Find minimum perimeter of the triangle circumscribing semicircle The following diagram shows triangle circumscribing a semi circle of unit radius. Find minimum perimeter of triangle
My try:
Letting $$AP=AQ=x$$
By power of a point we have:
$$BP^2=OB^2-1$$ where $O$ is center of the circle.
Also $$CQ^2=OC^2-1$$
Let $$OB=y$$
$$OC=z$$
Then $$BP=\sqrt{y^2-1}$$
$$CQ=\sqrt{z^2-1}$$
So the perimeter of triangle is:
$$P=2x+y+z+\sqrt{y^2-1}+\sqrt{z^2-1}$$
Any help from here?


Let us calculate the perimeter from the figure.
$OB = \dfrac{1}{\sin\alpha}, PB=\dfrac{1}{\tan\alpha},AP= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},AQ= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},QC=\dfrac{1}{\tan\theta}, OC =\dfrac{1}{\sin\theta}  $
So total perimeter is $ p = \dfrac{1}{\sin\alpha}+\dfrac{1}{\tan\alpha}+2\dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})}+\dfrac{1}{\tan\theta}+\dfrac{1}{\sin\theta}$
differentiating p wrt alpha and theta and setting them to zero we get.
$-{\csc}^2\alpha +\sec^2\dfrac{\alpha+\theta}{2}- \csc\alpha\cot\alpha = 0$
$-{\csc}^2\theta +\sec^2\dfrac{\alpha+\theta}{2}- \csc\theta\cot\theta = 0$
from above we get 
${\csc}^2\alpha + \csc\alpha\cot\alpha ={\csc}^2\theta + \csc\theta\cot\theta$
After squaring both side and simplification/cancelling terms, we get
${(\cos\alpha - \cos\theta)}^2=0$ that gives $\alpha= \theta$,   using this back to 
$-{\csc}^2\alpha +\sec^2\dfrac{\alpha+\theta}{2}- \csc\alpha\cot\alpha = 0$
 we get
$\tan^2\alpha = 1 + \cos\alpha$  from which we get
$\cos^3\alpha+2\cos^2\alpha-1=0$ or $\cos^3\alpha+\cos^2\alpha+\cos^2\alpha+\cos^1\alpha-\cos^1\alpha-1=0$
or $(\cos\alpha+1)(\cos^2\alpha+\cos^1\alpha-1) = 0$
the only possible solution from above is $\cos\alpha = \dfrac{(\sqrt 5 -1)}{2}$ or $\alpha=\theta = 51.827$ (approx)
