thedudemanguyperson

Use geometric probability! We are choosing random $x,y,z\ge 0$ such that $x+y+z=5$. We want the probability that $x,y,z\le 2$. The first equation is a plane, the second is a cube. Using the lower bounds on x,y,z, we get a pyramid. The volume of the pyramid is $\frac{125}{6}$. Therefore, the answer is

$$\frac{8}{\frac{125}{6}}=\frac{48}{125}$$ 

Note

$$\cos C=\cos(90-B)=\sin B=\frac{4}{17}$$

Using mass points, let $m_P=1$, then $m_R=1$, $m_Q=1$, and we find $m_N=2$. Thus $ON=\frac{18}{3}=6$. Now, by the Pythagorean theorem, $OR=\sqrt{6^2+7^2}=\sqrt{85}$.

The valid points $G$ are inside the circular sector centered at $D$. This sector has radius $1$ and angle $60$. Hence it's area is $\frac{\pi}{6}$. The area of the entire triangle is $\frac{9\sqrt{3}}{4}$, so by the principle of geometric probability, the answer is $\frac{4\pi}{54\sqrt{3}}=\frac{2\sqrt{3}\pi}{81}$

Th

The problem has 2 interpretations, one is EP's (which is ambiguous, BAC can be anything -- use trig ceva). The other is the above (intended) one.

$2\angle BAC+180-72=180\implies \angle BAC = 36$. Easy.

Approach 1: Manipulation

Let sum = S.

$$S=\frac{1}{3^1}+\frac{1}{3^2}+\frac{2}{3^2}+\frac{1}{3^3}+\frac{3}{3^3}+...=\left(\frac{1}{3^1}+\frac{1}{3^2}+...\right)+\frac{S}{3}=\frac{1}{2}+\frac{S}{3}$$

Hence $S=\frac{S}{3}+\frac{1}{2}\implies \frac{2S}{3}=\frac{1}{2}\implies \boxed{S=\frac{3}{4}}$.

Approach 2: Calculus

Take the series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$. Differentiating yields $\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty} nx^{n-1}\implies \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty} nx^n$. Putting $x=\frac{1}{3}$, the answer is $\frac{\frac{1}{3}}{\frac{2}{3}\frac{2}{3}}=\frac{3}{4}$.

The problem is trivialized by https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle.

The probability that you get those exact three is $\frac{1}{\binom{20}{3}}$. The probability that you get the SuperBall is $\frac{1}{10}$. Multiply them.

Factoring the denominator:

$$\frac{x^2+5x-\alpha}{(x+11)(x-4)}$$

Thus $(x+11)\mathrm{ or }(x-4)|(x^2+5x-\alpha)$.

Hence either $x^2+5x-\alpha=(x+11)(x+t)$ or $x^2+5x-\alpha = (x+t)(x-4)$.

The first case, sum of roots is $-5=-11-t\implies 5=11+t\implies t=-6$ and hence $(x+11)(x-6)=x^2+5x-66$ thus $\alpha = 66$. 

The second case, sum of roots is $-5=4-t\implies t=9$ and hence $\alpha=36$.

The answer is then $66+36=102$.