17

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.

mathlove
  • 139,939

5 Answers5

4

${\bf 1\ }$ 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 $$\inf\nolimits_{a\in Q_{\rm OP}}\Phi(a)=1-2\sqrt{2}\ .$$

Christian Blatter
  • 226,825
  • I understand that the function is not min at $A=B=C=D=E$. I have one question. I think this function doesn't have the minimum because $0^\circ$ is not allowed. Is this right? – mathlove Jul 19 '13 at 14:45
  • @mathlove: Yes. But my argument was necessary to arrive at this conclusion – if you really want to exclude $0$ and $\pi$ as allowed angles. – Christian Blatter Jul 19 '13 at 16:10
  • Compare to the other answer where the same principle about sines is articulated (reducing the problem to checking a finite set of possibilities), the same quintuples of potential optima are checked, and the same result is reached about the set of values of the sum of cosines. Both answers can be simplified by the observation that two different angles with a sum of $180$ degrees cannot appear in a true local extremum. – zyx Jul 19 '13 at 19:33
3

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.

Al S
  • 139
  • I just realized after posting that I was assuming the poster was familiar with multi-variable calculus. OP, if you're unfamiliar with partial derivatives I will reword my answer. In short, you will wend up with a set of equations which will lead you to what dtdarek commented above. – Al S Jul 18 '13 at 15:34
  • I understood, but I need another help. Can we get $A=B=C=D=E=108^\circ$ because $sinA=sinB=sinC=sinD=sinE$? – mathlove Jul 18 '13 at 15:52
  • Just because the $\sin$ of each variable is equal does not mean the variables themselves are equal (remmber $\sin x=\sin (\pi-x)$). Though you have the extra constraints that each angle is between $0^\circ$ and $180^\circ$. Because of those constraints you get lucky:) – Al S Jul 18 '13 at 16:13
  • Sami makes a good argument (using an epsilon proof) on why all angles being equal correctly minimized the function. – Al S Jul 18 '13 at 16:17
2

We can argue using the notion of compacity and continuity of the $\cos$ function that the minimum exists and attained.

Suppose that the minimum is given by the angles $A,B,C,D$ and $E$. Now WLG suppose that $A>B$ and take from $B$ an arbitrary small (angle) $\epsilon$ and add it to $A$, we have $$\cos(A+\epsilon)+\cos(B-\epsilon)=\cos(A)\cos(\epsilon)+\cos(B)\cos(\epsilon)+\sin(B)\sin(\epsilon)-\sin(A)\sin(\epsilon)$$ and using the approximation $\cos(\epsilon)\approx 1$ and $\sin(\epsilon)\approx \epsilon$ we have $$\cos(A+\epsilon)+\cos(B-\epsilon)\approx \cos(A)+\cos(B)+\underbrace{\epsilon(\sin(B)-\sin(A))}_{<0}$$ hence we find a contradiction since the minimum is attained at $A,B,C,D$ and $E$.

