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let $a,b,c$ such that $$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$

find the value $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$

is true?

Yes, I tink this problem can prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$

so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$ How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?

and I see this easy problem http://zhidao.baidu.com/question/260913315.html

math110
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3 Answers3

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First, we simplify the initial equation by multiplying out by the denominator. Let

$$f(a,b,c) = c^2(a^2+b^2-c^2)^2 + a^2(b^2+c^2-a^2)^2 + b^2(c^2+a^2-b^2)^2 - 12a^2b^2c^2$$

In Ron Gordon's deleted post, he realized that $f(a,b,c) = 0 $ if $a+b=c$, $b+c=a$, $c+a=b$. This strongly suggests that $a+b-c, b+c-a, c+a-b$ are factors (sort of Remainder Factor Theorem). And indeed they are. We have

$$f(a,b,c) = -(a+b-c)(a-b+c)(-a+b+c)(a+b+c)(a^2+b^2+c^2)$$

which you can check in Wolfram.

Since the denominators are non-zero, we have $(a^2+b^2+c^2)>0$. Thus $$f(a,b,c) = 0 \Leftrightarrow (a+b-c)(a-b+c)(-a+b+c)(a+b+c) = 0 $$

We now split into cases.

Case 1. $(a+b-c)(a-b+c)(-a+b+c) = 0$

(Once again, multiply by denominators, and using Ron's observation.) Defining

$$g(a,b,c) = c(a^2+b^2-c^2) + a(b^2+c^2-a^2) + b(c^2+a^2-b^2) - 2abc $$

gives us

$$g(a,b,c) = - (a+b-c)(a-b+c)(-a+b+c) = 0 $$

Hence, the answer is 1.

Case 2. $a+b+c = 0$.

Then each term can be simplified in the form $\frac{ (a+b)^2 - c^2 - 2ab}{2ab} = -1$, hence the answer is -3.

Calvin Lin
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I think the answer must be $1$ if we impose a triangle inequality. The individual terms are all cosines of angles of a triangle. Thus, the sum of those angles must be $\pi$. But the sum of the squares of their cosines is $3$; therefore each cosine must be $\pm 1$. But the sum of the angles is, again, $\pi$, so that two of the angles must be zero and the third $\pi$. Thus, the sum of the cosines, which is sought, must be $1+1-1=1$.

Ron Gordon
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    why are you imposing that $a,b,c$ must be sides of a triangle ? – lab bhattacharjee Jul 03 '13 at 15:13
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    I am undeleting and taking the hit on rep so that people can see the kernel of the idea I had which @CalvinLin took to completion. – Ron Gordon Jul 03 '13 at 15:30
  • But if there is a triangle with one angle $\pi$ then the triangle is simply a straight line with $a+c=b$, if B is $\pi$. That is why even I deleted my post. – Rohinb97 Jul 04 '13 at 06:14
  • @Rohinb97: of course, but see Calvin Lin's observation about the implications of that for the roots. – Ron Gordon Jul 04 '13 at 06:17
  • Ive seen it, @RonGordon. But the thing is he took complete cases and generalized the term to bring him this answer. But our answers jump straight to the conclusion, that too by assuming the cosines rule. That is why I deleted my answer. I saw this mistake and thought of correcting it, but didn't want to because it would still not be appropriate. – Rohinb97 Jul 04 '13 at 06:23
  • @Rohinb97: I too had deleted my answer, but on the suggestion of Calvin, I undeleted it because this answer complements his. I asked the question of your answer because you had the same idea and came to a different conclusion - not totally incorrect, but less useful. You predicted that the answer could be $3$; I predicted correctly that the answer could only be $1$. – Ron Gordon Jul 04 '13 at 06:27
  • @Rohinb97: by the way, you are free to undelete your answer, I only wanted to know if you had seen these answers prior to posting yours. – Ron Gordon Jul 04 '13 at 06:30
  • Actually I saw your answer and I saw you did not talk about the actual formula. I thought since someone might be confused so that's why I wrote an answer incorporating the actual formula and anything which came in the way. Undeleting my answer, and making sone edits right away... – Rohinb97 Jul 04 '13 at 07:10
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I think you might be talking about a triangle, with $a$, $b$, $c$ as it's sides of a triangle. If the angles opposite to side a, b, c are A, B,C, then $$\cos A=\frac{b^2+c^2-a^2}{2cb}$$ and similarly, $$\cos B=\frac{a^2+c^2-b^2}{2ac}$$ and $$\cos C=\frac{b^2+a^2-c^2}{2ab}$$ If it is true then $\cos A = \cos B = \cos C =1$ from the first term (since minimum and maximum of cos x with any x is plus minus 1). So the answer is either -3, -2,-1 ,1,2 or 3.

Form here, we are getting a variety of answers, but since here only one of the angle can be negative(obtuse), the answer could only be 1 or 3. (since the sum could only be 1+1-1 or 1+1+1).

Now since all three angles of the triangle cannot be 0, so they must be one angle 180 and the rest 0. So 1 is the only answer.

NOTE: The above is only true when a,b and c are angles of a triangle. And the above is the called the cosines rule of a triangle. Although the triangle discussed above is in fact a straight line, but it can be noticed that the sum of the two sees containing the 180 angle is equal to the third side.

Rohinb97
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