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Math Help - Resolution of triangles?

  1. #1
    Super Member fardeen_gen's Avatar
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    Resolution of triangles?

    If in a triangle ABC, length of the sides a,b & c are in Arithmetic Progression, then it is necessary that:

    A)2/3 < b/c < 2
    B)1/3 < b/c < 2/3
    C)2/3 < b/a < 2
    D)1/3 < b/a < 2/3

    More than one options are correct. How to solve this problem?
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    What about something using the law of sines ?

    And is there more information about a,b and c ? Because propositions A) and C) are equivalent, and so are B) and D)
    Last edited by Moo; July 18th 2008 at 11:02 AM.
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  3. #3
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    Hello, fardeen_gen!

    I have a start on this problem, but I haven't finished it yet.
    . . Maybe this will inspire you . . .


    If in a triangle ABC, length of the sides a,b,c are in Arithmetic Progression,
    then it is necessary that:

    . . \begin{array}{cccccc}(A)& \dfrac{2}{3} & < & \dfrac{b}{c} & < & 2 \\ \\[-2mm]<br />
(B)& \dfrac{1}{3} &<& \dfrac{b}{c} &<& \dfrac{2}{3} \\ \\[-2mm]<br />
(C)& \dfrac{2}{3} &<& \dfrac{b}{a} &<& 2 \\ \\[-2mm]<br />
(D)& \dfrac{1}{3} &<& \dfrac{b}{a} &<& \dfrac{2}{3} \end{array}

    More than one option is correct. How to solve this problem?

    Assume that a,b,c are ordered from smallest to largest: .  a \,<\, b \,<\, c

    Since the form an arithmetic progression, there is a common difference d.

    Then the three sides are: . b-d,\;b,\;b+d


    The Triangle Inequality says: . a + b \:>\:c

    . . which gives us: . (b-d) + b \:>\:b+d\quad\Rightarrow\quad b \:>\:2d \quad\rightarrow\quad d \:<\:\frac{b}{2}

    Add b to both sides: . b + d \:< \:b + \frac{b}{2} \quad\Rightarrow\quad b+d \:<\:\frac{3b}{2}

    Take reciprocals: . \frac{1}{b+d} \:>\:\frac{2}{3b}

    \text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b+d}_{\text{This is }c}} \:<\:\frac{2}{3}

    . . Therefore: . \frac{b}{c}\:<\:\frac{2}{3}

    We have half of statement (B); I'm still working on the other half.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We have: . d \:<\:\frac{b}{2}

    Multiply by -1: . -d \:>\:-\frac{b}{2}

    Add b to both sides: . b - d \:>\:b -\frac{b}{2} \quad\Rightarrow\quad b - d \:>\:\frac{b}{2}

    Take reciprocals: . \frac{1}{b-d} \:<\:\frac{2}{b}

    \text{Multiply by }b\!:\;\;\frac{b}{\underbrace{b-d}_{\text{This is }a}} \:<\:2

    . . Therefore: . \frac{b}{a} \:<\:2

    And we have half of statment (C).

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