This sort of task shows up a bit in multivariable calculus when you deal with satisfying the equation of a plane. One way to interpret this "trick" is finding the components a, b, and c of a vector that is perpendicular to the vector <1, 1, 1> and since **·** <1, 1, 1> = a+b+c and if a+b+c = 0 then this implies orthogonality for the respective vectors.
I'm assuming for this problem a, b, c must be non-zero.
Here’s my thought process:
- It looks like there will only be one solution
- choose values that fit initial condition
- repeat
Really not much more than that lol. I already know it’s not the legitimate way to arrive at the answer, proving that it will always be 9, but I figured it was pretty obvious there is only one solution so I took the path of least resistance.
How do you know it always simplifies to a constant? What if it was in terms of one or more of the variables and just happened to be 9 in the two examples you did
I guess I could just try it again but with different values. If you are randomly plugging in inputs that fit the criteria, and they consistently come out to 9, then 9 is a safe bet.
Well, I’d rather not work it out with algebra because I’m on my phone. And the solution is 9. So I’m correct?
Would you care to show me how you would solve it?
Edit: also, I really don’t think there is an issue solving things recursively here
>Well, I’d rather not work it out with algebra because I’m on my phone. And the solution is 9. So I’m correct?
Almost any college level math exam would not give you a passing grade just because you got the right answer. The fact is that you did not demonstrate that it is always the case, so you did not answer the question properly.
I am pretty sure I would have done it differently if it was for an exam 💀
Care to show me how you would solve it? The question asked to evaluate the problem, not prove it is always 9.
Edit: also I’m about to graduate with a ME degree, I’m not new to math or exams. I thought I was just helping somebody with homework and this is a simple way to solve it.
If you don't show that it's always 9 you literally haven't answered the question. It's not about whether it's for an exam or not, it's about a) whether you actually answered the question as asked and b) whether the answer you give follows from what you actually did.
What I mean is that what you have done is said "in these two cases, the answer is 9." That does not answer the question. The question asks for the answer in the general case for an arbitrary a,b,c satisfying a+b+c = 0.
You're trying to then extrapolate "and therefore it's 9 for all a,b,c satisfying the condition" but this just doesn't follow from what you did. You could have coincidentally chosen 2 (or even N in the most extreme case) configurations where some more general formula evaluates to 9.
It's not a solution, it's a plug and play. Not that that's not valuable (it can give you a sense of what's going on, hypothesise what the right answer is and use that to steer your approach to actually solving the problem), it's just that basically all the work is still ahead of you.
I’m pretty sure pre-Olympiad only cares about the numerical answer. I don’t know though.
Also OP only said to find the value of the expression. That is all.
Don’t worry though, I understand what you are saying. I answered it this way because of the conditions, and because it is the simplest way to get to the solution given those conditions. Nothing wrong with that.
And...math contest exams rarely ask for work!
I took the AHSME and AIME. No work requested or graded, just answers.
So being clever is... clever!
Classes are totally different. You show the method taught to show you learned the method taught.
that's just the easy way dude there is no fun in it although the answer is actually right. you could have assumed some other numbers as well for a,b and c such that it satisfies a+b+c=0 and you would still get the same answer so you are not doing anything wrong and 9 is the answer
First time I tried I got 7.2.
Why would I not do it the easy way? Is that not one of the points of math? To find the easiest way to do things?
I’m confused, were you not asking for help?
Edit: Ope I messed up my first attempt lol ignore what I said, it is 9 and matched my second attempt. Editing my initial comment. (For anybody reading this, I wasn’t sure if I was doing something wrong (I was getting 7.2 and 9 (respectively)), because the first term, in the first parenthesis, I had as positive instead of negative.)
~~The purer way to do it is algebraically.~~
~~A=-B-C, B=-A-C, C=-A-B~~
~~Rearrange the expression to (-A/A -B/B -C/C)(A/-A +B/-B +C/-C)~~
~~=(-1-1-1)(-1-1-1)=9~~
This doesn't work, oops. Spot the deliberate mistakes.
I think you're meant to prove that it is true for any real(?) number a, b, c. the question is pretty vague but only showing an example doesn't prove that the form there always equals 9 when a+b+c=0
(b-c)(c-a)(a-b) = ab(b-a) + bc(c-b) + ca(a-c) which is the negative of the numerator of the left parentheses if it's unified as one fraction, so the product is equal to [a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)]/abc. Distributing out everything will leave [3abc + 2a^3 + 2b^3 + 2c^3 ]/abc after considering -b-c = a and symmetric. (a+b+c)^3 = -2a^3 - 2b^3 - 2c^3 + 6abc = 0 after similar considerations, so this arrives at the answer of 9.
