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Posted on 2016-09-25
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I hope I have not published this one before.

Prove SIMULTANEOUSLY that

(x-b)(x-c)/((a-b)(a-c)) + (x-c)(x-a)/((b-c)(b-a)) + (x-a)(x-b)/((c-a)(c-b) = 1

and

a(x-b)(x-c)/((a-b)(a-c)) + b(x-c)(x-a)/((b-c)(b-a)) +c (x-a)(x-b)/((c-a)(c-b) = x
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Question by:BigRat
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by:Stefan Hoffmann
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Looks like homework.

Rearrange the terms. Expand the sub-terms. Use a common denominator to get one fraction. Multiply by that denomiator and reduce the terms by using the balance method. In the end you should get 1=1 or 0=0  as result.
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by:BigRat
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Not at my age, sonny. And that is NOT the way to do it.
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by:Stefan Hoffmann
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???
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Ste5an,  have a search for some of BigRat''s 'difficult' questions. Presumably this one is easier. I think BigRat sets homework instead of doing it. This will have probably have a nasty twist and an elegant solution.
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by:BigRat
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@ Thibault St john Cholmondeley-ffeatherstonehaugh the 2nd

I presume from your name that your family has been around for a number of years like me. Somehow I like the idea that an almost seventy-year old does homework questions. Well, I suppose that might be happen someday, but not now, since my grandchild is not yet three.

And yes it does have an elegant solution.
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It wouldn't let me fit in 'viscount of Worcestercestercestershire' i'm just testing the profile limits. It seems to stop at 40 characters. The rest of it is just to annoy or amuse people.
I will look at your equation tomorrow and probably fail, but there is always hope.
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by:CompProbSolv
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I'm very curious to see the solution.  I'm not getting much further than being able to claim that a<>b, b<>c, and a<>c.  Expanding didn't seem to get anywhere useful.
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by:BigRat
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We are assuming that a,b, and c are all different - otherwise the puzzle would be rather trivial. Expanding is the obvious way to go, but you can't expand BOTH expressions simultaneously. The question therefore suggests that whatever "method of solution" one applies to the first must also apply to the second.
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by:CompProbSolv
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I'm interpreting "simultaneously" differently from how you are.  I could easily be wrong but I took it to mean that there is a set of a, b, c, d, and x for which both equations are true.  Clearly they are not true for all possible numbers.

I also didn't see anything to suggest that a, b, and c being all different is to be assumed.  It can be reasoned from either equation but I didn't take that as an initial assumption.

Of course, this is coming from someone who isn't able to solve it!
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by:BigRat
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The question clearly states "prove SIMULTANEOUSLY that ..." which implies that ONE argument should be used for BOTH cases. And indeed there is one. One must use a theorem on polynominals - if you need a clue.
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by:d-glitch
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Well that is a good hint.  Both LHS expressions are Lagrange polynomials for quadratics.  
After simplification, both sides will have the form:   Ax²  +  Bx  +  C

The first polynomial is defined by the points:   (a, 1)    (b, 1)    (c, 1)     So  A = B = 0   and  C = 1

The second is defined by the points:   (a, a)    (b, b)    (c, c)     So  A = C = 0   and  B = 1
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by:CompProbSolv
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But.....  if a=b then the denominator in the first two terms will be 0.  That's why I argued that a, b, and c have different values.
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by:d-glitch
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Exception noted.
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by:BigRat
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Well d-glitch, that is a good start. I don't quite see why Guiseppe Lodovico has to come into it, they are both just simply quadratic expressions - that is to say the left hand sides are.
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by:d-glitch
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My first post was an observation rather than a proof.

All three LHS terms in the first equation are quadratics in x.  
The same terms appear in the second equation as coefficients of a, b, and c.
Note belatedly that the terms in the first equation have a implicit factor of 1.

After simplification, both LHS expressions will have the form:   Ax²  +  Bx  +  C

Evaluate both LHS expressions at    x = a,   x = b,   x = c
In each case, two of the quotients go to zero and the third to one.

The first function includes the points:   (a, 1)    (b, 1)    (c, 1)    
This must be a straight line   f1(x) = 1   ==>  a degenerate parabola with  A = B = 0   and  C = 1

The second function includes the points:   (a, a)    (b, b)    (c, c)
Another straight line   f2(x) = x   ==> a degenerate parabola with   A = C = 0   and  B = 1
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by:BigRat
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My first post was an observation rather than a proof.

Yes, observation is the key.

All three LHS terms in the first equation are quadratics in x.

But we are to consider BOTH equations SIMULTANEOUSLY

Evaluate both LHS expressions at    x = a,   x = b,   x = c

Good.

In each case, two of the quotients go to zero and the third to one.

Not true.

And the rest is completely off topic.
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by:d-glitch
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In each case, two of the quotients go to zero and the third to one.

Not true.
In the first equation, two of the terms (which are the coefficients of a, b, and c in the second) go to zero and the third to one.

But now I am at several of my limits.
I do observe that the first equation must be the derivative of the second.  But that would be a truly horrible calculation with no obvious benefit.
And that matrix methods are one way of dealing with equations simultaneously.  
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by:BigRat
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The "not true" referred to the phrase "In each case".

Again the word "simultaneously" refers to "one argument fits both cases". It has NOTHING to do with matrix methods or any other complicated method.
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by:BigRat
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Well it is now almost 2 weeks since the last post and if nobody is going to post the correct answer, I'll have to post it myself.

But as one last attempt, I'll give a hint. Subtract the R.H.S from the L.H.S so that you get an equation and then tell me something about the equations.
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BigRat earned 0 total points
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Now it's almost two weeks more and a wet Sunday in Brittany, so what better than to close this question with the answer.

By inspection we see that the two expressions on the LHS are totally symmetric in a,b and c. If we consider them as equations we note that the highest exponent of x is x², so they are quadratic equations in x. We also note that x=a is a solution to both equations and by symmetry also x=b and x=c. But a quadratic equation can have at most only two roots not three, so they are NOT quadratic equations but identities.
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by:CompProbSolv
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Interesting answer.  I'd go along with it as long as you acknowledge that a, b, and c are unequal to each other.  Otherwise, imagine that a=b but a<>c.  You'd have two solutions as x=a and x=b are the same solution.
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by:BigRat
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@CompProbSolv: You raised that question in ID 41818639 and I confirmed it in the subsequent post.

In fact in all puzzles of this type you can assume that. Indeed there will be another one shortly.
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Author Closing Comment

by:BigRat
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No solution has been posted. It is time to close
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