[u]Round 5[/u]


[b]p13.[/b]  Two concentric circles have radii $1$ and $3$. Compute the length of the longest straight line segment that can be drawn froma point on the circle of radius $1$ to a point on the circle of radius $3$ if the segment cannot intersect the circle of radius $1$.


[b]p14.[/b]  Find the value of $\frac{1}{3} + \frac29+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+...$


[b]p15.[/b] Bob is trying to type the word "welp". However, he has a $18$ chance ofmistyping each letter and instead typing one of four adjacent keys, each with equal probability. Find the probability that he types "qelp" instead of "welp".


[u]Round 6[/u]


[b]p16.[/b]  How many ways are there to tile a $2\times 12$ board using an unlimited supply of $1\times 1$ and $1\times 3$ pieces?


[b]p17.[/b]  Jeffrey and Yiming independently choose a number between $0$ and $1$ uniformly at random. What is the probability that their two numbers can formthe sidelengths of a triangle with longest side of length $1$?


[b]p18.[/b] On $\vartriangle ABC$ with $AB = 12$ and $AC = 16$, let $M$ be the midpoint of $BC$ and $E$,$F$ be the points such that $E$ is on $AB$, $F$ is on $AC$, and $AE = 2AF$. Let $G$ be the intersection of $EF$ and $AM$. Compute $\frac{EG}{GF}$ .


[u]Round 7[/u]


[b]p19.[/b] Find the remainder when $2019x^{2019} -2018x^{2018}+ 2017x^{2017}-...+x$ is divided by $x +1$.


[b]p20.[/b] Parallelogram $ABCD$ has $AB = 5$, $BC = 3$, and $\angle ABC = 45^o$. A line through C intersects $AB$ at $M$ and $AD$ at $N$ such that $\vartriangle BCM$ is isosceles. Determine the maximum possible area of $\vartriangle MAN$.


[b]p21[/b]. Determine the number of convex hexagons whose sides only lie along the grid shown below.
[img]https://cdn.artofproblemsolving.com/attachments/2/9/93cf897a321dfda282a14e8f1c78d32fafb58d.png[/img]

[u]Round 8[/u]


[b]p22.[/b]  Let $\vartriangle ABC$ be a triangle with side lengths $AB = 4$, $BC = 5$, and $C A = 6$. Extend ray $\overrightarrow{AB}$ to a point $D$ such that $AD = 12$, and similarly extend ray  $\overrightarrow{CB}$ to point $E$ such that $CE = 15$. Let $M$ and $N$ be points on the circumcircles of $ABC$ and $DBE$, respectively, such that line $MN$ is tangent to both circles. Determine the length of $MN$.


[b]p23.[/b] A volcano will erupt with probability $\frac{1}{20-n}$ if it has not erupted in the last $n$ years. Determine the expected number of years between consecutive eruptions.


[b]p24.[/b]  If $x$ and $y$ are integers such that $x+ y = 9$ and $3x^2+4x y = 128$, find $x$.


PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c3h3165997p28809441]here [/url] and 9-12 [url=https://artofproblemsolving.com/community/c3h3166099p28810427]here[/url].Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let our desired sum be \( S \), i.e.
   \[
   S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots + \frac{n}{3^n} + \ldots
   \]
2. Multiply both sides by \(\frac{1}{3}\):
   \[
   \frac{1}{3}S = \frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \ldots + \frac{n-1}{3^n} + \ldots
   \]
3. Subtract this equation from the first equation:
   \[
   S - \frac{1}{3}S = \left( \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots \right) - \left( \frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \ldots \right)
   \]
   Simplifying the right-hand side:
   \[
   \frac{2}{3}S = \frac{1}{3} + \left( \frac{1}{9} + \frac{1}{9} \right) + \left( \frac{1}{27} + \frac{1}{27} + \frac{1}{27} \right) + \ldots
   \]
   This can be written as:
   \[
   \frac{2}{3}S = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots
   \]
4. Recognize the series on the right-hand side as an infinite geometric series with the first term \(a = \frac{1}{3}\) and common ratio \(r = \frac{1}{3}\):
   \[
   \frac{2}{3}S = \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n
   \]
   The sum of an infinite geometric series is given by:
   \[
   \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
   \]
   Here, \(a = \frac{1}{3}\) and \(r = \frac{1}{3}\), so:
   \[
   \frac{2}{3}S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
   \]
5. Solving for \(S\):
   \[
   \frac{2}{3}S = \frac{1}{2} \implies S = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}
   \]
   However, this contradicts the earlier steps. Let's re-evaluate the infinite series sum:
   \[
   \frac{2}{3}S = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
   \]
   Correcting the final step:
   \[
   \frac{2}{3}S = \frac{3}{2} \implies S = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4}
   \]

The final answer is \(\boxed{\frac{9}{4}}\).