16. China has a long history of stone and seal culture, and seals are a representative of this culture. Seals are usually shaped as rectangular prisms, cubes, or cylinders, but the seal of the official Dugu Xin from the Northern and Southern Dynasties is shaped as a ``semi-regular polyhedron'' (Figure 1). A semi-regular polyhedron is a polyhedron formed by two or more types of regular polygons. Semi-regular polyhedra embody the symmetry of mathematics. Figure 2 shows a semi-regular polyhedron with 48 edges. All its vertices lie on the surface of a cube with edge length 1. This semi-regular polyhedron has $\_\_\_\_$ faces and edge length $\_\_\_\_$ . (The first blank is worth 2 points, the second blank is worth 3 points.)
[Figure]
Figure 1
[Figure]
Figure 2

III. Solution Questions: 70 points total. Show your work, proofs, or calculation steps. Questions 17-21 are required for all students. Questions 22 and 23 are optional; students answer according to requirements.

(A) Required Questions: 60 points total.

16. Students participate in labor practice at a factory using 3D printing technology to make models. As shown in the figure, the model is a rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ with a square pyramid $O - E F G H$ removed, where $O$ is the center of the rectangular prism, and $E , F , G , H$ are the midpoints of their respective edges. $A B = B C = 6 \mathrm {~cm} , ~ A A _ { 1 } = 4 \mathrm {~cm}$ . The density of the 3D printing material is $0.9 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ . Disregarding printing waste, the mass of material needed to make this model is $\_\_\_\_$ g. [Figure]
III. Solution Questions: Total 70 points. Solutions should include written explanations, proofs, or calculation steps. Questions 17--21 are required questions that all candidates must answer. Questions 22 and 23 are optional questions; candidates answer according to requirements.
(A) Required Questions: Total 60 points.

16. Students participate in labor practice at a factory using 3D printing technology to make models. As shown in the figure, the model is the solid obtained by removing the square pyramid $O - E F G H$ from the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ , where $O$ is the center of the rectangular prism, and $E , F , G , H$ are the midpoints of their respective edges. Given $A B = B C = 6 \mathrm {~cm} , ~ A A _ { 1 } = 4 \mathrm {~cm}$ , and the density of the 3D printing material is $0.9 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ . Disregarding printing waste, the mass of material needed to make this model is $\_\_\_\_$ g. [Figure]

III. Solution Questions: Total 70 points. Solutions should include written explanations, proofs, or calculation steps. Questions 17-21 are required for all students. Questions 22 and 23 are optional; students answer according to requirements.

(A) Required Questions: Total 60 points.

Since the reform and opening up, people's payment methods have undergone tremendous changes. In recent years, mobile payment has become one of the main payment methods. To understand the usage of two mobile payment methods, $A$ and $B$, among students at a certain school last month, 100 students were randomly selected from the entire school. It was found that 5 people in the sample used neither method. The distribution of payment amounts for students in the sample who used only method $A$ and only method $B$ is as follows:

Payment Amount (yuan)	$( 0,1000 ]$	$( 1000,2000 ]$	Greater than 2000
Payment Method
Using only $A$	18 people	9 people	3 people

(I) Randomly select 1 student from the entire school. Estimate the probability that this student used both payment methods $A$ and $B$ last month; (II) Randomly select 1 student each from the sample students who used only $A$ and only $B$. Let $X$ denote the number of people among these 2 people whose payment amount last month exceeded 1000 yuan. Find the probability distribution and mathematical expectation of $X$; (III) It is known that the payment methods of sample students did not change this month. Now, 3 students are randomly selected from the sample students who used only method $A$, and it is found that their payment amounts this month all exceeded 2000 yuan. Based on the sampling results, can we conclude that the number of students using only method $A$ in the sample whose payment amount this month exceeded 2000 yuan has changed? Explain the reasoning.

