Welcome 2016! Part 6

Algebra Level 4

$\dfrac{ab}{a+b}=3 \ , \ \dfrac{bc}{b+c}=4 \ , \ \dfrac{ca}{c+a}=5$

Let $a,b$ and $c$ be real numbers satisfying all 3 equations above. Given that $\dfrac{abc}{ab+ac+bc}$ is equal to $\dfrac mn$ , where $m$ and $n$ are coprime positive integers, find the value of $m+n$ .

The answer is 167.

2 solutions

Aareyan Manzoor
Dec 12, 2015

reciprocate all three $\dfrac{a+b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{3}$ $\dfrac{b+c}{bc}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{4}$ $\dfrac{c+a}{ca}=\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{1}{5}$ add these to get $2(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}$ $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ca}{abc}=\dfrac{47}{120}$ reciprocate again $\dfrac{abc}{ab+bc+ca}=\dfrac{120}{47}$ $120+47=167$

Ok, then where I made the mistake. I did like this.
$\frac{abc}{ab+bc+ca}=\frac{m}{n}$

$\frac{ab+bc+ca}{abc}=\frac{n}{m}$

$(\frac{1}{c}+\frac{1}{a})+\frac{1}{b}=\frac{n}{m}$

$\frac{a+c}{ac}+\frac{1}{b}=\frac{n}{m}$

$\frac{1}{5}+\frac{1}{b}=\frac{n}{m}$ ..... $(1)$

Now Finding $\frac{1}{b}$ .
$\frac{1}{b}+\frac{1}{a}=\frac{1}{3}$ [Given]

$\frac{1}{c}+\frac{1}{b}=\frac{1}{4}$ [Given]

$\Rightarrow \frac{1}{3}-\frac{1}{a}=\frac{1}{4}-\frac{1}{c}=\frac{1}{b}$
$\frac{1}{b}=\frac{1}{12}$
Putting in $(1)$
$\frac{12+5}{12×5}=\frac{n}{m}$
$\frac{17}{60}=\frac{n}{m}$
$\frac{m}{n}=\frac{60}{17}$
$m+n=\boxed{77}$

A Former Brilliant Member - 5 years, 6 months ago

can yoy show what you did after 1/3-1/a=b..., the mistake is probably there @Abhay Kumar

Aareyan Manzoor - 5 years, 6 months ago

how did u get 1/b=1/12?

Mehbub Litu - 5 years, 6 months ago

$\Rightarrow\frac{1}{3}-\frac{1}{4}=\frac{1}{a}-\frac{1}{c}=\frac{1}{b}$
$=\frac{1}{12}=\frac{1}{a}-\frac{1}{c}=\frac{1}{b}$
$\Rightarrow \boxed{\frac{1}{b}=\frac{1}{12}}$ .

A Former Brilliant Member - 5 years, 6 months ago

@A Former Brilliant Member – nope, you have to subtract 1/4 and 1/c from all three side not only 2!

Aareyan Manzoor - 5 years, 6 months ago

@Aareyan Manzoor – Use latex and then tell the mistake.We can also do it with 2.

A Former Brilliant Member - 5 years, 6 months ago

@A Former Brilliant Member – we have eqns $\frac{1}{3}-\frac{1}{a}=\frac{1}{b}$ $\frac{1}{4}-\frac{1}{c}=\frac{1}{b}$ now try doing what you did in these two

a fundemental example is $3+1=2+2=1+3$ $3-2=2-1\neq 1+3$

Aareyan Manzoor - 5 years, 6 months ago

@Aareyan Manzoor – Ok, thank you very much found that mistake.

A Former Brilliant Member - 5 years, 6 months ago

best approach ... it striked me instantly

Ganesh Ayyappan - 5 years, 6 months ago

Pulkit Gupta
Dec 12, 2015

We first note the LCM of 3,4 & 5 which is 5 * 4 * 3 = 60.

Multiplying the first, second and third equations by 20,15 and 12 respectively, we obtain

$\frac{20ab}{a+b}$ = 60, $\frac{15bc}{b+c}$ = 60, $\frac{12ac}{a+c}$ = 60

Next, we multiply first, second and third equations above by a,b & c respectively

$\frac{20abc}{a+b}$ = 60c, $\frac{15abc}{b+c}$ = 60a, $\frac{12abc}{a+c}$ = 60b

Simplifying, we get

47abc = 120 ( ab + bc + ca) Therefore, $\frac{abc}{ab+bc+ca}$ = 120/47

Welcome 2016! Part 6

The answer is 167.

2 solutions

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