# 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)

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 $$C$$.