[b]p1.[/b] Let $X,Y ,Z$ be nonzero real numbers such that the quadratic function $X t^2 - Y t + Z = 0$ has the unique root $t = Y$ . Find $X$.


[b]p2.[/b] Let $ABCD$ be a kite with $AB = BC = 1$ and $CD = AD =\sqrt2$. Given that $BD =\sqrt5$, find $AC$.


[b]p3.[/b] Find the number of integers $n$ such that $n -2016$ divides $n^2 -2016$. An integer $a$ divides an integer $b$ if there exists a unique integer $k$ such that $ak = b$.


[b]p4.[/b] The points $A(-16, 256)$ and $B(20, 400)$ lie on the parabola $y = x^2$ . There exists a point $C(a,a^2)$ on the parabola $y = x^2$ such that there exists a point $D$ on the parabola $y = -x^2$ so that $ACBD$ is a parallelogram. Find $a$.


[b]p5.[/b] Figure $F_0$ is a unit square. To create figure $F_1$, divide each side of the square into equal fifths and add two new squares with sidelength $\frac15$ to each side, with one of their sides on one of the sides of the larger square. To create figure $F_{k+1}$ from $F_k$ , repeat this same process for each open side of the smallest squares created in $F_n$. Let $A_n$ be the area of $F_n$. Find $\lim_{n\to \infty} A_n$.
[img]https://cdn.artofproblemsolving.com/attachments/8/9/85b764acba2a548ecc61e9ffc29aacf24b4647.png[/img]


[b]p6.[/b] For a prime $p$, let $S_p$ be the set of nonnegative integers $n$ less than $p$ for which there exists a nonnegative integer $k$ such that $2016^k -n$ is divisible by $p$. Find the sum of all $p$ for which $p$ does not divide the sum of the elements of $S_p$ .


[b]p7. [/b] Trapezoid $ABCD$ has $AB \parallel CD$ and $AD = AB = BC$. Unit circles $\gamma$ and $\omega$ are inscribed in the trapezoid such that circle $\gamma$ is tangent to $CD$, $AB$, and $AD$, and circle $\omega$ is tangent to $CD$, $AB$, and $BC$. If circles $\gamma$ and $\omega$ are externally tangent to each other, find $AB$.


[b]p8.[/b] Let $x, y, z$ be real numbers such that $(x+y)^2+(y+z)^2+(z+x)^2 = 1$. Over all triples $(x, y, z)$, find the maximum possible value of $y -z$.


[b]p9.[/b] Triangle $\vartriangle ABC$ has sidelengths $AB = 13$, $BC = 14$, and $CA = 15$. Let $P$ be a point on segment $BC$ such that $\frac{BP}{CP} = 3$, and let $I_1$ and $I_2$ be the incenters of triangles $\vartriangle ABP$ and $\vartriangle ACP$. Suppose that the circumcircle of $\vartriangle I_1PI_2$ intersects segment $AP$ for a second time at a point $X \ne P$. Find the length of segment $AX$.


[b]p10.[/b] For $1 \le i \le 9$, let Ai be the answer to problem i from this section. Let $(i_1,i_2,... ,i_9)$ be a permutation of $(1, 2,... , 9)$ such that $A_{i_1} < A_{i_2} < ... < A_{i_9}$. For each $i_j$ , put the number $i_j$ in the box which is in the $j$th row from the top and the $j$th column from the left of the $9\times 9$ grid in the bonus section of the answer sheet. Then, fill in the rest
of the squares with digits $1, 2,... , 9$ such that
$\bullet$ each bolded $ 3\times 3$ grid contains exactly one of each digit from $ 1$ to $9$,
$\bullet$ each row of the $9\times 9$ grid contains exactly one of each digit from $ 1$ to $9$, and
$\bullet$ each column of the $9\times 9$ grid contains exactly one of each digit from $ 1$ to $9$.


PS. You had better use hide for answers.
1. Given the quadratic function \( X t^2 - Y t + Z = 0 \) has a unique root \( t = Y \), we need to find \( X \).

   1.1. Since \( t = Y \) is a unique root, it must be a double root. Therefore, the quadratic can be written as:
   \[
   X(t - Y)^2 = 0
   \]
   Expanding this, we get:
   \[
   X(t^2 - 2Yt + Y^2) = 0 \implies X t^2 - 2XY t + XY^2 = 0
   \]
   Comparing this with the original quadratic equation \( X t^2 - Y t + Z = 0 \), we get:
   \[
   -2XY = -Y \quad \text{and} \quad XY^2 = Z
   \]
   From \( -2XY = -Y \), we can solve for \( X \):
   \[
   -2XY = -Y \implies X = \frac{1}{2}
   \]

   1.2. To confirm, substitute \( X = \frac{1}{2} \) back into the equation \( XY^2 = Z \):
   \[
   \frac{1}{2} Y^2 = Z
   \]
   This confirms that the quadratic function \( \frac{1}{2} t^2 - Y t + \frac{1}{2} Y^2 = 0 \) has a double root at \( t = Y \).

   Therefore, the value of \( X \) is \( \boxed{\frac{1}{2}} \).