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?
On
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 $C$.