Assumption on a geometry problem

[dunkelheit]

I was reading the proof of "If [MATH]A,B,C[/MATH] are the angles of a triangle, find the maximum of [MATH]\cos A \cos B \cos C[/MATH]", and there are two assumptions that are "obvious" from an intuitive point of view but I'm not sure that I've proven them rigorously; these assumptions are done so there isn't a loss of generality, so I would like to understand them well so I can improve my skills with proofs.
The first is "we may assume [MATH]A \leq B \leq C[/MATH]": this is intuitive, but I don't know precisely how to prove it.
The first thing that comes to my mind is the fact that [MATH]A+B+C=\pi[/MATH], and I would like to use the fact that in geometry angles are positive for convention (I'm not sure about this, can someone confirm this? It is important because the second part of the question use this, so I hope it is true).
The second is "we may assume [MATH]C\leq \frac{\pi}{2}[/MATH]": my attempt to prove it is to divide in two cases, because in this case an angle could be less or equal of [MATH]\frac{\pi}{2}[/MATH] or between [MATH]\frac{\pi}{2}[/MATH] and [MATH]{\pi}[/MATH]: if [MATH]C \leq \frac{\pi}{2}[/MATH] we have done, the angle I'm searching for is [MATH]C[/MATH] itself.
So let, for example, [MATH]C \geq \frac{\pi}{2}[/MATH]: we have from [MATH]A+B+C=\pi[/MATH] that [MATH]A+B+\frac{\pi}{2} \leq A+B+C=\pi[/MATH] and so it follows that [MATH]A+B\leq \frac{\pi}{2}[/MATH]; and since, for example, [MATH]A\geq 0[/MATH], it follows that [MATH]B\leq A+B\leq\frac{\pi}{2}[/MATH] and so I've shown the existence of an angle less or equal to [MATH]\frac{\pi}{2}[/MATH] (in this case that angle is [MATH]B[/MATH], but it is arbitrary so it is without loss of generality).
Is this correct? Thanks for your time.

 

[Steven G]

If you have three positive numbers, A, B and C that add up to finite sum, like pi (so the numbers are not infinite) you are wondering why the numbers can be put in increasing order? Or are you thinking that that maybe B<C<A instead of A<B<C? if it is the latter, then just relabel the vertices of the triangle so A<B<C. That is what is meant by WOLOG (without a loss of generality). there is nothing to prove here!

Have you ever heard of an obtuse triangle?? Why would you say that C must not be bigger than pi/2???

You say that you were reading the proof. How did that go? Why are you trying to prove it yourself? I'm not saying that is a bad idea.

EDIT: you say that you want to show that C<pi/2.
Then you say that if C<pi/2 you are done---OK
Then you say that if C>pi/2, then B (and A!)<pi/2. This is true. The problem is that you set out to show that C<pi/2, NOT B<pi/2. If you assume, like you did in this case, that C>pi/2, then you did not show that C<pi/2.

 

[dunkelheit]

Thanks for your answer: the proof I'm referring to is the answer of Christian Blatter in this question.
As you can see, he does these assumptions I've mentioned in my question; so the first assumption is just because [MATH]A,B[/MATH] and [MATH]C[/MATH] are arbitrary angles and so I can reorder them in the right order without losing generality?

Have you ever heard of an obtuse triangle?? Why would you say that C must not be bigger than pi/2???

Sure, I know obtuse triangles, I hope that with the link I've inserted of the proof my question will be clearer.

 

[dunkelheit]

EDIT: you say that you want to show that C<pi/2.
Then you say that if C<pi/2 you are done---OK
Then you say that if C>pi/2, then B (and A!)<pi/2. This is true. The problem is that you set out to show that C<pi/2, NOT B<pi/2. If you assume, like you did in this case, that C>pi/2, then you did not show that C<pi/2.

