If $A+B+C+D+E = 540^\circ$ what is $\min (\cos A+\cos B+\cos C+\cos D+\cos E)$?

Let each of $A, B, C, D, E$ be an angle that is less than $180^\circ$ and is greater than $0^\circ$. Note that each angle can be neither $0^\circ$ nor $180^\circ$.

If $A+B+C+D+E = 540^\circ,$ what is the minimum of the following function? $$\cos A+\cos B+\cos C+\cos D+\cos E$$

I suspect the minimum is achieved when $A=B=C=D=E$, but I can't prove it. I need your help.

Solution 1:

${\bf 1\ }$ For the moment a configuration $a=(\alpha_1,\alpha_2,\ldots,,\alpha_5)$ of angles $\alpha_i$ is admissible when $$0\leq\alpha_1\leq\alpha_2\leq\alpha_3\leq\alpha_4\leq\alpha_5\leq\pi,\qquad \sum_{i=1}^5\alpha_i=3\pi\ .$$ Let $Q$ be the compact set of admissible configurations $a$, and put $$\Phi(a):=\sum_{i=1}^5\cos\alpha_i\quad(a\in Q)\ ,\qquad \mu:=\min_{a\in Q}\Phi(a)\ .$$ When $\Phi(a)=\mu$ then $a$ is called an optimal configuration.

${\bf 2\ }$ Let $a\in Q$ be an optimal configuration. Then $$0\leq\alpha_1\leq\ldots\leq\alpha_r<{\pi\over2}\leq \alpha_{r+1}\leq\ldots\leq\alpha_5\leq\pi$$ for some $r\geq0$. Since the function $\cos$ is properly convex in the interval $\bigl[{\pi\over2},\pi\bigr]$ it follows from Jensen's inequality that $$\alpha_{r+1}=\ldots=\alpha_5=\alpha\tag{1}$$ for a certain $\alpha\in\bigl[{\pi\over2},\pi\bigr]$. We then have $$3\pi=\alpha_1+\ldots+\alpha_r+(5-r)\alpha<r{\pi\over2}+(5-r)\alpha=(5-r)\left(\alpha-{\pi\over2}\right)+5{\pi\over2}\ ,$$ or $$(5-r)\left(\alpha-{\pi\over2}\right)>{\pi\over2}\ .$$ It follows that $r\leq3$ (whence $r+2\leq5$) and that $\alpha>{\pi\over2}$.

${\bf 3\ }$ Let $a\in Q$ still be optimal and assume that $0<\alpha_i\leq\alpha_j<\pi$ for two entries $\alpha_i$, $\alpha_j$. Then one would have $${d\over d\epsilon}\bigl(\cos(\alpha_i+\epsilon)+\cos(\alpha_j-\epsilon)\bigr)\biggr|_{\epsilon=0}=0\ ,$$ or $\sin\alpha_i=\sin\alpha_j\>$. This implies that all $\alpha_i\notin\{0,\pi\}$ have the same sine.

${\bf 4\ }$ We first consider the case ${\pi\over2}<\alpha<\pi$. I claim that $\sin\alpha_r=\sin\alpha$ is impossible. Proof: We then would have $\cos\alpha_r=-\cos\alpha$ and could replace $\alpha_r$, $\alpha_{r+1}$ by $$\alpha_r'=\alpha_{r+1}':={\pi\over2}\quad <\alpha=\alpha_{r+2}$$ and still have $\Phi(a')=\mu$. This would contradict property $(1)$ of optimal configurations.

It follows that $\alpha_1=\ldots=\alpha_r=0$. This leaves us with the following configurations: $$\eqalign{&r=0:\quad a=({3\pi\over5},{3\pi\over5},{3\pi\over5},{3\pi\over5},{3\pi\over5}), \quad \Phi(a)=5\cos{3\pi\over5}\doteq-1.54508,\cr &r=1:\quad a=(0,{3\pi\over4},{3\pi\over4},{3\pi\over4},{3\pi\over4}), \quad \Phi(a)=1+4\cos{3\pi\over4}\doteq-1.82843.\cr}$$ It is easily checked that $r=2$ and $r=3$ are impossible in this case.

