Geometry Probability – 4

(1) Rotating $\triangle{ACD}$ around $C$ counter-clockwise $60^\circ$ so that $D$ is moved to point $D’$, and $A$ to $B$, we have $\triangle{DCD’}$ as an equilateral triangle, and $\triangle{ABD}\cong\triangle{ABD’}$. Therefore, $AD=BD’$, $CD=DD’$. Thus $AD$, $BD$, and $CD$ form $\triangle{DBD’}$. Therefore, the probability is $1$.

(2) If $\angle{BDC}=150^\circ$, $\angle{BDD’}=\angle{BDC}-\angle{CDD’}=150^\circ-60^\circ=90^\circ$. Therefore, if $D$ is on the red arcs in the following diagram, $AD$, $BD$, and $CD$ will form a right triangle. Therefore, the probability is $0$.

(3) If $D$ is inside the regions formed by arcs and the sides of $\triangle{ABC}$, then $AD$, $BD$, and $CD$ will form an obtuse triangle. The probability is $$\dfrac{3\cdot\Big(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\Big)}{\dfrac{\sqrt{3}}{4}}=\dfrac{2\pi\sqrt{3}}{3}-3\approx 0.6276$$

(4) The probability is $$1-\Big(\dfrac{2\pi\sqrt{3}}{3}-3\Big)=4-\dfrac{2\pi\sqrt{3}}{3}\approx 0.3724$$