Sorry, maybe there is a typo because I've used the letter [MATH]C[/MATH] in both the argument and in the sentence "we may assume...".
I'll try to explain with other words my doubt. My doubt is the following: he starts the proof by assuming that a chosen angle (he calls it [MATH]\gamma[/MATH]) is less or equal of [MATH]\frac{\pi}{2}[/MATH], and my original doubt is: "Why can he assume that? The proof can't be dependent of this or it is not general".
So I start to think: "it is reasonable that it will not influence the proof (or, better said, WLOG) if I can always assume that; this will make the proof general".
So I need to prove that always in every kind of triangle, I can find an angle that is less or equal of [MATH]\frac{\pi}{2}[/MATH], and this is the same thing as proving that, given a triangle, there exists an angle less or equal of [MATH]\frac{\pi}{2}[/MATH].
And my attempt so for showing this is the following: there are only two possibilities, an angle or it is bigger than or is less or equal of [MATH]\frac{\pi}{2}[/MATH] (and less of [MATH]\pi[/MATH] of course).
In the first case I've done, because the angle I'm searching for is the one that I've supposed to be less or equal than [MATH]\frac{\pi}{2}[/MATH].
If it is bigger than [MATH]\frac{\pi}{2}[/MATH] I've shown with the argument of my first post that exists another angle which is less or equal of [MATH]\frac{\pi}{2}[/MATH], so I can suppose in my proof that the other angle is the one I can suppose less or equal of [MATH]\frac{\pi}{2}[/MATH].
I hope this is clear, thanks for your patience.

 

[Steven G]

Yes, you can relabel the vertices of a triangle.
I'll look at the link now.

 

[Steven G]

I liked this one

If ?=cos?cos?cos?

2?=cos?[2cos?cos?]=cos?{cos(?−?)+cos(?+?)}

As ?+?=?−?, cos(?+?)=−cos?

On rearrangement we have
cos²?−cos?cos(?−?)+2?=0

As C is real, so will be cos⁡C ⟹⟹ the discriminant

cos²(?−?)−8?≥0⟺?≤[cos²(?−?)]/8≤1/8

The equality occurs if cos²(?−?)=1⟺sin²(?−?)=0

⟹?−?=?? where n is any integer

As 0<?,?<?, ?=0⟺?=? and consequently

cos²?−cos?+2⋅1/8=0⟹cos?=1/2⟹?=?/3

⟹?=?=(?+?)/2 =(?−?)/2=?/3=?

 

[Dr.Peterson]

The second is "we may assume [MATH]C\leq \frac{\pi}{2}[/MATH]": ...

It was very helpful that you gave your source, because context is all-important!

This statement's context was:

In the search of p_max we may assume γ≤π/2, hence 0≤cosγ≤1.​

Since we are at this point looking for a maximum value of the product of the cosines, we can restrict the search to cases where that product is positive! That can only happen if all three cosines are positive (since no more than one can be negative), so that the angles are all acute.

In other words, if the triangle were obtuse, then the quantity under study would be negative, and could not attain its maximum.

 

[dunkelheit]

@Jomo: I liked that one too, but I'm actually "suspicious" because he let [MATH]y=\cos A \cos B \cos C[/MATH], so [MATH]y[/MATH] is depending on [MATH]\cos⁡C[/MATH], why this works? I mean, if we're solving for [MATH]\cos⁡C[/MATH] then if [MATH]y[/MATH] is depending on [MATH]\cos⁡C[/MATH] it shouldn't appear in the discriminant. Am I missing something? Thanks.

@Dr.Peterson: Thanks! Now I see very well why that assumption was made, so the left inequality [MATH]0 \leq \cos \gamma[/MATH] comes from the same reason?

So what I say in the post #4 of this topic is meaningless? Sorry but that post appeared later because I still hadn't enough messages and it needed approval.

 

[Dr.Peterson]

@Dr.Peterson: Thanks! Now I see very well why that assumption was made, so the left inequality [MATH]0 \leq \cos \gamma[/MATH] comes from the same reason?

So what I say in the post #4 of this topic is meaningless? Sorry but that post appeared later because I still hadn't enough messages and it needed approval.

Correct on both counts.

 

[dunkelheit]

Thanks! Can I ask you to explain me way it was meaningless?

 

[Dr.Peterson]

Well, that was your conclusion, so I assume you agree.

You said this:

My doubt is the following: he starts the proof by assuming that a chosen angle (he calls it [MATH]\gamma[/MATH]) is less or equal of [MATH]\frac{\pi}{2}[/MATH], and my original doubt is: "Why can he assume that? The proof can't be dependent of this or it is not general".
So I start to think: "it is reasonable that it will not influence the proof (or, better said, WLOG) if I can always assume that; this will make the proof general".
So I need to prove that always in every kind of triangle, I can find an angle that is less or equal of [MATH]\frac{\pi}{2}[/MATH], and this is the same thing as proving that, given a triangle, there exists an angle less or equal of [MATH]\frac{\pi}{2}[/MATH].
And my attempt so for showing this is the following: there are only two possibilities, an angle or it is bigger than or is less or equal of [MATH]\frac{\pi}{2}[/MATH] (and less of [MATH]\pi[/MATH] of course).
In the first case I've done, because the angle I'm searching for is the one that I've supposed to be less or equal than [MATH]\frac{\pi}{2}[/MATH].
If it is bigger than [MATH]\frac{\pi}{2}[/MATH] I've shown with the argument of my first post that exists another angle which is less or equal of [MATH]\frac{\pi}{2}[/MATH], so I can suppose in my proof that the other angle is the one I can suppose less or equal of [MATH]\frac{\pi}{2}[/MATH].
I hope this is clear, thanks for your patience.

