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.
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 $b+c>a,a+b>c,c+a>b$