<s>[INST] 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?
 [/INST] The problem with that is that there are only really two degrees of freedom; the variables $x,y,z$ are related in some way. We could try to find what that relation is - or we could find another way to express things, one with easier relations to spot.
I'd go with the angles of the triangle. Labeling $O$ as you did, $OP$ is perpendicular to $AB$ and $OQ$ is perpendicular to $AC$. Therefore, from right triangle $OBP$, $OB=\csc B$ and $BP=\cot B$. Similarly, $OC=\csc C$ and $CQ=\cot C$. At vertex $A$, note that triangles $AOP$ and $AOQ$ are congruent - so $AO$ bisects angle $A$, and $AP=AQ=\cot\frac{A}{2}$.
We now seek to minimize the perimeter $p$ subject to the constraints $A+B+C=\pi$ and $A,B,C > 0$. There's one more thing to note: $\csc \theta + \cot \theta = \cot\frac{\theta}{2}$. Applying this identity,
$$p(A,B,C) = \csc B+\cot B +\csc C+\cot C+2\cot\frac A2 = 2\cot \frac A2+ \cot \frac B2+\cot \frac C2$$
Now, we solve this constrained optimization problem. First, the boundary constraints $A,B,C\ge 0$. Since each of the half-angles $\frac A2, \frac B2, \frac C2$ is in $(0,\pi)$, all of those cotangents are positive. As any of these angles goes to zero, its cotangent goes to $\infty$ and so too does the perimeter. No minimum there.
In the interior triangular region, we apply Lagrange multipliers. Differentiating,
$$\nabla p(A,B,C) = -\frac12\left(2\csc^2\frac A2, \csc^2 \frac B2, \csc^2 \frac C2\right)$$
must be a constant multiple of $(1,1,1)$, the gradient of the constraint. Since $\csc^2$ decreases monotonically on $(0,\frac{\pi}{2})$, we must have $B=C < A$. If the angles summed to $2\pi$, we would have a nice solution $\frac A2=\frac{\pi}{2}, \frac B2=\frac C2 = \frac{\pi}{4}$ - but no, we're not that lucky. Instead, we apply $\frac A2 = \frac{\pi}{2}-B$ to eliminate $A$:
$$1-\cos B=2\sin^2 \frac B2 = \sin^2 \frac A2 = \cos^2 B$$
so $\cos^2 B + \cos B - 1 = 0$ and $\sin \frac A2 = \cos B = \dfrac{\sqrt{5}-1}{2}$. Then
\begin{align*}\cos\frac A2 &= \sqrt{1-\left(\frac{\sqrt{5}-1}{2}\right)^2} = \sqrt{\frac{\sqrt{5}-1}{2}}\\
\sin\frac B2 &= \sqrt{\frac{3-\sqrt{5}}{4}} = \frac1{\sqrt{2}}\cdot \frac{\sqrt{5}-1}{2}\\
\cos\frac B2 &= \sqrt{\frac{1+\sqrt{5}}{4}} = \frac1{\sqrt{2}}\cdot\sqrt{\frac{\sqrt{5}+1}{2}}\\
\cot\frac A2 &= \sqrt{\frac{2}{\sqrt{5}-1}} = \sqrt{\frac{\sqrt{5}+1}{2}}\\
\cot\frac B2 &= \sqrt{\frac{\sqrt{5}+1}{3-\sqrt{5}}} = \sqrt{\sqrt{5}+2}\\
p &= 2\cot\frac A2 + 2\cot\frac B2 = 2\sqrt{\frac{\sqrt{5}+1}{2}}+2\sqrt{\sqrt{5}+2}\approx 6.66\end{align*}
A truly fiendish problem.
Numerically, the minimizing angles are about $76^\circ$ at $A$ and $52^\circ$ at $B$ and $C$.
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