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Prove a,b,c are in AP

  1. Nov 22, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If a,b,c are real numbers satisfying the equation [itex]25(9a^2+b^2)+9c^2-15(5ab+bc+3ca)=0[/itex], then prove that a,b,c are in A.P.

    2. Relevant equations

    3. The attempt at a solution
    Rearranging the question I can say that I have to prove this
    2b=a+c

    I tried simplifying the terms and I arrived at this
    [itex](15a+5b+3c)^2=45(5ab+bc+3ac)[/itex]

    Any ideas how should I proceed?
     
  2. jcsd
  3. Nov 22, 2012 #2
    [itex]25(9a^2+b^2)+9c^2-15(5ab+bc+3ca)=0[/itex]

    Distribute.
    [itex]225a^{2} + 25b^{2} + 9c^{2} - 75ab - 15bc - 45ac = 0[/itex]

    Now comes the weird part. You don't need to know the values of a, b, and c if you want to prove that [itex]2b = a + c[/itex]. You just need to know the ratios between them.

    Solve for a, b, and c in terms of each other. I decided to solve b and c in terms of a.

    This way, we get [itex]b = 3a[/itex] and [itex]c = 5a - \frac{5\sqrt{-9 a^2 + 18a^{2} - (3a)^{2}}}{2\sqrt{3}} = 5a - \frac{5\sqrt{9a^{2} - 9a^{2}}}{2\sqrt{3}} = 5a[/itex]

    Thus, [itex]2(3a) = a + 5a[/itex], ∴ [itex]2b = a + c[/itex]. QED.
     
  4. Nov 22, 2012 #3

    Ray Vickson

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    You are not supposed to do a person's homework for them; hints only.

    RGV
     
  5. Nov 22, 2012 #4

    utkarshakash

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    I simply can't understand what you're up to. Are you making a quadratic in b and then solving it? If yes then how did you get such a nice value? Mine becomes very long and complicated. The other possible situation I can think of is this.

    Are you expanding it like this

    [itex]15(15a^2-bc)+25b(b-3a)+9c(c-5a)=0[/itex]

    and setting each of term equal to zero (Since sum of +ve numbers can't be 0 each term must be zero) and then obtaining values of b and c in terms of a. If this is the case then I've got it.
     
    Last edited: Nov 22, 2012
  6. Nov 22, 2012 #5

    haruspex

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    You can greatly simplify the expressions by making some simple substitutions for a, b and c.
    Notice how wherever a appears there's a factor of 15 around? Similarly the other two variables.
     
  7. Nov 23, 2012 #6

    utkarshakash

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    Give me an example.
     
  8. Nov 23, 2012 #7

    ehild

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    Rewrite the equation as (15a)2+(5b)2+9c2-(15a)(5b)-3(5b)c-3(15a)c=0
    You can choose x=15a/c and y=5b/c as new variables.

    The new equation is

    x2+y2+9-xy-3y-3x=0

    or x2-x(y+3)+y2-3y+9=0

    It is a quadratic equation in x. The discriminant is D=(y+3)2-4(y2-3y+9)=-3(y2-6y+9), negative or zero. To get a solution, D must be zero, so y=3. That involves x= 3.

    ehild
     
  9. Nov 23, 2012 #8

    haruspex

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    I had in mind something very similar: t = 15a, u =5b, v = 3c producing t2 + u2 + v2 = uv + vt + tu.
    This can be turned into (t-u)2 + (u-v)2 + (v-t)2 = 0, which has only the solution t = u = v.
     
  10. Nov 24, 2012 #9

    utkarshakash

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    I always find your solutions to be better than others. You are a life-saver. Thank you so much for helping me!
     
  11. Nov 24, 2012 #10

    ehild

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    To be fair, it was Haruspex's idea and his solution is the best:smile:.

    ehild
     
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