Question : If $2s = a + b + c$, then the value of $s(s - c) + (s - a)(s - b)$ is:
Option 1: $ab$
Option 2: $abc$
Option 3: $0$
Option 4: $\frac{a+b+c}{2}$
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Correct Answer: $ab$
Solution : Given: $2s = a + b + c$ ⇒ $s = \frac{a\:+\:b\:+\:c}{2}$ Now, $s(s - c) = (\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:+\:c}{2}-c)$ ⇒ $s(s - c) =(\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:-\:c}{2})$ Also, $(s - a)(s - b) = (\frac{a\:+\:b\:+\:c}{2}-a)(\frac{a\:+\:b\:+\:c}{2}-b)$ ⇒ $(s - a)(s - b) = (\frac{b\:+\:c\:-\:a}{2})(\frac{a\:+\:c\:-\:b}{2})$ So, $s(s - c) + (s - a)(s - b)$ $= (\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:–\:c}{2})+(\frac{b\:+\:c\:–\:a}{2})(\frac{a\:+\:c\:–\:b}{2})$ $= \frac{1}{4}[(a+b)^2-c^2+c^2-(a-b)^2]$ $= \frac{1}{4}[(a+b)^2-(a-b)^2]$ $= \frac{1}{4}(4ab)$ $=ab$ Hence, the correct answer is $ab$.
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Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$.
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