Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures $15$ degrees, and sides AB and AC both have length $1$.
Amare starts at point B and wants to ultimately arrive on side AC. However, the queen of his colony has asked him to make several stops along the way. Specifically, his path must:
$\cdot$ Start at point B.
$\cdot$ Second, touch a point — any point — on side AC.
$\cdot$ Third, touch a point — any point — back on side AB.
$\cdot$ Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).
What is the shortest distance Amare can travel to complete the queen’s desired path?
Amare the evidently male ant knows that that his colony has to travel along the isoceles triagle ABC with smallest angle $15$ degrees, but he wants to solve the general problem knowing that almost all queens have the same obscure rules and if he cannot mate with his queen perhaps he can abscond to some other colony to mate with that queen dazzling her with an optimal route, or something, I'm not an myrmecologist. So instead, he will solve the case where the smallest angle is some $\theta \in [0, \frac{\pi}{3}).$
Amare decides to set up an optimization problem by labeling setting the point A as the origin and B at the point $(1,0),$ thus surmising that $C$ is the point $(\cos \theta, \sin \theta).$ He decides to model his first point on the AC side of the triangle by the variable $(\alpha \cos \theta, \alpha \sin \theta)$ then the point he visits along the AB side as $(\beta, 0)$ and finally returning to the AC side at $(\gamma \cos \theta, \gamma \sin \theta)$ for some $0 \leq \alpha, \beta, \gamma \leq 1.$ Let $d_1(\alpha) = \sqrt{\alpha^2 - 2 \cos \theta \alpha + 1}$ be the length of Amare's first transit from AB to AC, let $d_2(\alpha, \beta) = \sqrt{\alpha^2 - 2 \alpha \beta \cos \theta + \beta^2}$ be the length of his passage from AC back to AB and $d_3(\beta, \gamma) = \sqrt{\beta^2 - 2 \beta \gamma \cos \theta + \gamma^2}$ the length of the final leg of his journey. Then Amare's total distance traveled is \begin{align*} d(\alpha, \beta, \gamma) &= d_1(\alpha) + d_2(\alpha, \beta) + d_3(\beta, \gamma) \\ &=\sqrt{\alpha^2 - 2 \cos \theta \alpha + 1} + \sqrt{ \alpha^2 - 2 \alpha \beta \cos \theta + \beta^2} + \sqrt{ \beta^2 - 2 \beta \gamma \cos \theta + \gamma^2}.\end{align*}
Taking the gradient, Amare gets that $$\nabla d = \begin{pmatrix} \frac{\alpha - \cos \theta}{d_1(\alpha)} + \frac{\alpha - \beta \cos \theta}{d_2(\alpha, \beta)} \\ \frac{\beta - \alpha \cos \theta}{d_2(\alpha, \beta)} + \frac{\beta - \gamma \cos \theta}{d_3(\beta, \gamma)} \\ \frac{\gamma - \beta \cos \theta }{d_3(\beta, \gamma)} \end{pmatrix}.$$ For the time being, Amare just wants to find when $\nabla d = 0,$ so he first notices that if this is the case then $\gamma = \beta \cos \theta,$ which in turn means that $$d_3(\beta, \gamma) = \sqrt{ \beta^2 - 2 \beta \gamma \cos \theta + \gamma^2} = \beta \sin \theta.$$
Therefore, if $\nabla d = 0,$ then $\gamma = \beta \cos \theta$ and plugging back into $\frac{\partial}{\partial \beta} d=0,$ we also have the equation $$\frac{\beta - \alpha \cos \theta}{d_2(\alpha, \beta)} + \frac{\beta - \gamma \cos \theta}{d_3(\beta, \gamma)} = \frac{\beta - \alpha \cos \theta}{d_2(\alpha, \beta)} + \sin \theta = 0,$$ or equivalently, $$\beta = -d_2 \sin \theta + \alpha \cos \theta.$$ Plugging this value of $\beta,$ back into $d_2(\alpha, \beta)^2$, we get \begin{align*} d_2^2 &= \alpha^2 - 2 \alpha (-d_2\sin \theta + \alpha \cos \theta) \cos \theta + (- d_2 \sin \theta + \alpha \cos \theta)^2 \\ &= \alpha^2 - 2 \alpha^2 \cos^2 \theta + 2 \alpha d_2 \sin \theta \cos \theta + d_2^2 \sin^2 \theta - 2 \alpha d_2 \sin \theta \cos \theta + \alpha^2 \cos^2 \theta \\ &= (1- \cos^2 \theta) \alpha^2 + d_2^2 \sin^2 \theta\end{align*} or equivalently $$d_2(\alpha, \beta) = \alpha \tan \theta.$$ Plugging this value back into the formula for $\beta,$ we get $$\beta = -d_2 \sin \theta + \alpha \cos \theta = \alpha \bigl( \cos \theta - \sin \theta \tan \theta\bigr) = \alpha \frac{\cos 2\theta}{\cos \theta}.$$
