Show that there is no polynomial $p ( x )$ for which $\cos ( \theta ) = p ( \sin \theta )$ for all angles $\theta$ in some nonempty interval.
Hint: Note that $x$ and $| x |$ are different functions but their values are equal on an interval (as $x = | x |$ for all $x \geq 0$). You may want to show as a first step that this cannot happen for two polynomials, i.e., if polynomials $f$ and $g$ satisfy $f ( x ) = g ( x )$ for all $x$ in some interval, then $f$ and $g$ must be equal as polynomials, i.e., in each degree they must have the same coefficient.

A prime $p$ is an integer $\geq 2$ whose only positive integer factors are 1 and $p$.
(a) For any prime $p$ the number $p ^ { 2 } - p$ is always divisible by 3.
(b) For any prime $p > 3$ exactly one of the numbers $p - 1$ and $p + 1$ is divisible by 6.
(c) For any prime $p > 3$ the number $p ^ { 2 } - 1$ is divisible by 24.
(d) For any prime $p > 3$ one of the three numbers $p + 1 , p + 3$ and $p + 5$ is divisible by 8.

[11 points] Given $\triangle XYZ$, the following constructions are made: mark point $W$ on segment $XZ$, point $P$ on segment $XW$ and point $Q$ on segment $YZ$ such that
$$\frac{WZ}{YX} = \frac{PW}{XP} = \frac{QZ}{YQ} = k$$
Extend segments $QP$ and $YX$ to meet at the point $R$ as shown. Prove that $XR = XP$.
Hint (use this or your own method): A suitable construction may help in calculations.

[11 points] In the XY plane, draw horizontal and vertical lines through each integer on both axes so as to get a grid of small $1 \times 1$ squares whose vertices have integer coordinates.
(i) Consider the line segment $D$ joining $(0,0)$ with $(m,n)$. Find the number of small $1 \times 1$ squares that $D$ cuts through, i.e., squares whose interiors $D$ intersects. (Interiors consist of points for which both coordinates are non-integers.) For example, the line segment joining $(0,0)$ and $(2,3)$ cuts through 4 small squares, as you can check by drawing.
(ii) Now $L$ is allowed to be an arbitrary line in the plane. Find the maximum number of small $1 \times 1$ squares in an $n \times n$ grid that $L$ can cut through, i.e., we want $L$ to intersect the interiors of maximum possible number of small squares inside the square with vertices $(0,0)$, $(n,0)$, $(0,n)$ and $(n,n)$.

[15 points] Two distinct real numbers $r$ and $s$ are said to form a good pair $(r, s)$ if
$$r^3 + s^2 = s^3 + r^2$$
(i) Find a good pair $(a, \ell)$ with the largest possible value of $\ell$. Find a good pair $(s, b)$ with the smallest possible value $s$. For every good pair $(c, d)$ other than the two you found, show that there is a third real number $e$ such that $(d, e)$ and $(c, e)$ are also good pairs.
(ii) Show that there are infinitely many good pairs of rational numbers.
Hints (use these or your own method): The function $f(x) = x^3 - x^2$ may be useful. If $(r, s)$ is a good pair, can you express $s$ in terms of $r$? You may use that there are infinitely many right triangles with integer sides such that no two of these triangles are similar to each other.

[14 points] In $\triangle A B C , \angle B A C = 2 \angle A C B$ and $0 ^ { \circ } < \angle B A C < 120 ^ { \circ }$. A point $M$ is chosen in the interior of $\triangle A B C$ such that $B A = B M$ and $M A = M C$. Prove that $\angle M C B = 30 ^ { \circ }$.
Hint (use this or your own method): Draw a suitable segment $C D$ of appropriate length making an appropriate angle with $C A$.

[14 points] Suppose an integer $n > 1$ is such that $n + 1$ is not a multiple of 4 (i.e., such that $n$ is not congruent to $3 \bmod 4$). Prove that there exist $1 \leq i < j \leq n$ such that the following is a perfect square.
$$\frac { 1 ! 2 ! \cdots n ! } { i ! j ! }$$
Hint (use this or your own method): Make cases and first treat the case $n = 4k$.

Suppose a rectangle $EBFD$ is given and a rhombus $ABCD$ is inscribed in it so that the point $A$ is on side $ED$ of the rectangle. The diagonals of $ABCD$ intersect at point $G$.
Statements
(5) Triangles $CGD$ and $DFB$ must be similar. (6) It must be true that $\frac { AC } { BD } = \frac { EB } { ED }$. (7) Triangle $CGD$ cannot be similar to triangle $AEB$. (8) For any given rectangle $EBFD$, a rhombus $ABCD$ as described above can be constructed.

