Aspire Faculty ID #10216 · Topic: NIMCET 2015 · Just now
NIMCET 2015

$a, b, c$ are positive integers such that $a^{2}+2b^{2}-2bc=100$ and $2ab-c^{2}=100$. Then the value of $\frac{a+b}{c}$ is

Solution

Given the two equations: 
(1) $a^2 + 2b^2 - 2bc = 100$ 
(2) $2ab - c^2 = 100$ 

Equating equation (1) and equation (2): $a^2 + 2b^2 - 2bc = 2ab - c^2$ 

Rearranging all terms to one side: $a^2 - 2ab + 2b^2 - 2bc + c^2 = 0$ 
Split $2b^2$ into $b^2 + b^2$: 
$(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) = 0$ 
$(a - b)^2 + (b - c)^2 = 0$ 
Since $a, b, c$ are real numbers (positive integers), the sum of two non-negative squares can only be zero if each term is zero: 
$a - b = 0 \implies a = b$ 
$b - c = 0 \implies b = c$ 
Thus: $a = b = c$ Substitute $a = b = c$ into equation (2): 

$2(c)(c) - c^2 = 100$ 
$2c^2 - c^2 = 100$ 
$c^2 = 100$ 
Since $c$ is a positive integer: $c = 10 \implies a = b = c = 10$ 
 Now, compute the required value: $\frac{a + b}{c} = \frac{10 + 10}{10} = \frac{20}{10} = 2$
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