So finally, plugging back into $\frac{\partial}{\partial \alpha} d = 0$ we get \begin{align*}0 &= \frac{\alpha - \cos \theta}{d_1} + \frac{\alpha - \beta \cos \theta}{d_2}\\ &= \frac{\alpha - \cos \theta}{d_1} + \frac{ \alpha - \alpha \frac{\cos 2 \theta}{\cos \theta} \cos \theta }{ \alpha \tan \theta}\\ &= \frac{\alpha - \cos \theta}{d_1} + \frac{1 - \cos 2\theta}{\tan \theta}\\ &= \frac{\alpha - \cos \theta}{d_1} + \sin 2\theta\end{align*} or equivalently $$d_1 = \frac{\cos \theta - \alpha}{ \sin 2\theta}.$$ Plugging this back into $d_1^2$ we get $$\alpha^2 - 2 \alpha \cos \theta + 1 = \frac{ \cos^2 \theta - 2 \alpha \cos \theta + \alpha^2 }{ \sin^2 2\theta }$$ or equivalently $$\cos^2 2\theta \alpha^2 - 2 \cos \theta \cos^2 2\theta \alpha + \cos^2 \theta - \sin^2 2\theta = 0.$$ Solving the quadratic equation we get \begin{align*}\alpha &= \frac{2 \cos \theta \cos^2 2\theta \pm \sqrt{ 4 \cos^2 \theta \cos^4 2\theta -4 (\cos^2 2\theta) (\cos^2 \theta - \sin^2 2\theta)}}{2 \cos^2 2\theta}\\ &= \cos \theta \pm \frac{ 2 \cos 2\theta \sqrt{\cos^2 \theta \cos^2 2\theta - \cos^2 \theta + \sin^2 2\theta}}{2\cos^2 2\theta} \\ &= \cos \theta \pm \frac{\sqrt{\cos^2 \theta ( \cos^2 2\theta - 1) + \sin^2 2\theta}}{\cos 2\theta} = \cos \theta \pm \frac{ \sqrt{ \sin^2 2\theta (1 - \cos^2 \theta) }}{\cos 2\theta} \\ &= \cos \theta \pm \frac{ \sin 2\theta \sin \theta }{\cos 2\theta} = \cos \theta \pm \sin \theta \tan 2\theta = \frac{\cos \theta \cos 2\theta \pm \sin \theta \sin 2\theta}{\cos 2\theta}\end{align*} that is either $$\alpha = \frac{\cos \theta}{\cos 2\theta} \,\,\, \text{ or } \,\,\, \alpha = \frac{\cos 3\theta}{\cos 2\theta}.$$ Since $\frac{\cos \theta}{\cos 2\theta} \not \in [0,1]$ for any $\theta \in [0,\frac{\pi}{3}),$ Amare will go with $$\alpha = \frac{\cos 3\theta}{\cos 2\theta}$$ whenever $\theta \in [0, \frac{\pi}{6})$ and in this case, $$d_1 = \frac{\cos \theta - \frac{\cos 3\theta}{\cos 2\theta}}{\sin 2\theta} = \frac{\sin \theta}{\cos 2\theta}.$$ Note that when $\theta \in [\frac{\pi}{6}, \frac{\pi}{3})$ we have $\alpha = 0$ and $d_1 = 1.$
So if $\theta \in [0, \frac{\pi}{6}),$ then \begin{align*}\hat{\alpha} &= \frac{\cos 3\theta}{\cos 2\theta}, \\\hat{\beta} &= \hat{\alpha} \frac{\cos 2\theta}{\cos \theta} = \frac{\cos 3\theta}{\cos \theta} \\ \hat{\gamma} &= \hat{\beta} \cos \theta = \cos 3\theta\end{align*} and the shortest distance is \begin{align*}\hat{d} &= d(\hat{\alpha}, \hat{\beta}, \hat{\gamma}) = d_1 + d_2 + d_3 = \\ &= \frac{\sin \theta}{\cos 2\theta} + \frac{\cos 3\theta}{\cos 2\theta} \tan \theta + \frac{\cos 3\theta}{\cos \theta} \sin \theta\\ &= \sin \theta \left( \frac{\cos \theta + \cos 3\theta + \cos 2\theta \cos 3\theta}{\cos \theta \cos 2\theta} \right) \\ &= \sin \theta \left( \frac{ \cos \theta + 2 \cos^2 \theta \cos 3\theta }{\cos \theta \cos 2\theta } \right) \\ &= \sin \theta \left( \frac{1 + 2 \cos \theta \cos 3\theta}{\cos 2 \theta} \right) \\ &= \sin \theta \left( \frac{1 + \cos 4\theta + \cos 2\theta}{\cos 2\theta} \right) \\ &= \sin \theta \left( \frac{2 \cos^2 2\theta + \cos 2\theta }{\cos 2\theta} \right)\\ &= \sin \theta ( 2 \cos 2\theta + 1 ) = \sin \theta (2 \cos^2 \theta + \cos 2\theta) \\ &= \sin 2\theta \cos \theta + \sin \theta \cos 2\theta = \sin 3\theta. \end{align*} In particular, when $\theta = \frac{\pi}{12},$ we get that Amare's shortest path for his home colony's queen is $\hat{d} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}.$ Whenever $\theta \in [\frac{\pi}{6}, \frac{\pi}{3})$ then $\hat{\alpha}=\hat{\beta}=\hat{\gamma} = 0$ and $\hat{d} = 1.$
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