Question

If r and R are radii of the in-circle and circum-circle of $△ABC$, then $8rR[{cos}^{2}A/2+{cos}^{2}B/2+{cos}^{2}C/2]$ is equal to

• A. $2bc+2ca+2ab$
• B. $2bc-2ca-2ab+a^2+b^2+c^2$
• C. $2bc+2ca+2ab-a^2-b^2-c^2$
• D. $2bc+2ca+2ab+a^2+b^2+c^2$

Answer 1

Answer: (C)

$2bc+2ca+2ab-a^2-b^2-c^2$

$8rR[\frac{S(s-a)}{bc}+\frac{S(s-b)}{ac}+\frac{S(s-c)}{ab}]$

$=8rR\left[\frac{aS(s-a)+bs(S-b)+cS(S-c)}{abc}\right]$

$=8rRS\left[\frac{aS-a^2-bS-b^2+cS-c^2}{abc}\right]$

$=8\times \frac{\Delta }{S}\times \frac{abc}{4\Delta }\times \frac{S}{abc}\left[S\right(a+b+c)-a^2-b^2-c^2]$

$=2\left[S\right(a+b+c)-a^2-b^2-c^2]$

$=(a+b+c)(a+b+c)-2a^2-2b^2-2c^2$

$=a^2+b^2+c^2+2ab+2bc+2ca-2a^2-2b^2-2c^2$

$=2ab+2bc+2ca-a^2-b^2-c^2$