, 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 $t > 0$.

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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 $AB = \frac{13}{7}$.