Question

If ABC is an equilateral triangle with $R=4$ then the distance between the circum-centre and the orthocentre is

• A. $4\sqrt{.8}$
• B. $4$
• C. $0$
• D. $4\sqrt{1.5}$

Answer 1

Answer: (C)

$0$

As mentioned in Above Concept
Answer is Distance between orthocentre $\left(O\right)$ and incentre $\left(I\right)$ is

$O^{\prime }{I}^{\prime }=R\sqrt{1-8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}=\sqrt{R^2-2Rr}$

Distance between excentres and orthocentre is

$O^{\prime }{I}_1^{\prime }=R\sqrt{1+8\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}=\sqrt{R^2+2R{r}_1}$

$O^{\prime }{I}_2^{\prime }=R\sqrt{1+8cos\frac{A}{2}\sin \frac{B}{2}cos\frac{C}{2}}=\sqrt{R^2+2R{r}_2}$

$O^{\prime }{I}_3^{\prime }=R\sqrt{1+8cos\frac{A}{2}cos\frac{B}{2}\sin \frac{C}{2}}=\sqrt{R^2+2R{r}_3}$

According to the Formula , Distance between circumcentre and orthocentre is

$=R\sqrt{1-8cosAcosBcosC}=R\sqrt{1-8cos{60}^{o}cos{60}^{o}cos{60}^o}=R\sqrt{1-1}=0$

Therefore, Answer is $C$