What is the probability that orthocentre lies inside the circumcircle If a triangle is chosen at random inscribed in a fixed circle and what is the probability that the orthocentre lies inside circumcircle?
I thought that if the triangle is acute then only the orthocentre will lie inside the circumcircle. And so thought that answer is 1/2 but answer given at back is 1/9.
Can anyone suggest how to get to right answer and what was wrong in my thinking.
 A: I do not think $\frac{1}{9}$ is the right answer unless I have understood the problem wrong.
If you randomly select 3 points A, B and C on the circumference, it is easy to see that the probability of one of the angles being obtuse is greater than $\frac{1}{2}$. It requires only one angle out of the three to be obtuse.
What is the probability of $\angle A$ being obtuse? Say, you take point C and draw a line through the circumcenter. For $\angle A$ to be obtuse, point A has to be on the same side of the semi-circle as B.
i) The probability of both A and B being on the same side = $\frac{1}{2}$
ii) The probability of A being between B and C, rather than B being in between C and A = $\frac{1}{2}$
So, the probability of $\angle A$ being obtuse = $\frac{1}{4}$
For any of the 3 angles being obtuse, the probability will be $\frac{3}{4}$.
So, the probability of the triangle being acute = $\frac{1}{4}$
The answer should be $\frac{1}{4}$.
A: Hint: There are three types of triangles with regard to the angles: acute, right, and obtuse. The orthocenter will always be inside of an acute triangle, it is on a right triangle, and it will be outside of an obtuse triangle.