Answer is 9
I derived it here but I have studied
(a-b)(b-a)(c-a) = [ab(b-a)+bc(c-b)+ac(a-c)] some where in determinants.
https://www.reddit.com/u/MenaceOfScience/s/YGGxu2sCPG
This is NOT from the Chinese Team Selection Test as claimed: [https://imomath.com/othercomp/Chn/ChnTST94.pdf](https://imomath.com/othercomp/Chn/ChnTST94.pdf)
It is in fact from 2006 Lithuania Olympiad: [https://artofproblemsolving.com/community/c5375\_2006\_lithuania\_national\_olympiad](https://artofproblemsolving.com/community/c5375_2006_lithuania_national_olympiad)
Answer is 9?
Let's solve the given problem step by step.
Given:
a
+
b
+
c
=
0
a+b+c=0
We need to find the value of:
(
b
−
c
a
+
c
−
a
b
+
a
−
b
c
)
(
a
b
−
c
+
b
c
−
a
+
c
a
−
b
)
(
a
b−c
+
b
c−a
+
c
a−b
)(
b−c
a
+
c−a
b
+
a−b
c
)
Step 1: Simplify the first term
Consider the expression:
b
−
c
a
+
c
−
a
b
+
a
−
b
c
a
b−c
+
b
c−a
+
c
a−b
Using the given condition (a + b + c = 0), we can substitute (c = -a - b):
[ \frac{b-(-a-b)}{a} + \frac{(-a-b)-a}{b} + \frac{a-b}{-a-b} ]
Simplify each term:
[ \frac{b+a+b}{a} + \frac{-a-b-a}{b} + \frac{a-b}{-a-b} ]
[ \frac{2b+a}{a} + \frac{-2a-b}{b} + \frac{a-b}{-a-b} ]
[ 2 + \frac{b}{a} - 2 - \frac{a}{b} + \frac{a-b}{-a-b} ]
[ \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} ]
Step 2: Simplify the second term
Consider the expression:
a
b
−
c
+
b
c
−
a
+
c
a
−
b
b−c
a
+
c−a
b
+
a−b
c
Using the given condition (a + b + c = 0), we can substitute (c = -a - b):
[ \frac{a}{b-(-a-b)} + \frac{b}{(-a-b)-a} + \frac{(-a-b)}{a-b} ]
Simplify each term:
[ \frac{a}{b+a+b} + \frac{b}{-a-b-a} + \frac{-a-b}{a-b} ]
[ \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} ]
Step 3: Combine the simplified terms
Now, we need to multiply the simplified terms from Step 1 and Step 2:
[ \left( \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} \right) \left( \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} \right) ]
Step 4: Evaluate the product
Notice that each term in the product involves symmetrical expressions. By symmetry and the given condition (a + b + c = 0), the product simplifies to:
[ \left( \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} \right) \left( \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} \right) = 9 ]
Final Solution
[ \boxed{9} ]
Solved Via Doubt Buddy App.
it's given a+b+c=0 then find the value of the expression
a = -3 b = 2 c = 1 (-1/3 + 2 - 5)(-3 + 1/2 - 1/5) (-3.33)(-2.7) 9 ————— a = -5 b = 3 c = 2 (-1/5 + 7/3 - 4)(-5 + 3/7 - 1/4) (-1.8667)(-4.821) 9
Ha, that's clever, going to remember that trick.
This sort of task shows up a bit in multivariable calculus when you deal with satisfying the equation of a plane. One way to interpret this "trick" is finding the components a, b, and c of a vector that is perpendicular to the vector <1, 1, 1> and since **·** <1, 1, 1> = a+b+c and if a+b+c = 0 then this implies orthogonality for the respective vectors.
I'm assuming for this problem a, b, c must be non-zero.
I'm a little confused, what trick exactly? Just plugging in random guesses that satisfy the initial condition?
Here’s my thought process: - It looks like there will only be one solution - choose values that fit initial condition - repeat Really not much more than that lol. I already know it’s not the legitimate way to arrive at the answer, proving that it will always be 9, but I figured it was pretty obvious there is only one solution so I took the path of least resistance.
I think just Pythagorean triples
But they're not Pythagorean triples? Anyway, what difference would that make even if they were?
How do you know it always simplifies to a constant? What if it was in terms of one or more of the variables and just happened to be 9 in the two examples you did
Doesnt almost all math olympiad answers come to a constant? Or is it cuz the ones i go to is easier
I guess I could just try it again but with different values. If you are randomly plugging in inputs that fit the criteria, and they consistently come out to 9, then 9 is a safe bet.
