Geometry Probability – 3

Solution: First pick up the first point A randomly and uniformly from the interior of the unit circle centered at $O$ and orientate $OA$ be on positive side of the $x$-axis, with the distance between $O$ and $A$ as $t$:

Draw another circle with $OA$ as its diameter. Draw a chord $DE \perp OA$ and link $OD$ and $OE$.

If point $B$ is selected from the left side of the circle, $\triangle{OAB}$ would be obtuse. Therefore, in order for the another point $B$ randomly and uniformly selected from the interior of the circle so that $\triangle{OAB}$ is acute, point B must be on the right half of the unit circle.

Additionally, point $B$ must be not inside the smaller circle with its diameter as $OA$, and not on the right side of line $DE$.

Therefore, the probability $\triangle{OAB}$ is obtuse when $B$ is on the right half of the unit circle for $t$ would be:

$$f(t)=\dfrac{area\ of\ the\ smaller\ circle\ + \ area\ of\ unit\ circle\ on\ the\ rigit\ side\ of \ line\ DE}{area\ of\ the\ semi\ unit\ circle}$$ $$=\dfrac{\dfrac{1}{4}\pi t^2 + cos^{-1}(t)-t\sqrt{1-t^2}}{\dfrac{1}{2}\pi} \tag{1}$$

Therefore the probability of $\triangle{OAB}$ is obtuse on the right half of the unit circle would be:

$$P=\int_{0}^{1}f(\sqrt{x})dx \tag{2}$$

$\sqrt{x}$ is used to ensure that point $A$ is uniformly distributed across whole unit circle area. The square root function compensates for the quadratic increase in area, spreading the points evenly.

Simplifying $(2)$ we have:

$$P=\dfrac{1}{2}\int_{0}^{1} x dx + \dfrac{2}{\pi}\int_{0}^{1}cos^{-1}(\sqrt{x}) dx -\dfrac{2}{\pi}\int_{0}^{1}\sqrt{x(1-x)} dx$$

$$=\dfrac{1}{2}\cdot \dfrac{1}{2}+\dfrac{2}{\pi}\cdot \dfrac{\pi}{4}-\dfrac{2}{\pi}\cdot \dfrac{\pi}{8}=\dfrac{1}{2}$$

Therefore the probability of $\triangle{OAB}$ is acute on the right half of the unit circle is $1-P=\dfrac{1}{2}$.

Since the probability for $B$ to be in the right half of the unit circle is $\dfrac{1}{2}$, the overall probability is $\dfrac{1}{2}\cdot \dfrac{1}{2}=\boxed{\dfrac{1}{4}}$