17. (12 points) To understand the retention levels of two types of ions in mice, the following experiment was conducted: 200 mice were randomly divided into groups $A$ and $B$, with 100 mice in each group. Group $A$ mice were given a solution of ion type 1, and group $B$ mice were given a solution of ion type 2. Each mouse was given the same volume of solution with the same molar concentration. After a period of time, the percentage of ions retained in the mice's bodies was measured using a scientific method. Based on the experimental data, the following histograms were obtained: [Figure] [Figure]
Let $C$ be the event: ``the retention percentage of ion type 2 in the body is at least 5.5.'' Based on the histogram, the estimated value of $P ( C )$ is 0.70.
(1) Find the values of $a$ and $b$ in the histogram for ion type 2's retention percentage.
(2) Estimate the mean retention percentages for ions of type 1 and type 2 respectively (use the midpoint of each interval as the representative value for data in that interval).

17. (12 points) To understand the residual levels of two types of ions in mice, the following experiment was conducted: 200 mice were randomly divided into groups A and B, with 100 mice in each group. Group A mice were given a solution of ion type 甲, and group B mice were given a solution of ion type 乙. Each mouse was given the same volume of solution with the same molar concentration. After a period of time, a scientific method was used to measure the percentage of ions remaining in the mice's bodies. Based on the experimental data, the following histograms were obtained: [Figure]
Let $C$ be the event: ``the residual percentage of ion 乙 in the body is not less than 5.5''. Based on the histogram, the estimated value of $P ( C )$ is 0.70.
(1) Find the values of $a$ and $b$ in the histogram for the residual percentage of ion 乙;
(2) Estimate the mean residual percentage for ions 甲 and 乙 respectively (use the midpoint of each interval as the representative value for data in that interval).

\section*{18. (12 Let $CF = a$. In $\triangle AFM$, $AM = \sqrt{3}$, $FM = \sqrt{a^2 + 1}$, $AF = \sqrt{a^2 + 4}$.
Then $\frac{1}{2} \times \sqrt{3} \times \sqrt{a^2 + 1} = \frac{1}{2} \times \sqrt{a^2 + 4} \times \frac{\sqrt{30}}{5}$. Solving, we get $a = 1$.
Thus $V_{C-AFM} = V_{F-ACM} = \frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{6}$.

19. (12 points) Figure 1 is a planar figure composed of rectangle $A D E B$ , right triangle $A B C$ , and rhombus $B F G C$ , where $A B = 1 , B E = B F = 2$ , $\angle F B C = 60 ^ { \circ }$ . Fold it along $A B$ and $B C$ so that $B E$ and $B F$ coincide, and connect $D G$ , as shown in Figure 2.
(1) Prove: In Figure 2, points $A , C , G , D$ are coplanar, and plane $A B C \perp$ plane $B C G E$ .
(2) Therefore, from the known condition we have $( x - 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - a ) ^ { 2 } \geq \frac { ( 2 + a ) ^ { 2 } } { 3 }$ , equality holds if and only if $x = \frac { 4 - a } { 3 } , y = \frac { 1 - a } { 3 } , z = \frac { 2 a - 2 } { 3 }$ . Thus the minimum value of $( x - 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - a ) ^ { 2 }$ is $\frac { ( 2 + a ) ^ { 2 } } { 3 }$ .
From the given condition we have $\frac { ( 2 + a ) ^ { 2 } } { 3 } \geq \frac { 1 } { 3 }$ , solving gives $a \leq - 3$ or $a \geq - 1$ .

21. (12 points) Given the function $f ( x ) = ( x

During the COVID-19 pandemic prevention and control period, a supermarket opened online sales services and can complete 1200 orders per day. Due to a sharp increase in order volume, orders have accumulated. To solve this problem, many volunteers eagerly signed up to help with order fulfillment. It is known that the supermarket had 500 accumulated unfulfilled orders on a certain day, and the probability that the next day's new orders exceed 1600 is 0.05. Each volunteer can complete 50 orders per day. To ensure that the probability of completing accumulated orders and current day orders within two days is at least 0.95, the minimum number of volunteers needed is
A． 10 people
B． 18 people
C． 24 people
D． 32 people

Given the determinant $\left| \begin{array} { l l l } 1 & a & c \\ 2 & d & b \\ 3 & 0 & 0 \end{array} \right| = 6$, find the determinant $\left| \begin{array} { l l } a & c \\ d & b \end{array} \right| =$ $\_\_\_\_$

The figure on the right shows the three views of a polyhedron. One endpoint of a certain edge of this polyhedron corresponds to point $M$ in the front view and point $N$ in the top view. This endpoint corresponds to which point in the side view?
A．$E$
B．$F$
C．$G$
D．$H$

