[u]Round 1[/u]


[b]p1.[/b] What is the smallest number equal to its cube?


[b]p2.[/b] Fhomas has $5$ red spaghetti and $5$ blue spaghetti, where spaghetti are indistinguishable except for color. In how many different ways can Fhomas eat $6$ spaghetti, one after the other? (Two ways are considered the same if the sequence of colors are identical)


[b]p3.[/b] Jocelyn labels the three corners of a triangle with three consecutive natural numbers. She then labels each edge with the sum of the two numbers on the vertices it touches, and labels the center with the sum of all three edges. If the total sum of all labels on her triangle is $120$, what is the value of the smallest label?


[u]Round 2[/u]


[b]p4.[/b] Adam cooks a pie in the shape of a regular hexagon with side length $12$, and wants to cut it into right triangular pieces with angles $30^o$, $60^o$, and $90^o$, each with shortest side $3$. What is the maximum number of such pieces he can make?


[b]p5.[/b] If $f(x) =\frac{1}{2-x}$ and $g(x) = 1-\frac{1}{x}$ , what is the value of $f(g(f(g(... f(g(f(2019))) ...))))$, where there are $2019$ functions total, counting both $f$ and $g$?


[b]p6.[/b] Fhomas is buying spaghetti again, which is only sold in two types of boxes: a $200$ gram box and a $500$ gram box, each with a fixed price. If Fhomas wants to buy exactly $800$ grams, he must spend $\$8:80$, but if he wants to buy exactly 900 grams, he only needs to spend $\$7:90$! In dollars, how much more does the $500$ gram box cost than the $200$ gram box?


[u]Round 3[/u]


[b]p7.[/b] Given that $$\begin{cases} a + 5b + 9c = 1 \\ 4a + 2b + 3c = 2 \\ 7a + 8b + 6c = 9\end{cases}$$ what is $741a + 825b + 639c$?


[b]p8.[/b] Hexagon $JAMESU$ has line of symmetry $MU$ (i.e., quadrilaterals $JAMU$ and $SEMU$ are reflections of each other), and $JA = AM = ME = ES = 1$. If all angles of $JAMESU$ are $135$ degrees except for right angles at $A$ and $E$, find the length of side $US$.


[b]p9.[/b] Max is parked at the $11$ mile mark on a highway, when his pet cheetah, Min, leaps out of the car and starts running up the highway at its maximum speed. At the same time, Max starts his car and starts driving down the highway at $\frac12$ his maximum speed, driving all the way to the $10$ mile mark before realizing that his cheetah is gone! Max then immediately reverses directions and starts driving back up the highway at his maximum speed, nally catching up to Min at the $20$ mile mark. What is the ratio between Max's max speed and Min's max speed?


[u]Round 4[/u]


[b]p10.[/b] Kevin owns three non-adjacent square plots of land, each with side length an integer number of meters, whose total area is $2019$ m$^2$. What is the minimum sum of the perimeters of his three plots, in meters?


[b]p11.[/b] Given a $5\times 5$ array of lattice points, how many squares are there with vertices all lying on these points?


[b]p12.[/b] Let right triangle $ABC$ have $\angle A = 90^o$, $AB = 6$, and $AC = 8$. Let points $D,E$ be on side $AC$ such that $AD = EC = 2$, and let points $F,G$ be on side $BC$ such that $BF = FG = 3$. Find the area of quadrilateral $FGED$.


PS. You should use hide for answers. Rounds 5-8 have been posted [url=https://artofproblemsolving.com/community/c3h2949413p26408203]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To determine the number of different ways Fhomas can eat 6 spaghetti, we need to consider the different combinations of red and blue spaghetti he can eat. We will use combinations to count the sequences.

1. **Case 1: Eating 5 of one color and 1 of the other color**
   - If Fhomas eats 5 red and 1 blue, the number of ways to choose the position of the blue spaghetti in the sequence of 6 is given by:
     \[
     \binom{6}{1} = 6
     \]
   - Similarly, if Fhomas eats 5 blue and 1 red, the number of ways to choose the position of the red spaghetti in the sequence of 6 is also:
     \[
     \binom{6}{1} = 6
     \]
   - Therefore, the total number of sequences for this case is:
     \[
     2 \times 6 = 12
     \]

2. **Case 2: Eating 4 of one color and 2 of the other color**
   - If Fhomas eats 4 red and 2 blue, the number of ways to choose the positions of the 2 blue spaghetti in the sequence of 6 is given by:
     \[
     \binom{6}{2} = 15
     \]
   - Similarly, if Fhomas eats 4 blue and 2 red, the number of ways to choose the positions of the 2 red spaghetti in the sequence of 6 is also:
     \[
     \binom{6}{2} = 15
     \]
   - Therefore, the total number of sequences for this case is:
     \[
     2 \times 15 = 30
     \]

3. **Case 3: Eating 3 of each color**
   - If Fhomas eats 3 red and 3 blue, the number of ways to choose the positions of the 3 red spaghetti in the sequence of 6 is given by:
     \[
     \binom{6}{3} = 20
     \]

4. **Summing up all cases**
   - The total number of different ways Fhomas can eat 6 spaghetti is:
     \[
     12 + 30 + 20 = 62
     \]

The final answer is \(\boxed{62}\).