Question

If $\mathrm{S}$ is circumcenter, $\mathrm{G}$ the centroid, $\mathrm{O}$ the orthocenter of $∆\mathrm{ABC}$, then $\mathrm{SA}+\mathrm{SB}+\mathrm{SC}$ is equal to:

• A. $SG$
• B. $OS$
• C. $SO$
• D. $OG$

Answer 1

The correct option is C

$SO$

Explanation for the correct option.

Step 1: Find the centroid

Given, $\mathrm{S}$ is the circumcenter, $\mathrm{G}$ the centroid, $\mathrm{O}$ the orthocenter of $∆\mathrm{ABC}$.

We solve this by using the concept of vectors.

Let $\mathrm{S}\left(0\right);\mathrm{A};\mathrm{B};\mathrm{C}$ be the vectors of circumcenter.
Then, $\mathrm{SA}+\mathrm{SB}+\mathrm{SC}$
Centroid $\mathrm{G}=\frac{\mathrm{SA}+\mathrm{SB}+\mathrm{SC}}{3}$

Step 2: Find the value of expression
We know the centroid divides the median in $2:1$.

$\mathrm{HG}:\mathrm{GS}=2:1$
So here $\mathrm{G}$ divides $\mathrm{HS}$ in $2:1$ ratio.
$$\begin{gathered} \Rightarrow \mathrm{SG}=\frac{\mathrm{SO}+2\mathrm{S}}{3} \\ \Rightarrow \mathrm{O}=3\mathrm{SG} \\ =\mathrm{SA}+\mathrm{SB}+\mathrm{SC} \\ \Rightarrow \mathrm{SA}+\mathrm{SB}+\mathrm{SC}=\mathrm{SO} \end{gathered}$$
Hence, option C is correct.