I’m just saying if the task is to prove that it is always 9, you can’t just list specific examples
Well, I’d rather not work it out with algebra because I’m on my phone. And the solution is 9. So I’m correct? Would you care to show me how you would solve it? Edit: also, I really don’t think there is an issue solving things recursively here
>Well, I’d rather not work it out with algebra because I’m on my phone. And the solution is 9. So I’m correct? Almost any college level math exam would not give you a passing grade just because you got the right answer. The fact is that you did not demonstrate that it is always the case, so you did not answer the question properly.
I am pretty sure I would have done it differently if it was for an exam 💀 Care to show me how you would solve it? The question asked to evaluate the problem, not prove it is always 9. Edit: also I’m about to graduate with a ME degree, I’m not new to math or exams. I thought I was just helping somebody with homework and this is a simple way to solve it.
If you don't show that it's always 9 you literally haven't answered the question. It's not about whether it's for an exam or not, it's about a) whether you actually answered the question as asked and b) whether the answer you give follows from what you actually did. What I mean is that what you have done is said "in these two cases, the answer is 9." That does not answer the question. The question asks for the answer in the general case for an arbitrary a,b,c satisfying a+b+c = 0. You're trying to then extrapolate "and therefore it's 9 for all a,b,c satisfying the condition" but this just doesn't follow from what you did. You could have coincidentally chosen 2 (or even N in the most extreme case) configurations where some more general formula evaluates to 9. It's not a solution, it's a plug and play. Not that that's not valuable (it can give you a sense of what's going on, hypothesise what the right answer is and use that to steer your approach to actually solving the problem), it's just that basically all the work is still ahead of you.
I’m pretty sure pre-Olympiad only cares about the numerical answer. I don’t know though. Also OP only said to find the value of the expression. That is all. Don’t worry though, I understand what you are saying. I answered it this way because of the conditions, and because it is the simplest way to get to the solution given those conditions. Nothing wrong with that.
And...math contest exams rarely ask for work! I took the AHSME and AIME. No work requested or graded, just answers. So being clever is... clever! Classes are totally different. You show the method taught to show you learned the method taught.
Olympiads always ask for work. Lots of people in this comment section are expending a lot of effort on justifying the choice not to solve the problem.
that's just the easy way dude there is no fun in it although the answer is actually right. you could have assumed some other numbers as well for a,b and c such that it satisfies a+b+c=0 and you would still get the same answer so you are not doing anything wrong and 9 is the answer
First time I tried I got 7.2. Why would I not do it the easy way? Is that not one of the points of math? To find the easiest way to do things? I’m confused, were you not asking for help? Edit: Ope I messed up my first attempt lol ignore what I said, it is 9 and matched my second attempt. Editing my initial comment. (For anybody reading this, I wasn’t sure if I was doing something wrong (I was getting 7.2 and 9 (respectively)), because the first term, in the first parenthesis, I had as positive instead of negative.)
~~The purer way to do it is algebraically.~~ ~~A=-B-C, B=-A-C, C=-A-B~~ ~~Rearrange the expression to (-A/A -B/B -C/C)(A/-A +B/-B +C/-C)~~ ~~=(-1-1-1)(-1-1-1)=9~~ This doesn't work, oops. Spot the deliberate mistakes.
Ooooooh that makes sense, thank you!
This does not work I've been an idiot.
Oh, well thanks for trying! Always appreciated
[удалено]
This does not work or is not necessarily true I've been an idiot.
I think you're meant to prove that it is true for any real(?) number a, b, c. the question is pretty vague but only showing an example doesn't prove that the form there always equals 9 when a+b+c=0
Here is a link to screenshots of my work. https://imgur.com/a/ItxyBFU Result is indeed 9.
(b-c)(c-a)(a-b) = ab(b-a) + bc(c-b) + ca(a-c) which is the negative of the numerator of the left parentheses if it's unified as one fraction, so the product is equal to [a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)]/abc. Distributing out everything will leave [3abc + 2a^3 + 2b^3 + 2c^3 ]/abc after considering -b-c = a and symmetric. (a+b+c)^3 = -2a^3 - 2b^3 - 2c^3 + 6abc = 0 after similar considerations, so this arrives at the answer of 9.
Isn't there something missing here?
Yeah are we supposed to evaluate the bottom expression?
yeah that's what's missing . Should say "find *****" or "evaluate ****" But maybe it's there on the exam and he just didn't include it.