Remarks

  • The case $A>B$ and $\sin(A)\leq \sin(B)$ is not a problem in the reasoning because in this case we take $\epsilon$ from $A$ and add it to $B$.
  • This reasoning is clearly valid if $ B \neq 0$ or $ A \neq180 $ and these few cases should be treated separately.
  • 2
    Your approximation has an error which could be larger than $\epsilon(\sin(B)-\sin(A))$, which makes the solution wrong. Note that if $A=90^o+\alpha$ and $B=90^o-\alpha$, then $\cos(A+\epsilon)+\cos(B-\epsilon)\approx \cos(A)+\cos(B)$ yet you prove that $\cos(A+\epsilon)+\cos(B-\epsilon) < \cos(A)+\cos(B)$. – N. S. Jul 18 '13 at 17:02
  • @N.S. I don't understand why there's an error in my approximation, can you explain more please? –  Jul 18 '13 at 17:09
  • 1
    Because $\cos(\epsilon)\neq 1$ and $\sin(\epsilon) \neq \epsilon$, those are approximations. And any approximation has an error (NOT mistake, error like difference between actual value and approximation).... Also, as it was pointed out, $A=B=180^o, C=90^o, D=E=45^o$ leads to a negative value, so $0$ cannot be the minimum.... – N. S. Jul 18 '13 at 17:13
  • 2
    Just to be a little clearer, $0.000000000000001 \sim -0.0000000000001$ is a good approximation but it cannot be used to conclude that $0.000000000000001<0$. You do something similar, you approximate the LHS by $\cos(A)+\cos(B)+$ something very small, but because you approximate, there is an error in the approximation, it is not equality. What if the actual approximation error is larger than "something very small".... – N. S. Jul 18 '13 at 17:14
  • @N.S. First I never said that the minimum is $0$ second I suppose that $A\neq B$ so you counterexample $A=90$ and $B=90$ isn't valid and third I can use approximation since the neglected term in the approximation $\cos\epsilon\approx 1$ is at order of $\epsilon^2=o(\epsilon)$ recall that I can choose $\epsilon$ as small as I want. –  Jul 18 '13 at 17:18
  • True, but you have to make sure that the error in the approximation can be made smaller than the actual difference in the inequalities. Using approximations to prove tight inequalities is not a good idea. I think the solution can be fixed to work, you just have to pick the right epsilon. – N. S. Jul 18 '13 at 17:38
  • Wait: $A=135^o, B=0^0, C=D=E=135^o$ leads to a lower value than you minimum. – N. S. Jul 18 '13 at 17:56
  • One last comment, in the first part of the proof you are basically trying to argue that for $A+B$ fixed the minimum of $\cos(A)+\cos(B)$ is obtained when $A=B$. But that is just the Jensen inequality for convex functions, and $\cos$ is convex ONLY on $[90^o, 180^o]$. – N. S. Jul 18 '13 at 18:40
  • @N.S. I think that the fault in my reasoning that I didn't treat the case $B=0$ since in this case I deal with $0-\epsilon$ which is not allowed. –  Jul 18 '13 at 19:10
  • @SamiBenRomdhane I see, I couldn't figure the mistake but my intuition told me something was wrong. But I think it is something more subtle which is hidden in the approximation and probably the signs of the trig functions, as the case $b=0.000001^o$ still doesn't work.. BTW: I didn't downvote. – N. S. Jul 18 '13 at 19:29
  • Sorry to have confused you. In this problem, I meant that each angle can be neither $0^\circ$ nor $180^\circ$. I've already edited. – mathlove Jul 19 '13 at 06:52
  • 2
    @N. S.: This answer is wrong. The minimum is $=1-4\sqrt{2}$. See my answer below. – Christian Blatter Jul 19 '13 at 09:36
2

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.

N. S.
  • 132,525
0

What the differentiation-based arguments in other answers are really proving is that

extrema (on the allowed domain plus its boundary) can occur only at points where a set of $k$ of the sines of the angles are equal, and the other set of $(5 - k)$ sines are zero (from points on the boundary), for some value of $k$ between $0$ and $5$.

Five equal sines ($k = 5$) happen at permutations of

$f(108,108,108,108,108)= 5 \cos 108 = -1.545... 0$ (from Ron Gordon's comment)

$f(120,120,120,120,60)= 4 \cos 120 + \cos 60 = 3 \cos 120 = -3/2 $

$f(180,180,180,0,0)=-1$

Four equal sines ($k=4$)

$f(135,135,135,135,0) = -2\sqrt{2}+1 = -1.8...$ (from N.S.' comment)

$f(90,90,90,90,180) = -1$

$f(x,x,180-x,180-x,180) = -1$

I stopped here, but it does not look like $k \leq 3$ can do better than $-\frac{3}{2}$. If that is true, then the result is

any value slightly higher than $a = -2\sqrt{2}+1$ occurs at $(4u,135-u,135-u,135-u,135-u)$ for a small positive $u$, and no value $\leq a$ is attained.

zyx
  • 35,436