Statements
(17) $4 < \sqrt { 5 + 5 \sqrt { 5 } }$. (18) $\log _ { 2 } 11 < \frac { 1 + \log _ { 2 } 61 } { 2 }$. (19) $( 2023 ) ^ { 2023 } < ( 2023 ! ) ^ { 2 }$. (20) $92 ^ { 100 } + 93 ^ { 100 } < 94 ^ { 100 }$.

(a) For non-negative numbers $a, b, c$ and any positive real number $r$ prove the following inequality and state precisely when equality is achieved. $$a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r(c-a)(c-b) \geq 0$$ Hint: Assuming $a \geq b \geq c$ do algebra with just the first two terms. What about the third term? What if the assumption is not true?
(b) As a special case obtain an inequality with $a^4 + b^4 + c^4 + abc(a+b+c)$ on one side.
(c) Show that if $abc = 1$ for positive numbers $a, b, c$, then $$a^4 + b^4 + c^4 + a^3 + b^3 + c^3 + a + b + c \geq \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} + 3.$$

An integer $d$ is called a factor of an integer $n$ if there is an integer $q$ such that $n = qd$. In particular the set of factors of $n$ contains $n$ and contains 1. You are given that $2024 = 8 \times 11 \times 23$.
Write the number of ordered pairs $(a,b)$ of positive integers such that $a^2 - b^2 = 2024^2$. If there are infinitely many such pairs, write the word infinite as your answer. [3 points]

6. Let $n \geq 3$ be an integer, and let $x _ { 1 } , x _ { 2 } , \ldots , x _ { n }$ be variables which take real values with $0 \leq x _ { i } \leq 1$ for all $1 \leq i \leq n$. Let
$$\begin{aligned} A & = x _ { 1 } + x _ { 2 } + \ldots + x _ { n } \\ B & = x _ { 1 } x _ { 2 } + x _ { 2 } x _ { 3 } + \ldots + x _ { n - 1 } x _ { n } + x _ { n } x _ { 1 } \end{aligned}$$
Which of the following statements is/are true.
(a) $A \geq B$ is always true.
(b) $B > A$ is true for some values of the $x _ { i }$ 's and $A > B$ is true for some values of the $x _ { i }$ 's.
(c) $A = B$ has a finite number of solutions
(d) $A = B$ has an infinite number of solutions.

A function $f ( x )$ continuous on the entire set of real numbers satisfies $$\{ f ( x ) \} ^ { 3 } - \{ f ( x ) \} ^ { 2 } - x ^ { 2 } f ( x ) + x ^ { 2 } = 0$$ for all real numbers $x$. When the maximum value of $f ( x )$ is 1 and the minimum value is 0, what is the value of $f \left( - \frac { 4 } { 3 } \right) + f ( 0 ) + f \left( \frac { 1 } { 2 } \right)$? [4 points]
(1) $\frac { 1 } { 2 }$
(2) 1
(3) $\frac { 3 } { 2 }$
(4) 2
(5) $\frac { 5 } { 2 }$

20. (Total Score: 14 points) Subquestion 1 is worth 6 points, Subquestion 2 is worth 8 points
A quadratic function $y = f _ { 1 } ( x )$ has its vertex at the origin and passes through the point $(1,1)$. A reciprocal function $y = f _ { 2 } ( x )$ has its graph intersecting the line $y = x$ at two points with distance 8 between them. Let $f ( x ) = f _ { 1 } ( x ) + f _ { 2 } ( x )$.
(1) Find the expression for function $f ( x )$;
(2) Prove: When $a > 3$, the equation $f ( x ) = f ( a )$ in $x$ has three real solutions.

21. (Total Score: 16 points) Subquestion 1 is worth 4 points, Subquestion 2 is worth 6 points, Subquestion 3 is worth 6 points As shown in the figure, $P - ABC$ is a regular triangular pyramid with base edge length 1. Points $D$, $E$, $F$ are on edges $PA$, $PB$, $PC$ respectively. The cross-section $DEF$ is parallel to the base $ABC$, and the sum of edge lengths of the frustum $DEF - ABC$ equals the sum of edge lengths of the pyramid $P - ABC$. (The sum of edge lengths is the sum of the lengths of all edges of a polyhedron)
(1) Prove: $P - ABC$ is a regular tetrahedron;
(2) If $PD = \frac { 1 } { 2 } PA$, find the dihedral angle $D - BC - A$. (Express the result using inverse trigonometric functions)
(3) Let the volume of the frustum $DEF - ABC$ be $V$. Does there exist a right parallelepiped with all edges equal and volume $V$ such that it has the same sum of edge lengths as the frustum $DEF - ABC$? If it exists, construct such a parallelepiped explicitly and provide a proof; if it does not exist, explain why. [Figure]

