1. ## Minimize perimeter

An isosceles triangle is circumscribed about a circle of radius $R$. Express the perimeter of the triangle as a function of its altitude, and show that the perimeter is a minimum when this altitude is $3R$.

This should not be that hard, but I cannot for the life of me figure out the equation in terms of the altitude and radius ONLY. I have too many variables. Thanks for you help.

2. This was kind of tricky. I made a diagram.

By the included diagram: R=OB=OC; y=AD, h=AB

AB is the height of the triangle, y is the length of the side.

Look at triangle ABC. $h^{2}+x^{2}=y^{2}$

Look at triangle AOC. $(h-R)^{2}=R^{2}+(y-x)^{2}$

Using these, we can solve for x. $x=\frac{Rh}{\sqrt{h^{2}-2hR}}$

The area of a triangle is 1/2xh.

$\frac{1}{2}\frac{Rh^{2}}{\sqrt{h^{2}-2hR}}$

There, R is a constant and it is in terms of h. Now, differentiate:

$\frac{dA}{dh}=\frac{Rh(h-3R)}{(h-2R)\sqrt{h(h-2R)}}$

Set to 0 and solve for h.

$Rh(h-3R)=0$

$h=3R$

3. 3 things:

you assumed cd=db.

The problem wants perimeter minimized.

I don't think that solution is correct

### minimize the perimeter of a tr

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