The figure on the right shows the three-view drawing of a certain solid. The surface area of this solid is
A. $6 + 4 \sqrt { 2 }$
B. $4 + 4 \sqrt { 2 }$
C. $6 + 2 \sqrt { 3 }$
D. $4 + 2 \sqrt { 3 }$

The figure on the right shows the three-view drawing of a certain geometric solid. The surface area of this solid is
A. $6 + 4 \sqrt { 2 }$
B. $4 + 4 \sqrt { 2 }$
C. $6 + 2 \sqrt { 3 }$
D. $4 + 2 \sqrt { 3 }$

It is known that $\triangle A B C$ is an equilateral triangle with area $\frac { 9 \sqrt { 3 } } { 4 }$ , and all its vertices lie on the surface of sphere $O$. If the surface area of sphere $O$ is $16 \pi$ , then the distance from $O$ to plane $A B C$ is
A．$\sqrt { 3 }$
B．$\frac { 3 } { 2 }$
C． 1
D．$\frac { \sqrt { 3 } } { 2 }$

0-1 periodic sequences have important applications in communication technology. If a sequence $a _ { 1 } a _ { 2 } \cdots a _ { n } \cdots$ satisfies $a _ { i } \in \{ 0,1 \} ( i = 1,2 , \cdots )$ and there exists a positive integer $m$ such that $a _ { i + m } = a _ { i } ( i = 1,2 , \cdots )$ , then it is called a 0-1 periodic sequence, and the smallest positive integer $m$ satisfying $a _ { i + m } = a _ { i } ( i = 1,2 , \cdots )$ is called the period of this sequence. For a 0-1 sequence $a _ { 1 } a _ { 2 } \cdots a _ { n } \cdots$ with period $m$ , $C ( k ) = \frac { 1 } { m } \sum _ { i = 1 } ^ { m } a _ { i } a _ { i + k } ( k = 1,2 , \cdots , m - 1 )$ is an important index describing its properties. Among the following 0-1 sequences with period 5, the one satisfying $C ( k ) \leqslant \frac { 1 } { 5 } ( k = 1,2,3,4 )$ is
A． $11010 \ldots$
B． $11011 \cdots$
C． $10001 \cdots$
D． $11001 \cdots$

A cone has a base radius of 1 and slant height of 3. The volume of the largest sphere that can be inscribed in this cone is $\_\_\_\_$ .

Consider the following four propositions:
$p _ { 1 }$ : Three lines that are pairwise intersecting and do not pass through the same point must lie in the same plane.
$p _ { 2 }$ : Through any three points in space, there is exactly one plane.
$p _ { 3 }$ : If two lines in space do not intersect, then these two lines are parallel.
$p _ { 4 }$ : If line $l \subset$ plane $\alpha$ and line $m \perp$ plane $\alpha$ , then $m \perp l$ .
The sequence numbers of all true propositions among the following statements are $\_\_\_\_$.
(1) $p _ { 1 } \wedge p _ { 4 }$
(2) $p _ { 1 } \wedge p _ { 2 }$
(3) $\neg p _ { 2 } \vee p _ { 3 }$
(4) $\neg p _ { 3 } \vee \neg p _ { 4 }$

If there exists $a \in \mathbb{R}$ with $a \neq 0$ such that for all $x \in \mathbb{R}$, the inequality $f ( x + a ) < f ( x ) + f ( a )$ always holds, then function $f ( x )$ is said to have property $P$. Given: $q _ { 1 }$: $f ( x )$ is monotonically decreasing and $f ( x ) > 0$ always holds; $q _ { 2 }$: $f ( x )$ is monotonically increasing and there exists $x _ { 0 } < 0$ such that $f \left( x _ { 0 } \right) = 0$. Which is a sufficient condition for $f ( x )$ to have property $P$? ( )
A. Only $q _ { 1 }$
B. Only $q _ { 2 }$
C. Both $q _ { 1 }$ and $q _ { 2 }$
D. Neither $q _ { 1 }$ nor $q _ { 2 }$

