Finding max perimeter of triangle of three circulating points

Question

I'm thinking a plane geometry problem, and it seems quite puzzling.

Here it is.

Question:

Consider three concentric circles with radius 3, 5 and 7 each. and construct a triangle by picking one point from each circle.

Question is, in which condition that triangle has maximum perimeter?

Maybe we can try to use law of cosine and calculus? ... but I want more geometric argument.
How can we find the condition by using some geometry?

Answer 1

Summary

1. the condition is: $O$ is the in-center for $\triangle ABC$.
2. the maximal perimeter $p \approx 26.47085983465459$

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• Let $a = 3$, $b = 5$ and $c = 7$.
• For any $r > 0$, let $\mathcal{C}_r$ be the circle centered at $O$ with radius $r$.
• Let $A, B, C$ be three points lie on the circle $\mathcal{C}_a$,$\mathcal{C}_b$ and $\mathcal{C}_c$ respectively.
• Furthermore, assume $A, B, C$ are positioned such that the perimeter $p \stackrel{def}{=} AB + BC + CA$ of $\triangle ABC$ , is maximized.

Let $\mathcal{E}$ be the locus of point $P$ such that $PB + PC = AB + AC$.

$\mathcal{E}$ an ellipse with foci $B$, $C$ intersecting $\mathcal{C}_a$ at $A$. $\mathcal{E}$ cannot intersect $\mathcal{C}_a$ transversely. Otherwise, we can a find a point $P$ on $\mathcal{C}_a$ near $A$ outside $\mathcal{E}$. This contradict with the condition that the perimeter of $\triangle ABC$ is maximized. As a result, $\mathcal{E}$ and $\mathcal{C}_a$ are tangent to each other at $A$.

As an ellipse, the normal line of $\mathcal{E}$ at $A$ is a angular bisector of the $\angle CAB$. Since $\mathcal{E}$ and $\mathcal{C}_a$ share the same normal line at $A$ and $O$ lies on this normal line. We find $O$ lies on the angular bisector for $\angle CAB$.

Apply a similar argument to $B$, $C$, we find $O$ lies on the three angular bisectors of $\triangle ABC$ and hence is the in-center.

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• Let $r$ be the in-radius for $\triangle ABC$.
• Let $\alpha = \frac12 \angle CAB$, $\beta = \frac12 \angle ABC$ and $\gamma = \frac12 \angle BCA$.

By elementary geometry, we know $a\sin\alpha = b\sin\beta = c\sin\gamma = r$ and $$ p = 2(a\cos\alpha+b\cos\beta+c\cos\gamma) = 2\left(\sqrt{a^2-r^2}+\sqrt{b^2-r^2}+\sqrt{c^2-r^2}\right) $$ Since $\alpha + \beta + \gamma = \frac{\pi}{2}$, it is easy to show they satisfy following trigonometric identity.

$$1 - \sin\alpha^2 - \sin\beta^2 - \sin\gamma^2 - 2\sin\alpha\sin\beta\sin\gamma = 0$$ This implies $r$ is a positive root of the cubic equation:

$$1 - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)r^2 - \frac{2}{abc} r^3 = 0 $$

Define $\rho$ and $t$ such that $$\frac{1}{\rho^2} = \frac13 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)\quad\text{ and }\quad r = \frac{\rho}{2t}$$

We can simplify above to $$4t^3 - 3t = \frac{\rho^3}{abc}$$ Compare this with the triple angle formula for cosine: $\cos(3\theta) = 4\cos\theta^3 - 3\cos\theta$, we get

$$r = \frac{\rho}{2\cos\phi}\quad\text{ where }\quad \phi = \frac13\left[ \cos^{-1}\left(\frac{\rho^3}{abc}\right) + 2\pi N\right] \quad\text{ for some }\quad N \in \{0,1,2\} $$ For the case at hand, $(a,b,c) = (3,5,7)$ and $\rho = 105\sqrt{\frac{3}{1891}}$. One can check that only $N = 0$ gives us a positive $r$. As a result, the inradius $r$ for the optimal configuration is $$r = \frac{ 105\sqrt{\frac{3}{1891}} }{ 2\cos\left\{ \frac13\left[ \cos^{-1}\left(105^2\left(\frac{3}{1891}\right)^{3/2}\right) \right] \right\}} \approx 2.167727449130353 $$ and the maximal perimeter equals to $$p = 2\left(\sqrt{3^2-r^2} + \sqrt{5^2-r^2}+\sqrt{7^2-r^2}\right) \approx 26.47085983465459 $$