Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.

My solution:

  1. $$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$

  2. $$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$

My query:

I am totally stuck on this problem. Please help.

Thanks a lot!

share|cite|improve this question
up vote 1 down vote accepted

Let $$x=b+c,$$ $$y=a+c,$$ $$z=a+b.$$ Then, $x+y+z=2(a+b+c)=28$ and $z:x:y=6:7:8$.

By substituting $z=6k$, $x=7k$, $y=8k$ into $x+y+z=28$, we have $21 k=28$, $k=\frac{4}{3}$, from where we get: $$z=6k=8,$$ $$x=7k=\frac{28}{3},$$ $$y=8k=\frac{32}{3}.$$ Finally, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$c=a+b+c-z=6.$$

share|cite|improve this answer
Thanks a lot. Clear and simple solution. – Gaurang Tandon Feb 15 '14 at 16:56


Similarly, each ratio is equal to $$\frac{a+b+(b+c)+(c+a)}{6+7+8}=\frac{2(a+b+c)}{21}$$

$$\implies \frac{2c}9=\frac{2(a+b+c)}{21}$$

Now, we have $a+b+c=14$

share|cite|improve this answer
@GaurangTandon, how about this? – lab bhattacharjee Feb 16 '14 at 15:33
I was wondering how did you get $$\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$. Please explain. Thanks. – Gaurang Tandon Feb 16 '14 at 16:46
@GaurangTandon, Ratio & Proportion formula says : $$\frac aA=\frac bB=\frac cC=\frac{pa+qb+rc}{pA+qB+rC}$$ etc. – lab bhattacharjee Feb 17 '14 at 5:52
Oh, I see. Thank you :) Got to know a new formula ! – Gaurang Tandon Feb 17 '14 at 5:59
@GaurangTandon, please derive this – lab bhattacharjee Feb 17 '14 at 6:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.