Question 28

If a + b + c = 2s, then $$\frac{(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}}{a^{2}+b^{2}+c^{2}}$$ is equal to

Solution

a + b + c = 2s

put a=b=c=1, then 2s = 3, s = 3/2.

then, s-a = s - b = s - c = 1/2

$$\frac{(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}}{a^{2}+b^{2}+c^{2}}$$

= $$\frac{(1/2)^{2}+(1/2)^{2}+(1/2)^{2}+(3/2)^{2}}{1^{2}+1^{2}+1^{2}}$$

= $$\frac{(3/4)+(9/4)}{3}$$

= $$\frac{12/4}{3}$$

= 1

only option C satisfies this.

so the answer is option C.

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