22. (Total Score: 18 points) Subquestion 1 is worth 6 points, Subquestion 2 is worth 8 points, Subquestion 3 is worth 4 points. Let $P _ { 1 } ( x _ { 1 } , y _ { 1 } )$, $P _ { 2 } ( x _ { 2 } , y _ { 2 } )$, $\cdots$, $P _ { n } ( x _ { n } , y _ { n } )$ ($n \geq 3$, $n \in \mathbb{N}$) be points on a conic section C, and let $a _ { 1 } = \left| OP _ { 1 } \right| ^ { 2 From $\left\{ \begin{array} { l } \frac { x ^ { 2 } } { 100 } + \frac { y ^ { 2 } } { 25 } = 1 \\ x _ { 3 } ^ { 2 } + y _ { 3 } ^ { 2 } = 70 \end{array} \right.$, we obtain $\left\{ \begin{array} { l } x _ { 3 } ^ { 2 } = 60 \\ y _ { 3 } ^ { 2 } = 10 \end{array} \right.$ $\therefore$ The coordinates of point $P_{3}$ can be $( 2 \sqrt { 15 } , \sqrt { 10 } )$.
(2) Solution 1: The minimum distance from the origin $O$ to each point on the conic curve $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ $(a > b > 0)$ is $b$, and the maximum distance is $a$. $\because a _ { 1 } = \left| O P _ { 1 } \right| ^ { 2 } = a ^ { 2 }$, $\therefore d < 0$, and $a _ { n } = \left| O P _ { n } \right| ^ { 2 } = a ^ { 2 } + ( n - 1 ) d \geq b ^ { 2 }$, $\therefore \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } \leq d < 0$. $\because n \geq 3$, $\frac { n ( n - 1 ) } { 2 } > 0$, $\therefore S _ { n } = n a ^ { 2 } + \frac { n ( n - 1 ) } { 2 } d$ is increasing on $\left[ \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } , 0 \right)$, thus the minimum value of $S _ { n }$ is $n a ^ { 2 } + \frac { n ( n - 1 ) } { 2 } \cdot \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } = \frac { n \left( a ^ { 2 } + b ^ { 2 } \right) } { 2 }$. Solution 2: For each natural number $k$ $(2 \leq k \leq n)$, from $\left\{ \begin{array} { l } x _ { k } ^ { 2 } + y _ { k } ^ { 2 } = a ^ { 2 } + ( k - 1 ) d \\ \frac { x _ { k } ^ { 2 } } { a ^ { 2 } } + \frac { y _ { k } ^ { 2 } } { b ^ { 2 } } = 1 \end{array} \right.$, we solve to get $y _ { k } ^ { 2 } = \frac { - b ^ { 2 } ( k - 1 ) d } { a ^ { 2 } - b ^ { 2 } }$ $\because 0 < y _ { k } ^ { 2 } \leq b ^ { 2 }$, we obtain $\frac { b ^ { 2 } - a ^ { 2 } } { k - 1 } \leq d < 0$, $\therefore \frac { b ^ { 2 } - a ^ { 2 } } { n - 1 } \leq d < 0$. The rest is the same as Solution 1.
(3) Solution 1: If the hyperbola $C: \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, and point $P _ { 1 } ( a , 0 )$, then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $d > 0$. $\because$ The distance from the origin $O$ to each point on the hyperbola $C$ is $h \in [ |a| , +\infty )$, and $\left| O P _ { 1 } \right| = a ^ { 2 }$, $\therefore$ Points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ exist if and only if $\left| O P _ { n } \right| ^ { 2 } > \left| O P _ { 1 } \right| ^ { 2 }$, i.e., $d > 0$. Solution 2: If the parabola $C : y ^ { 2 } = 2px$, and point $P _ { 1 } ( 0 , 0 )$, then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $d > 0$. The reasoning is the same as above. Solution 3: If the circle $C: ( x - a ) ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ $(a \neq 0)$, and $P _ { 1 } ( 0 , 0 )$, then for a given $n$, the necessary and sufficient condition for the existence of points $P _ { 1 } , P _ { 2 } , \cdots , P _ { n }$ is $0 < d \leq \frac { 4 a ^ { 2 } } { n - 1 }$. $\because$ The minimum distance from the origin $O$ to each point on the circle $C$ is $0$, and the maximum distance is $2 |a|$, and $\left| O P _ { 1 } \right| ^ { 2 } = 0$, $\therefore d > 0$ and $\left| O P _ { n } \right| ^ { 2 } = ( n - 1 ) d \leq 4 a ^ { 2 }$, i.e., $0 < d \leq \frac { 4 a ^ { 2 } } { n - 1 }$.

6. Given a quadrangular pyramid $P - A B C D$ with a square base of side length 6, the lateral edge $P A \perp$ base $A B C D$ , and $P A = 8$ , then the volume of this quadrangular pyramid is $\_\_\_\_$.

