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Homework Help: Quadratic equation and trigonometry

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    If the quadratic equation ax2+bx+c=0 has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
    $$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

    2. Relevant equations

    3. The attempt at a solution
    Since the given quadratic has equal roots, its discriminant is zero.
    The above is equivalent to:
    $$\sin^2 B-4\sin A\sin C=0 \,\, (*)$$
    Now I am thinking of writing (*) in terms of sinA and sinC. Substituting B with ##\pi##-(A+C) doesn't look to me a good idea. I am honestly stuck here.

    Any help is appreciated. Thanks!
    Last edited: Oct 30, 2013
  2. jcsd
  3. Oct 30, 2013 #2


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    Go back to your original condition from (1), and your result in (3), and eliminate the sin A * sin C in the denominator. This gives an expression in terms of sin^2 A, sin^2 B, sin^2 C ...
  4. Oct 31, 2013 #3
    As per your suggestion, I get
    How does this help?
  5. Oct 31, 2013 #4


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    Note that A,B,C are the angles of the triangle ... this provides additional constraints, and you can use any law that pertains to arbitrary triangles.
  6. Oct 31, 2013 #5
    And which constraints am I supposed to use? :confused:

    I already said that using A+B+C=##\pi## is not a good idea. I can't think of anything else.
  7. Oct 31, 2013 #6


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    Is that "the sum of those positive integers which are less than or equal to" the given quantity?

    I would be tempted to use the law of sines to do away with trig functions entirely. Thus [itex]a \sin C = c \sin A[/itex] and hence
    \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} = \frac ac + \frac ca = \frac{a^2 + c^2}{ac}[/tex]
    Also, by the triangle inequality, [itex]a + c \geq b[/itex].
  8. Oct 31, 2013 #7
    No, I have copied down the question word by word. Its exactly what I have written.

    $$\frac ac +\frac ca=\frac{a^2 + c^2}{ac}=\frac{(a+c)^2-2ac}{ac}=\frac{(a+c)^2}{ac}-2$$
    From the triangle inequality, a+c>=b.
    $$4\left(\frac{(a+c)^2}{b^2}\right) \geq 4 \Rightarrow 4\left(\frac{(a+c)^2}{b^2}\right)-2\geq 2$$
    Anyways, the above was obvious from AM-GM inequality.

    How do I determine the upper bound for the range? Just a guess, Cauchy-Schwarz?
  9. Nov 1, 2013 #8
    Can you please give me some more help? I am still stuck on this one.
  10. Nov 1, 2013 #9
    Well you can use weighted means.
    If that does not help, you can use Tchebychef's inequality or Weierstress' inequality ?
    I often see many books using the inequality to solve range you have to do.

    However I am not an adept in using them though and its not in JEE syllabus also.
  11. Nov 1, 2013 #10
    I doubt any of these inequalities would help. Can you show how would you even apply them here?
  12. Nov 1, 2013 #11
    No. Sorry just did not see the question properly.. :p

    Ok, so by AM-GM inequality you already found out that the expression is greater than or equal to 2.
    Hence its range is [2,∞)...

    Now to find out sum of all integers, you just have to evaluate the term, I think :

    lim n→∞Sn = 2+3+4+5+6+7.....+......+n


    Wait the above is not making sense.

    Did you make use of b^2-4ac=0 and sin^2B−4sinAsinC=0
    in using AM-GM inequality ?

    Also make an intuitive guesswork or find the range of sinA/sinC + sinC/sinA.. Obviously this will find additional conditions.....And replace sinAsinC=sin^2B/4

    Have you also tried cauchy-shwartz ?
    Last edited: Nov 1, 2013
  13. Nov 1, 2013 #12
    That isn't true. I have only proved that the lower bound for range is 2 but I still have to find the upper bound. I never stated that the upper bound is infinity.
  14. Nov 1, 2013 #13
    Is answer 9 ? I made two condition one from Pasmith's and other from UltrafastPED.

    I also gave you hints in Edit in previous post.. I hope its 9.. :p
  15. Nov 1, 2013 #14
    Sankalp, please come up with a proper proof for the answer you get. 9 is incorrect.
  16. Nov 1, 2013 #15
    Expression≥2 which you found using pasmith's hint.

    Sorry yes 9 is incorrect. You made use of equal roots in finding Expression≥2 or not ?

    You need to find the other inequality. I am already thinking of one:

    If A+B+C=pi,


    cosA + cosB + cosC ≤ 3/2

    You can prove this also.
  17. Nov 1, 2013 #16
    I honestly don't see that of any help here.
    Last edited: Nov 1, 2013
  18. Nov 1, 2013 #17
    Has anyone got some idea about this? I have made zero progress on this problem so far.

    As always, help is appreciated.
  19. Nov 2, 2013 #18
  20. Nov 2, 2013 #19

    Ray Vickson

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    Ultimately, it is your homework and you are responsible for solving the problem, or not. Sometimes the best thing to do is to say "I cannot do this problem" and just accept a less than perfect mark! I can guarantee that every one of the homework helpers here have encountered such situations themselves in the past!
  21. Nov 2, 2013 #20
    This is not a homework. Had this been a homework style problem, I would never put it here. Also, I am no longer a high school student that my work is graded. I am only solving problems and problems for a specific purpose.

    Instead of a paragraph long lecture, a few hints would have been great. I very well know its my responsibility to solve the problem and most of the times, I don't put it on forums until I have spent enough time on it.
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