${\bf 5\ }$ When $\alpha=\pi$ then necessarily $r=2$ or $r=3$. The case $r=2$ enforces the configuration $$a=(0,0,\pi,\pi,\pi),\quad \Phi(a)=-1\ ,$$ and the case $r=3$ together with $\alpha_r<{\pi\over2}$ enforces the configuration $$a=({\pi\over3},{\pi\over3},{\pi\over3},\pi,\pi),\quad \Phi(a)=-{1\over2}\ .$$ ${\bf 6\ }$ Comparing the obtained data we conclude that $$\mu=\Phi\bigl(0,{3\pi\over4},{3\pi\over4},{3\pi\over4},{3\pi\over4}\bigr)=1-2\sqrt{2}\doteq-1.82843\ .$$ In the formulation of the problem by the OP angles $\alpha_i\in\{0,\pi\}$ are forbidden. It follows from our analysis that the function $\Phi$ assumes no minimum on $Q_{\rm OP}$. All we can say is that $$\inf\nolimits_{a\in Q_{\rm OP}}\Phi(a)=1-2\sqrt{2}\ .$$

Solution 2:

What you're looking at is a constrained optimization problem.

Put another way, you are being asked to minimize $\cos A+\cos B+\cos C+\cos D+\cos E$ given the constraint $A+B+C+D+E=540$.

Using lagrange multipliers, we can rewrite this as the minimization of the function $L(A,B,C,D,E,\lambda)=\cos A+\cos B+\cos C+\cos D+\cos E+\lambda(540-A-B-C-D-E)$

To minimize the function, you want to set each partial derivative of L ($\frac{\partial L}{\partial A},\frac{\partial L}{\partial B},...,\frac{\partial L}{\partial \lambda})$ to zero. This should get you the value for $\lambda$ as well as values for each angle.

With that I think you can take the last step.

Solution 3:

Ugly Partial Solution

Note first that by Jensen Inequality, if $x_1,..,x_k$ are angles in $[90^o, 180^o]$ then

$$\cos(\frac{x_1+...+x_k}{k}) \leq \frac{\cos(x_1)+..+\cos(x_k)}{k}$$

This proves that the absolute minimum (which exists as Sami proved) is attained at a point where all the obtuse angles are equal.

We split now the problem in few cases:

Case 1: All $5$ angles are $\geq 90^o$. Then by Jensen

$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq 5 \cos(108^o) \sim -1.545 \,.$$

Case 2: Exactly $4$ angles are $\geq 90^o$. Without loss of generality $A =x <90^o$. Then by Jensen

$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq \cos(A)+ 4 \cos(\frac{B+C+D+E}{4}) $$ $$= \cos(x)+4 \cos(\frac{540^o-x}{4})\,.$$

Now, the only Critical number of $f(x)=\cos(x)+4 \cos(\frac{540^o-x}{4})$ on $[0^o,90^o]$ is $60^o$.

So the absolute minimum of $\cos(x)+4 \cos(\frac{540^o-x}{4})$ is one of $\cos(60^o)+4 \cos(120^o) \,;\, \cos(90^o)+4 \cos(112.5^o) \,;\, \cos(0^o)+4 \cos(135^o)$.

Then

$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq 1+4 \cos(135^o) \sim -1.828$$

It is easy to show (using the Jensen for the acute angles) that $3$ or more acute angles lead to higher minimum, so the only case left to study is :

Case 3: Exactly $3$ angles are $\geq 90^o$. Without loss of generality $A =x <90^o$ and $B =y <90^0. Then by Jensen

$$\cos(A)+\cos(B)+\cos(C)+\cos(D)+\cos(E) \geq \cos(A)+ \cos(B)+ 3 \cos(\frac{C+D+E}{4}) $$ $$= \cos(x)+\cos(y)+3 \cos(\frac{540^o-x-y}{3})\,.$$

The minimum of this function on $[0^o,90^o] \times [0^o,90^o]$ can be calculated with multivariable calculus. Yikes.