Pieces of the Puzzle

Start with the information about the hexagon inscribed in a parabola; the area Q can be determined from the given information, rather than the other way around.

Draw the lines above to create three triangles. The side lengths of the equilateral triangles are equal to $S$ , the side lengths of the hexagon. Knowing the properties of $30-60-90$ triangles, it is clear that the $x$ -coordinate of $A$ is $\frac{S}{2}$ , and the height of each triangle is $\frac { s\sqrt { 3 } }{ 2 }$ . Since points A and C lie on the parabola, their $y$ coordinates must equal the square of their $x$ -coordinates.

Using this, we can write a system of equations:

${ y }_{ A }={ \left( \frac { S }{ 2 } \right) }^{ 2 }=\frac { { S }^{ 2 } }{ 4 }$

${ y }_{ A }+\frac { S\sqrt { 3 } }{ 2 } ={ S }^{ 2 }$

Substituting ${ y }_{ A }$ in the second equation:

$\frac { { S }^{ 2 } }{ 4 } +\frac { S\sqrt { 3 } }{ 2 } ={ S }^{ 2 } \Rightarrow 3{ S }^{ 2 }-2\sqrt { 3 } S=0$

Since $S\neq 0$ ,

$S=\frac { 2 }{ \sqrt { 3 } }$

Plugging the value of S into the first equation:

${ y }_{ A }=\frac { 1 }{ 3 }$

The $y$ -coordinate of the center of the hexagon (the distance between the center and the origin) is given by:

${ y }_{ H }{ =y }_{ A }+\frac { S\sqrt { 3 } }{ 2 } =\frac { F }{ G } =\frac { 4 }{ 3 }$

We now know that $F=4$ and $G=3$ .

In addition, the area of the hexagon can be easily calculated:

$Q=\frac { Apothem\quad \times \quad Perimeter }{ 2 } =\frac { \frac { S\sqrt { 3 } }{ 2 } \times 6S }{ 2 } =2\sqrt { 3 } =Q$

Now that Q is known, let's go to the part about the clock.

Since the hour is $\frac{45}{60}=\frac{3}{4}$ of the way complete, and the angle between $12$ and $1$ is $\frac{2\pi}{12}=\frac{\pi}{6}$ :

$\theta = \frac{\pi}{6}\times \frac{3}{4}=\frac{\pi}{8}$

The angle of the total sector, then, is:

$\frac{\pi}{8}+\frac{\pi}{2}=\frac{5\pi}{8}$

The radius of the clock can be found using the formula for the area of a circular sector, since the area Q is already known:

$Q=2\sqrt { 3 } =\frac { 1 }{ 2 } \left( \frac { 5\pi }{ 8 } \right) \left( { r }^{ 2 } \right) \Rightarrow r=\sqrt{\frac { 32\sqrt { 3 } }{ 5\pi } }$

r is one leg of a $30-60-90$ triangle, shown in the diagram. The other leg is half of P, the side length of the equilateral triangle. This fact can be used to calculate P:

$\frac { P }{ 2 } =r\sqrt { 3 } =\sqrt { \frac { 96\sqrt { 3 } }{ 5\pi } } =4\sqrt { \frac { 6\sqrt { 3 } }{ 5\pi } } \Rightarrow P=A\sqrt { \frac { B\sqrt { C } }{ D\pi } } =8\sqrt { \frac { 6\sqrt { 3 } }{ 5\pi } }$

We now know that $A=8$ , $B=6$ , $C=3$ , and $D=5$ . This will be enough information to look at the part about grids.

The prime factorization of $234$ is $2 \times 3^{2} \times 13$ . Therefore, $1$ , $2$ , and $3$ are factors of $234$ , but $4$ is not. There is a $\frac{1}{4}$ chance that the first number is not a factor of $234$ .

The prime factorization of $2340$ is $2^{2} \times 3^{2} \times 5 \times 13$ . Therefore, the following are factors of $2340$ less than or equal to $25$ :

$\left\{ 1,2,3,4,5,6,9,10,12,13,15,18,20 \right\}$

There are $13$ factors out of $25$ total numbers. The probability that the second number is a factor of $2340$ is $\frac{13}{25}$ . Therefore, the probability of both outcomes occurring consecutively is:

$\frac{1}{4}\times \frac{13}{25}=\frac{13}{100}=\frac{H}{100}$

We now know that $H = 13$ . This gives us enough information to look at the problem about the 4-digit integer.

Since $\overline{WX}=\overline{ZY}$ , $\overline{WXYZ}=\overline{WXXW}$ .

Since $W+X+X+W=2(W+X)$ is a power of $2$ , $W+X$ must also be a power of two (a power of $2$ divided by $2$ is still a power of $2$ ).

Since $W \times X \times X \times W = (W\times X)^{2}$ is a power of $3$ , $W\times X$ is also a power of $3$ . It is easy to show that there are only three 4-digit integers for which $W+X$ is a power of $2$ and $W\times X$ is a power of $3$ :

$\left\{1111,1331,3113 \right\}$

The problem states that $\overline{WX}\neq \overline{XW}$ and that $\overline{WXXW}$ has $H-D-B=2$ distinct prime factors. This leaves $3113$ as the only possibility. Therefore, $E=31$ .

We can now go to the polynomial.

$f(x) = DHx^{5} - D^{2}EH^{2}x^{4} + FG^{F}x^{3} - C^{F}DEFHx^{2} + EHx - DE^{2}H^{2}.$

Substituting in all known variables:

$f(x) = 65x^{5} - 130975x^{4} + 324x^{3} - 652860x^{2} + 403x - 812045.$

This polynomial has only one real root $N$ : $\boxed{2015}$