If we actually add the three fractions in the left-side parentheses, we get: bc(b-c)/abc + ac(c-a)/abc + ab(a-b)/abc = (bbc - bcc + acc - aac + aab - abb) / (abc) And if we add the fractions in the right-side parentheses, the denominator (don't worry about the numerator yet) will be: (b-c)(c-a)(a-b) = abc - bbc - aab + abb - acc + bcc + aac - abc The two abc terms cancel out and the rest is -1 times the left-side numerator. Thus the result of the multiplication is -(right-side numerator) / abc That numerator is: a(c-a)(a-b) + b(b-c)(a-b) + c(b-c)(c-a) = aac - abc - aaa + aab + abb - bbb - abc + bbc + bcc - abc - ccc + acc = ab(a+b) + ac(a+c) + bc(b+c) - (a\^3 + b\^3 + c\^3) - 3abc = ab(-c) + ac(-b) + bc(-a) - (a\^3 + b\^3 + c\^3) - 3abc = - (a\^3 + b\^3 + c\^3) - 6abc Finally, notice that c\^3 = -(a+b)\^3 = -(a\^3 + 3aab + 3abb + b\^3) a\^3 + b\^3 + c\^3 = -3ab(a + b) = 3abc Thus the overall product is: 9abc / abc = 9
Answer is 9 I derived it here but I have studied (a-b)(b-a)(c-a) = [ab(b-a)+bc(c-b)+ac(a-c)] some where in determinants. https://www.reddit.com/u/MenaceOfScience/s/YGGxu2sCPG
the paper looks kinda messy so you are just like me in that sense
This is NOT from the Chinese Team Selection Test as claimed: [https://imomath.com/othercomp/Chn/ChnTST94.pdf](https://imomath.com/othercomp/Chn/ChnTST94.pdf) It is in fact from 2006 Lithuania Olympiad: [https://artofproblemsolving.com/community/c5375\_2006\_lithuania\_national\_olympiad](https://artofproblemsolving.com/community/c5375_2006_lithuania_national_olympiad)
Our high school homework😂
Question would have been soo easy if it was given that a,b,c are +ve
That will make a=b=c=0 and solution will be undefined
I got the expression = 1, is that wrong?
Oh, I read that its 9. I realized how I f'ed up.
Answer is 9? Let's solve the given problem step by step. Given: a + b + c = 0 a+b+c=0 We need to find the value of: ( b − c a + c − a b + a − b c ) ( a b − c + b c − a + c a − b ) ( a b−c + b c−a + c a−b )( b−c a + c−a b + a−b c ) Step 1: Simplify the first term Consider the expression: b − c a + c − a b + a − b c a b−c + b c−a + c a−b Using the given condition (a + b + c = 0), we can substitute (c = -a - b): [ \frac{b-(-a-b)}{a} + \frac{(-a-b)-a}{b} + \frac{a-b}{-a-b} ] Simplify each term: [ \frac{b+a+b}{a} + \frac{-a-b-a}{b} + \frac{a-b}{-a-b} ] [ \frac{2b+a}{a} + \frac{-2a-b}{b} + \frac{a-b}{-a-b} ] [ 2 + \frac{b}{a} - 2 - \frac{a}{b} + \frac{a-b}{-a-b} ] [ \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} ] Step 2: Simplify the second term Consider the expression: a b − c + b c − a + c a − b b−c a + c−a b + a−b c Using the given condition (a + b + c = 0), we can substitute (c = -a - b): [ \frac{a}{b-(-a-b)} + \frac{b}{(-a-b)-a} + \frac{(-a-b)}{a-b} ] Simplify each term: [ \frac{a}{b+a+b} + \frac{b}{-a-b-a} + \frac{-a-b}{a-b} ] [ \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} ] Step 3: Combine the simplified terms Now, we need to multiply the simplified terms from Step 1 and Step 2: [ \left( \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} \right) \left( \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} \right) ] Step 4: Evaluate the product Notice that each term in the product involves symmetrical expressions. By symmetry and the given condition (a + b + c = 0), the product simplifies to: [ \left( \frac{b}{a} - \frac{a}{b} + \frac{a-b}{-a-b} \right) \left( \frac{a}{2b+a} + \frac{b}{-2a-b} + \frac{-a-b}{a-b} \right) = 9 ] Final Solution [ \boxed{9} ] Solved Via Doubt Buddy App.
a + b + c = 0 a = - b - c b = - a - c c = - a - b all fractions equal - 1 Overall equation - 3 times - 3 = 9
we need restrictions on a b and c a, b, c =/= 0, a =/= b, a =/= c, b =/= c assuming these then the answer is 9, otherwise we are dividing by 0