The fact that some angle is acute does not imply that every angle (specifically, the largest angle, C) is acute. Existence does not mean universality.

And, of course, what you wrote has nothing to do with the real reason for the assumption, which is based on the assumption that the product is a maximum.

 

[dunkelheit]

You're very helpful, thanks. Indeed I wasn't sure about that reasoning, so you've made me understand why it was wrong.
One last thing: the product [MATH]\cos \alpha \cos \beta \cos \gamma[/MATH] is positive when all the terms are positive but it is positive or when alternating two of them negative and one positive; but you've said this

we can restrict the search to cases where that product is positive! That can only happen if all three cosines are positive (since no more than one can be negative), so that the angles are all acute.

This is because the case with two cosines negative means two of the angles are obtuse angle and this is not possible because then the sum of the three would be greater then [MATH]\pi[/MATH]?

 

[Dr.Peterson]

Correct.

 

[Steven G]

Sorry, maybe there is a typo because I've used the letter [MATH]C[/MATH] in both the argument and in the sentence "we may assume...".
I'll try to explain with other words my doubt. My doubt is the following: he starts the proof by assuming that a chosen angle (he calls it [MATH]\gamma[/MATH]) is less or equal of [MATH]\frac{\pi}{2}[/MATH], and my original doubt is: "Why can he assume that? The proof can't be dependent of this or it is not general".
So I start to think: "it is reasonable that it will not influence the proof (or, better said, WLOG) if I can always assume that; this will make the proof general".
So I need to prove that always in every kind of triangle, I can find an angle that is less or equal of [MATH]\frac{\pi}{2}[/MATH], and this is the same thing as proving that, given a triangle, there exists an angle less or equal of [MATH]\frac{\pi}{2}[/MATH].
And my attempt so for showing this is the following: there are only two possibilities, an angle or it is bigger than or is less or equal of [MATH]\frac{\pi}{2}[/MATH] (and less of [MATH]\pi[/MATH] of course).
In the first case I've done, because the angle I'm searching for is the one that I've supposed to be less or equal than [MATH]\frac{\pi}{2}[/MATH].
If it is bigger than [MATH]\frac{\pi}{2}[/MATH] I've shown with the argument of my first post that exists another angle which is less or equal of [MATH]\frac{\pi}{2}[/MATH], so I can suppose in my proof that the other angle is the one I can suppose less or equal of [MATH]\frac{\pi}{2}[/MATH].
I hope this is clear, thanks for your patience.

Even if a triangle has an angle greater than pi/2, there will two angles in that triangle less than pi/2. I'm sure that you know that, but what you wrote above goes against that.

It would have been nice if you told us which proof you were referring to. Fortunately I know which one you are talking about. What the poster said was that if [MATH]\gamma[/MATH]> pi/2, then cos([MATH]\gamma[/MATH])<0 and the product that you want to maximize would be negative (since if [MATH]\gamma[/MATH]>pi/2 that the other two angles would have be less than pi/2 and the cosines of those two angles would have been positive.) So the poster assumed that [MATH]\gamma[/MATH]<pi/2 and went on with their proof.

 

[Steven G]

@Jomo: I liked that one too, but I'm actually "suspicious" because he let [MATH]y=\cos A \cos B \cos C[/MATH], so [MATH]y[/MATH] is depending on [MATH]\cos⁡C[/MATH], why this works? I mean, if we're solving for [MATH]\cos⁡C[/MATH] then if [MATH]y[/MATH] is depending on [MATH]\cos⁡C[/MATH] it shouldn't appear in the discriminant. Am I missing something? Thanks.

I don't see a problem there but will think about what you are saying and will get back to you.

My immediate thought would be if y = x^2 -4x + C, then the discriminate would be 16-4C. Nobody would question that result. What if you later found out that C was a function of y, would you change your mind about the discriminant? As I said I have to think about this later.