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Question:

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!

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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, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$c=a+b+c-z=6.$$

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  • $\begingroup$ Thanks a lot. Clear and simple solution. $\endgroup$ – Gaurang Tandon Feb 15 '14 at 16:56
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$$\frac{a+b}6=\frac{b+c}7=\frac{c+a}8=\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$

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 $a+b+c=14$

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

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