, 2022

The probability that the integer $n$ is chosen is $\log_{100}\left(1 + \dfrac{1}{n}\right)$.

The probability that an integer between 81 and 99, inclusive, is chosen
equals the sum of the probabilities that the integers 81, 82, $\ldots$, 98, 99 are selected, which equals $$\log_{100}\left(1 + \dfrac{1}{81}\right) + \log_{100}\left(1 + \dfrac{1}{82}\right) + \cdots + \log_{100}\left(1 + \dfrac{1}{98}\right) + \log_{100}\left(1 + \dfrac{1}{99}\right)$$ Since the second probability equals 2 times the first probability, the following equations are equivalent: $$\begin{aligned} \hspace{-2cm} \log_{100}\left(1 + \dfrac{1}{81}\right) + \log_{100}\left(1 + \dfrac{1}{82}\right) + \cdots + \log_{100}\left(1 + \dfrac{1}{98}\right) + \log_{100}\left(1 + \dfrac{1}{99}\right) & = 2\log_{100}\left(1 + \dfrac{1}{n}\right) \\ \log_{100}\left(\dfrac{82}{81}\right) + \log_{100}\left(\dfrac{83}{82}\right) + \cdots + \log_{100}\left(\dfrac{99}{98}\right) + \log_{100}\left(\dfrac{100}{99}\right) & = 2\log_{100}\left(1 + \dfrac{1}{n}\right)\end{aligned}$$ Using logarithm laws, these equations are further equivalent to $$\begin{aligned} \log_{100}\left(\dfrac{82}{81} \cdot \dfrac{83}{82}\cdot \cdots \cdot \dfrac{99}{98} \cdot \dfrac{100}{99}\right) & = \log_{100}\left(1 + \dfrac{1}{n}\right)^2 \\ \log_{100}\left(\dfrac{100}{81}\right) & = \log_{100}\left(1 + \dfrac{1}{n}\right)^2\end{aligned}$$ Since logarithm functions are invertible, we obtain $\$\dfrac{100}{81} = \left(1 + \dfrac{1}{n}\right)^2$$. Since$n>0$, then$1 + $\dfrac{1}{n} = \sqrt{\dfrac{100}{81}} = \dfrac{10}{9}$$, and so$$\dfrac{1}{n} = \dfrac{1}{9}$$, which gives$n = 9$.
Since $\dfrac{AC}{AD} = \dfrac{3}{4}$, then we let $AC = 3t$ and $AD = 4t$ for some real number $t > 0$. [[IMAGE0]] Using the cosine law in $\$\triangle$
ACD, the following equations are equivalent:

$\begin{aligned} AD^2 & = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(\angle ACD) \\ (4t)^2 & = (3t)^2 + 1^2 - 2(3t)(1)(-\tfrac{3}{5}) \\ 16t^2 & = 9t^2 + 1 + \tfrac{18}{5}t \\ 80t^2 & = 45t^2 + 5 + 18t \\ 35t^2 - 18t - 5 & = 0 \\ (7t-5)(5t + 1) & = 0\end{aligned}$$Since$t > 0$, then$t = $\frac{5}{7}$$. Thus,$AC = 3t = $\frac{15}{7}$. Using the cosine law in

$\triangle$
ACB$and noting that$$\cos(\angle ACB) = \cos(180^\circ - \angle ACD) = -\cos(\angle ACD) = \tfrac{3}{5}$ the following equations are equivalent:

$\begin{aligned} AB^2 & = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle ACB) \\ & = \left(\tfrac{15}{7}\right)^2 + 2^2 - 2(\tfrac{15}{7})(2)(\tfrac{3}{5}) \\[1mm] & = \tfrac{225}{49} + 4 - \tfrac{36}{7} \\[1mm] & = \tfrac{225}{49} + \tfrac{196}{49} - \tfrac{252}{49} \\[1mm] & = \tfrac{169}{49}\end{aligned}$$Since$AB>0$, then$AB = $\frac{13}{7}\$.$