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This is the question which I am referring to

Each of two triplets of numbers (log a, log b, log c) and (log a-log2b, log 2b-log 3c, log 3c-log a) is an AP.prove that a, b, c can be lengths of sides of a triangle.also find a:b:c.

My try:

  • first of all we know that length of sides of a triangle should follow this condition to form a triangle. a+b> c, a+c> b, b+c> a Here a, b, c are length of sides of a triangle
  • Then I applied the condition of AP 2q=p+r , where p, q, r are in AP I got the following relation b^2=ac and 8b^3=27c^3 from 1st and 2nd AP's respectively.

After that I am not getting any clue to move towards the result also I have a doubt weather it is a pythagorean triplet or a prime triplet, I also want to know how you thought to solve the problem.

Kartik Watwani
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  • Find first the ratio of sides, then it's easier to prove it can be side of triangle. That is i think, 9:6:4 , rough calculation. – Someone Jun 25 '15 at 08:00
  • But how did u thought about it that this can be easier – Kartik Watwani Jun 25 '15 at 08:02
  • Let $a=9k$ ,$ b=6k $, $c=4k$. $\forall k \in \mathbb{R}$ – Someone Jun 25 '15 at 08:03
  • But how and why did u got this result – Kartik Watwani Jun 25 '15 at 08:06
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    $8b^3=27c^3$ $\implies $ $2b=3c$ $\implies$ $\frac{b}{c}=\frac{3}{2}$ also $b^2=ac$ $\implies$ $\frac{b}{c}=\frac{a}{b}$ , usually in triangles as you know, the ratio of sides is actually more important in defining whether a triangle can exist or not. For e.g., the trigonometric ratios law of sines and cosines etc. Most work on ratios, not on actual length. Hence finding ratio of sides is a good way to start. – Someone Jun 25 '15 at 08:09

1 Answers1

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As $b^2=ac\implies\dfrac cb=\dfrac ba=k(\ne0)$(say)

$\implies b=ak, c=ak^2$

Now $8b^3=27b^3\implies8(ak)^3=27(ak^2)^3\iff a^3k^3(27k^3-8)=0$

As $ak\ne0, k^3=\dfrac8{27}\implies k=\dfrac23$ as $k$ is real

$\implies b=\dfrac23a, c=\left(\dfrac23\right)^2a$

Show that $b+c>a,a+b>c,c+a>b$

lab bhattacharjee
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