about to only mathematics

Options

A. 1
B. 2
C. `3//2`
D. 4

Answer

3

Explanation

Let the circle be `x^(2)+y^(2)=1`.
Let the vertices of the triagnel be `( cos theta_(1),sin theta_(1)),i=1,2,3.`
Then, the orthocenter is
`(cos theta_(1)+cos theta_(2)+cos theta_(3),sin theta_(1)+sin theta_(2)+sin theta_(3))`
Therefore, the distance between the orthocenter and the circumcenter is
`sqrt((cos theta_(1)+cos theta_(2)+cos theta_(3))^(2)+(sin theta_(1)+sin theta_(2)+sin theta_(3))^(2))`
`=sqrt((cos^(2)theta_(1)+sin^(2)theta_(1))+(cos^(2)theta_(2)+sin^(2)theta_(2))+(cos^(2)theta_(3)+sin^(2)theta_(3))+2[cos (theta_(1)-theta_(2))+(cos (theta_(2)-theta_(2))+cos(theta_(3)-theta_(1))]`
`=sqrt(3+2[cos (theta_(1)-theta_(2))+cos (theta_(2)-theta_(3))+cos (theta_(3)-theta_(1))])lt3`