[u]Round 5[/u]


[b]p13.[/b] Five different schools are competing in a tournament where each pair of teams plays at most once. Four pairs of teams are randomly selected and play against each other. After these four matches, what is the probability that Chad's and Jordan's respective schools have played against each other, assuming that Chad and Jordan come from different schools?


[b]p14.[/b] A square of side length $1$ and a regular hexagon are both circumscribed by the same circle. What is the side length of the hexagon?


[b]p15.[/b] From the list of integers $1,2, 3,...,30$ Jordan can pick at least one pair of distinct numbers such that none of the $28$ other numbers are equal to the sum or the difference of this pair. Of all possible such pairs, Jordan chooses the pair with the least sum. Which two numbers does Jordan pick?


[u]Round 6[/u]


[b]p16.[/b] What is the sum of all two-digit integers with no digit greater than four whose squares also have no digit greater than four?


[b]p17.[/b] Chad marks off ten points on a circle. Then, Jordan draws five chords under the following constraints:
$\bullet$ Each of the ten points is on exactly one chord.
$\bullet$ No two chords intersect.
$\bullet$ There do not exist (potentially non-consecutive) points $A, B,C,D,E$, and $F$, in that order around the circle, for which $AB$, $CD$, and $EF$ are all drawn chords.
In how many ways can Jordan draw these chords?


[b]p18.[/b] Chad is thirsty. He has $109$ cubic centimeters of silicon and a 3D printer with which he can print a cup to drink water in. He wants a silicon cup whose exterior is cubical, with five square faces and an open top, that can hold exactly $234$ cubic centimeters of water when filled to the rim in a rectangular-box-shaped cavity. Using all of his silicon, he prints a such cup whose thickness is the same on the five faces. What is this thickness, in centimeters?


[u]Round 7[/u]


[b]p19.[/b] Jordan wants to create an equiangular octagon whose side lengths are exactly the first $8$ positive integers, so that each side has a different length. How many such octagons can Jordan create?


[b]p20.[/b] There are two positive integers on the blackboard. Chad computes the sum of these two numbers and tells it to Jordan. Jordan then calculates the sum of the greatest common divisor and the least common multiple of the two numbers, and discovers that her result is exactly $3$ times as large as the number Chad told her. What is the smallest possible sum that Chad could have said?


[b]p21.[/b] Chad uses yater to measure distances, and knows the conversion factor from yaters to meters precisely. When Jordan asks Chad to convert yaters into meters, Chad only gives Jordan the result rounded to the nearest integer meters. At Jordan's request, Chad converts $5$ yaters into $8$ meters and $7$ yaters into $12$ meters. Given this information, how many possible numbers of meters could Jordan receive from Chad when requesting to convert $2014$ yaters into meters?


[u]Round 8[/u]


[b]p22.[/b] Jordan places a rectangle inside a triangle with side lengths $13$, $14$, and $15$ so that the vertices of the rectangle all lie on sides of the triangle. What is the maximum possible area of Jordan's rectangle?


[b]p23.[/b] Hoping to join Chad and Jordan in the Exeter Space Station, there are $2014$ prospective astronauts of various nationalities. It is given that $1006$ of the astronaut applicants are American and that there are a total of $64$ countries represented among the applicants. The applicants are to group into $1007$ pairs with no pair consisting of two applicants of the same nationality. Over all possible distributions of nationalities, what is the maximum number of possible ways to make the $1007$ pairs of applicants? Express your answer in the form $a \cdot b!$, where $a$ and $b$ are positive integers and $a$ is not divisible by $b + 1$.

Note: The expression $k!$ denotes the product $k \cdot (k - 1) \cdot ... \cdot  2 \cdot  1$.


[b]p24.[/b]  We say a polynomial $P$ in $x$ and $y$ is $n$-[i]good [/i] if $P(x, y) = 0$ for all integers $x$ and $y$, with $x \ne y$, between $1$ and $n$, inclusive. We also define the complexity of a polynomial to be the maximum sum of exponents of $x$ and $y$ across its terms with nonzero coeffcients. What is the minimal complexity of a nonzero $4$-good polynomial? In addition, give an example of a $4$-good polynomial attaining this minimal complexity.


PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c3h2915803p26040550]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let the two positive integers on the blackboard be \(a\) and \(b\), with the greatest common divisor \(d\). We can express \(a\) and \(b\) as:
   \[
   a = da', \quad b = db'
   \]
   where \(a'\) and \(b'\) are coprime integers.

2. The sum of the greatest common divisor and the least common multiple of \(a\) and \(b\) is given by:
   \[
   \gcd(a, b) + \text{lcm}(a, b) = d + \frac{ab}{d}
   \]
   Substituting \(a = da'\) and \(b = db'\), we get:
   \[
   d + \frac{da' \cdot db'}{d} = d + d a' b'
   \]
   Simplifying, we have:
   \[
   d + d a' b' = d(1 + a' b')
   \]

3. According to the problem, this sum is three times the sum of \(a\) and \(b\):
   \[
   d(1 + a' b') = 3(a + b) = 3(da' + db') = 3d(a' + b')
   \]
   Dividing both sides by \(d\), we obtain:
   \[
   1 + a' b' = 3(a' + b')
   \]

4. Rearranging the equation, we get:
   \[
   1 + a' b' = 3a' + 3b' \implies a' b' - 3a' - 3b' = -1 \implies (a' - 3)(b' - 3) = 8
   \]

5. We need to find pairs \((a', b')\) such that \((a' - 3)(b' - 3) = 8\). The integer solutions to this equation are:
   \[
   (a' - 3, b' - 3) = (1, 8), (8, 1), (2, 4), (4, 2)
   \]
   Thus, the pairs \((a', b')\) are:
   \[
   (a', b') = (4, 11), (11, 4), (5, 7), (7, 5)
   \]

6. The minimum possible value of \(a + b\) for a given \(d\) is:
   \[
   d(4 + 11) = 15d, \quad d(5 + 7) = 12d
   \]
   Clearly, this will be minimized when \(d = 1\), so the answer is:
   \[
   12 \cdot 1 = 12
   \]

The final answer is \(\boxed{12}\)