[b]p1.[/b] How many lines of symmetry does a square have?


[b]p2.[/b] Compute$ 1/2 + 1/6 + 1/12 + 1/4$.


[b]p3.[/b] What is the maximum possible area of a rectangle with integer side lengths and perimeter $8$?


[b]p4.[/b] Given that $1$ printer weighs $400000$ pennies, and $80$ pennies weighs $2$ books, what is the weight of a printer expressed in books?


[b]p5.[/b] Given that two sides of a triangle are $28$ and $3$ and all three sides are integers, what is the sum of the possible lengths of the remaining side?


[b]p6.[/b] What is half the sum of all positive integers between $1$ and $15$,  inclusive, that have an even number of positive divisors?


[b]p7.[/b] Austin the Snowman has a very big brain. His head has radius $3$, and the volume of his torso is one third of his head, and the volume of his legs combined is one third of his torso. If Austin's total volume is $a\pi$ where $a$ is an integer, what is $a$?


[b]p8.[/b] Neethine the Kiwi says that she is the eye of the tiger, a fighter, and that everyone is gonna hear her roar. She is standing at point $(3, 3)$. Neeton the Cat is standing at $(11,18)$, the farthest he can stand from Neethine such that he can still hear her roar. Let the total area of the region that Neeton can stand in where he can hear Neethine's roar be $a\pi$ where $a$ is an integer. What is $a$?


[b]p9.[/b] Consider $2018$ identical kiwis. These are to be divided between $5$ people, such that the first person gets $a_1$ kiwis, the second gets $a_2$ kiwis, and so forth, with $a_1 \le a_2 \le a_3 \le a_4 \le a_5$. How many tuples $(a_1, a_2, a_3, a_4, a_5)$ can be chosen such that they form an arithmetic sequence?


[b]p10.[/b] On the standard $12$ hour clock, each number from $1$ to $12$ is replaced by the sum of its divisors. On this new clock, what is the number of degrees in the measure of the non-reflex angle between the hands of the clock at the time when the hour hand is between $7$ and $6$ while the minute hand is pointing at $15$?


[b]p11.[/b] In equiangular hexagon $ABCDEF$, $AB = 7$, $BC = 3$, $CD = 8$, and $DE = 5$. The area of the hexagon is in the form $\frac{a\sqrt{b}}{c}$ with $b$ square free and $a$ and $c$ relatively prime. Find $a+b+c$ where $a, b,$ and $c$ are integers.


[b]p12.[/b] Let $\frac{p}{q} = \frac15 + \frac{2}{5^2} + \frac{3}{5^3} + ...$ . Find $p + q$,  where $p$ and $q$ are relatively prime positive integers.


[b]p13.[/b] Two circles $F$ and $G$ with radius $10$ and $4$ respectively are externally tangent. A square $ABMC$ is inscribed in circle $F$ and equilateral triangle $MOP$ is inscribed in circle $G$ (they share vertex $M$). If the area of pentagon $ABOPC$ is equal to $a + b\sqrt{c}$, where $a$, $b$, $c$ are integers $c$ is square free, then find $a + b + c$.


[b]p14.[/b] Consider the polynomial $P(x) = x^3 + 3x^2 + ax + 8$. Find the sum of all integer $a$ such that the sum of the squares of the roots of $P(x)$ divides the sum of the coecients of $P(x)$.


[b]p15.[/b] Nithin and Antonio play a number game. At the beginning of the game, Nithin picks a prime $p$ that is less than $100$. Antonio then tries to find an integer $n$ such that $n^6 + 2n^5 + 2n^4 + n^3 + (n^2 + n + 1)^2$ is a multiple of $p$. If Antonio can find such a number n, then he wins, otherwise, he loses. Nithin doesn't know what he is doing, and he always picks his prime randomly while Antonio always plays optimally. The probability of Antonio winning is $a/b$ where $a$ and $b$ are relatively prime positive integers. Find$a + b$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve this problem, we need to find the area of the pentagon \(ABOPC\) which is formed by inscribing a square \(ABMC\) in circle \(F\) and an equilateral triangle \(MOP\) in circle \(G\). The circles \(F\) and \(G\) are externally tangent and share the vertex \(M\).

1. **Calculate the side length of the square \(ABMC\):**
   - The radius of circle \(F\) is 10.
   - The diagonal of the square \(ABMC\) is equal to the diameter of circle \(F\), which is \(20\).
   - Using the relationship between the side length \(s\) of a square and its diagonal \(d\), \(d = s\sqrt{2}\), we get:
     \[
     s\sqrt{2} = 20 \implies s = \frac{20}{\sqrt{2}} = 10\sqrt{2}
     \]
   - The area of the square \(ABMC\) is:
     \[
     [ABMC] = s^2 = (10\sqrt{2})^2 = 200
     \]

2. **Calculate the side length of the equilateral triangle \(MOP\):**
   - The radius of circle \(G\) is 4.
   - The side length of the equilateral triangle inscribed in a circle is equal to the radius multiplied by \(\sqrt{3}\):
     \[
     MP = 4\sqrt{3}
     \]
   - The area of the equilateral triangle \(MOP\) is:
     \[
     [MOP] = \frac{\sqrt{3}}{4} \times (4\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 48 = 12\sqrt{3}
     \]

3. **Calculate the area of the triangles \(OMB\) and \(OMC\):**
   - The angle between \(OM\) and \(BM\) is \(105^\circ\) (since \(60^\circ + 45^\circ = 105^\circ\)).
   - Using the sine rule for the area of a triangle:
     \[
     \sin 105^\circ = \sin (60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}
     \]
   - The area of triangle \(OMB\) is:
     \[
     [OMB] = \frac{1}{2} \times 10\sqrt{2} \times 4\sqrt{3} \times \frac{\sqrt{6} + \sqrt{2}}{4} = 30 + 10\sqrt{3}
     \]
   - Since \(OMC\) is congruent to \(OMB\), its area is also \(30 + 10\sqrt{3}\).

4. **Calculate the total area of the pentagon \(ABOPC\):**
   - The total area is the sum of the areas of the square \(ABMC\), the equilateral triangle \(MOP\), and the two triangles \(OMB\) and \(OMC\):
     \[
     [ABOPC] = [ABMC] + [MOP] + 2 \times [OMB] = 200 + 12\sqrt{3} + 2 \times (30 + 10\sqrt{3}) = 200 + 12\sqrt{3} + 60 + 20\sqrt{3} = 260 + 32\sqrt{3}
     \]

5. **Determine the values of \(a\), \(b\), and \(c\):**
   - The area of the pentagon is in the form \(a + b\sqrt{c}\), where \(a = 260\), \(b = 32\), and \(c = 3\).

6. **Calculate \(a + b + c\):**
   \[
   a + b + c = 260 + 32 + 3 = 295
   \]

The final answer is \( \boxed{ 295 } \).