3. Let $\mathrm { p } : x < 3 , \mathrm { q } : - 1 < x < 3$. Then $p$ is a condition for $q$ to hold that is
(A) necessary and sufficient
(B) sufficient but not necessary
(C) necessary but not sufficient
(D) neither sufficient nor necessary

Let $\alpha , \beta$ be two planes. Then a necessary and sufficient condition for $\alpha \parallel \beta$ is
A. There are infinitely many lines in $\alpha$ that are parallel to $\beta$
B. There are two intersecting lines in $\alpha$ that are parallel to $\beta$
C. $\alpha$ and $\beta$ are both parallel to the same line
D. $\alpha$ and $\beta$ are both perpendicular to the same plane

Let $l , m$ be two different lines outside plane $\alpha$. Three propositions are given: (1) $l \perp m$; (2) $m \| \alpha$; (3) $l \perp \alpha$.
Using two of these propositions as conditions and the remaining one as the conclusion, write out a correct proposition: $\_\_\_\_$.

23. Solution: (1) Since $a^2 + b^2 \geq 2ab$, $b^2 + c^2 \geq 2bc$, $c^2 + a^2 \geq 2ac$, and $abc = 1$, we have $a^2 + b^2 + c^2 \geq ab + bc + ca = \frac{ab + bc + ca}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Therefore $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq a^2 + b^2 + c^2$.
(2) Since $a, b, c$ are positive numbers and $abc = 1$, we have $(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 3\sqrt[3]{(a+b)^3(b+c)^3(a+c)^3}$ $= 3(a+b)(b+c)(a+c)$ $\geq 3 \times (2\sqrt{ab}) \times (2\sqrt{bc}) \times (2\sqrt{ac})$ $= 24$. Therefore $(a+b)^3 + (b+c)^3 + (c+a)^3 \geq 24$.

23. Solution: (1) Since $[(x-1) + (y+1) + (z+1)]^2$
$= (x-1)^2 + (y+1)^2 + (z+1)^2 + 2[(x-1)(y+1) + (y+1)(z+1) + (z+1)(x-1)]$
$\leq 3[(x-1)^2 + (y+1)^2 + (z+1)^2]$,
from the given condition we have $(x-1)^2 + (y+1)^2 + (z+1)^2 \geq \frac{4}{3}$,
with equality if and only if $x = \frac{5}{3}$, $y = -\frac{1}{3}$, $z = -\frac{1}{3}$.
Therefore, the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$ is $\frac{4}{3}$.
(2) Since
$[(x-2) + (y-1) + (z-a)]^2$
$= (x-2)^2 + (y-1)^2 + (z-a)^2 + 2[(x-2)(y-1) + (y-1)(z-a) + (z-a)(x-2)]$
$\leq 3[(x-2)^2 + (y-1)^2 + (z-a)^2]$,
from the given condition $(x-2)^2 + (y-1)^2 + (z-a)^2 \geq \frac{(2+a)^2}{3}$,
with equality if and only if $x = \frac{4-a}{3}$, $y = \frac{1-a}{3}$, $z = \frac{2a-2}{3}$.
Therefore, the minimum value of $(x-2)^2 + (y-1)^2 + (z-a)^2$ is $\frac{(2+a)^2}{3}$.
From the given condition, $\frac{(2+a)^2}{3} \geq \frac{1}{3}$, solving gives $a \leq -3$ or $a \geq -1$.

The Great Pyramid of Khufu in Egypt is one of the ancient wonders of the world. Its shape can be viewed as a regular square pyramid. The area of a square with side length equal to the height of the pyramid equals the area of one lateral triangular face of the pyramid. Then the ratio of the height of the lateral triangle to the base of the square is
A. $\frac { \sqrt { 5 } - 1 } { 4 }$
B. $\frac { \sqrt { 5 } - 1 } { 2 }$
C. $\frac { \sqrt { 5 } + 1 } { 4 }$
D. $\frac { \sqrt { 5 } + 1 } { 2 }$

As shown in the figure, $D$ is the apex of the cone, $O$ is the center of the base of the cone, $\triangle A B C$ is an equilateral triangle inscribed in the base, and $P$ is a point on $D O$ with $\angle A P C = 90 ^ { \circ }$ .
(1) Prove that plane $P A B \perp$ plane $P A C$ ;
(2) Given $D O = \sqrt { 2 }$ and the lateral surface area of the cone is $\sqrt { 3 } \pi$ , find the volume of the triangular pyramid $P - A B C$ .

[Elective 4-5: Inequalities] Let $a , b , c \in \mathbf { R } , a + b + c = 0 , a b c = 1$ .
(1) Prove: $a b + b c + c a < 0$;
(2) Let $\max \{ a , b , c \}$ denote the maximum value among $a , b , c$. Prove: $\max \{ a , b , c \} \geqslant \sqrt[3]{\frac{3}{2}}$.