What's the sum?

Algebra Level 3

If $a, b, c$ are positive integers such that $a^2+2b^2-2ab=169$ and $2bc-c^2=169$ , then what is the value of $a+b+c ?$

1 solution

Toshit Jain
Mar 15, 2017

$a^{2} \space + \space 2b^{2} \space -2ab \space=\space 169 \space \space \space (1)$

$2bc \space - \space c^{2} \space=\space 169 \space\space\space (2)$

$Subtracting \space (2) \space from \space(1)$

$\rightarrow \space a^{2} \space +2 \space b^{2} \space - \space 2ab \space -2bc \space + \space c^{2} \space=\space 169\space-\space169$

$After \space arranging \space them$

$\rightarrow \space a^{2} \space+\space b^{2} \space - \space 2ab \space + \space b^{2} \space + \space c^{2} \space-\space 2bc \space =\space 0$

$\rightarrow \space (a-b)^{2} \space + \space (b-c)^{2} \space = \space 0$

$\Rightarrow \space (a-b)^{2} \space =\space 0 \space, \space (b-c)^{2} \space = \space 0$

$\therefore \space a \space=\space b \space = \space c$

$\Rightarrow \space a^{2} \space+\space 2a^{2} \space - \space 2a^{2}\space=\space 169$

$\Rightarrow \space a \space = \space \sqrt{169} \space = \space 13$

$\therefore \space a \space=\space b \space = \space c \space = \space 13$

$\Rightarrow \space a \space + \space b \space + \space c \space=\space 3 \space \times \space 13$

$\boxed{ \therefore \space a \space + \space b \space + \space c \space=\space 39}$