannot see one suitable.
thanks for helping.

 

[LCKurtz]

I notice that only 5 people (6 counting me) have viewed your document. Most people will not go to the trouble of opening a Word document to see a simple figure. Had more people looked, you might have learned sooner that you are trying to work the wrong problem. The question asks for a triangle circumscribed about the circle, not inscribed as you have pictured.

If you think about it the minimum area inscribed such "triangle" would have area 0.

 

[John O' Meara]

Thanks for the reply. I hope the new word attachment has the situtation described correctly. I still have the problem of finding the minimized area of this triangle though. The area of the triangle is T = \frac{a}{2} b \sin(C) I have a,b and sinC to express in terms of the radius of the circle R. or what should my approach be? Maybe I should differentiate implicitly the area, T, first, and then substitute in expressions for a, b, sinC. Finding suitable expressions is the problem for me. Thanks in advance.

 

[John O' Meara]

I do not know why this new attachment didn't post in my last post.

 

[John O' Meara]

T=\frac{a}{2} c \sin(B) \mbox{therefore we have } T = \frac{a}{2} c \frac{x+2R}{c} The c's cancel out, now the problem is to find an expression for a from the figure in terms of x and R only? Or is there a more simple way to do this?

 

[John O' Meara]

Ok, I found another formula for the area of this triangle area =R*s, where R =radius of the circle just touching each of the sides and lies inside the triangle, and s = \frac{a+b+c}{2}, but if you differentiate this w.r.t., R and put result equal to 0, you get s=0 i.e., \frac{a+2b}{2}=0 Then a=-2b clearly wrong. Please help and Thanks very much for the help.