A factory accepted a processing contract. The processed products (unit: pieces) are classified into four grades: A, B, C, and D according to standards. According to the contract: for grade A, B, and C products, the customer pays processing fees of 90 yuan, 50 yuan, and 20 yuan per piece respectively; for grade D products, the factory must compensate 50 yuan per piece for raw material loss. The factory has two branch factories, Factory A and Factory B, that can undertake the processing contract. Factory A has a processing cost of 25 yuan per piece, and Factory B has a processing cost of 20 yuan per piece. To decide which branch factory should undertake the contract, the factory conducted trial processing of 100 pieces of this product at each branch factory and recorded the grades of these products, as shown below:
Frequency Distribution Table of Product Grades for Factory A:

Grade	A	B	C	D
Frequency	40	20	20	20

Frequency Distribution Table of Product Grades for Factory B:

Grade	A	B	C	D
Frequency	28	17	34	21

(1) Estimate the probability that a product from Factory A and Factory B respectively is grade A;
(2) Find the average profit for 100 products from Factory A and Factory B respectively. Based on average profit, which branch factory should the factory choose to undertake the contract?

A square $ABCD$ with side length 1 is rotated around $BC$ to form a cylinder.
(1) Find the surface area of the cylinder;
(2) The square $ABCD$ is rotated counterclockwise by $\frac { \pi } { 2 }$ around $BC$ to position $A _ { 1 } B C D _ { 1 }$. Find the angle between $A D _ { 1 }$ and plane $ABCD$.

1. B

Solution: Note that $4 \notin A , 3 \in A \cap B$, so the answer is $B$.

3. B

Solution: Let the slant height of the cone be $l$. According to the property that the arc length of a semicircle equals the circumference of the cone's base, we have $\pi l = 2 \sqrt { 2 } \pi$, so $l = 2 \sqrt { 2 }$.

4. The BeiDou-3 Global Navigation Satellite System is an important achievement of China's space program. In satellite navigation systems, geostationary satellites orbit in the plane of Earth's equator at an orbital altitude of 36,000 km (orbital altitude is the distance from the satellite to Earth's surface). Consider Earth as a sphere with center $O$ and radius $r = 6400$ km. The latitude of a point $A$ on Earth's surface is defined as the angle that $OA$ makes with the equatorial plane. The maximum latitude at which a geostationary satellite can be directly observed from Earth's surface is $a$. The surface area covered by the satellite signal on Earth's surface is $S = 2 \pi r ^ { 2 } ( 1 - \cos a )$ (in $\mathrm { km } ^ { 2 }$). What percentage of Earth's total surface area does $S$ represent? ( )
A. $26 \%$
B. $34 \%$
C. $42 \%$
D. $50 \%$
【Answer】C 【Solution】 【Analysis】From the given information, use the provided surface area formula and the formula for the surface area of a sphere to calculate the result. 【Detailed Solution】From the given information, the percentage of $S$ relative to Earth's surface area is approximately: $\frac { 2 \pi r ^ { 2 } ( 1 - \cos a ) } { 4 \pi r ^ { 2 } } = \frac { 1 - \cos a } { 2 } = \frac { 1 - \frac { 6400 } { 6400 + 36000 } } { 2 } \approx 0.42 = 42 \%$ . Therefore, the answer is: C.

5. A right square frustum has upper and lower base edges of lengths 2 and 4 respectively, and lateral edge length 2. Its volume is ( )
A. $20 + 12 \sqrt { 3 }$
B. $28 \sqrt { 2 }$
C. $\frac { 56 } { 3 }$
D. $\frac { 28 \sqrt { 2 } } { 3 }$
【Answer】D 【Solution】 【Analysis】Using the geometric properties of the frustum, calculate the height of the solid and the areas of the upper and lower bases, then use the volume formula for a frustum.
【Detailed Solution】Draw a figure and connect the centers of the upper and lower bases of the right square frustum, as shown in the diagram. [Figure]
Since the upper and lower base edges of the frustum are 2 and 4 respectively, and the lateral edge length is 2, the height of the frustum is $h = \sqrt { 2 ^ { 2 } - ( 2 \sqrt { 2 } - \sqrt { 2 } ) ^ { 2 } } = \sqrt { 2 }$, the area of the lower base is $S _ { 1 } = 16$, and the area of the upper base is $S _ { 2 } = 4$, so the volume of the frustum is $V = \frac { 1 } { 3 } h \left( S _ { 1 } + S _ { 2 } + \sqrt { S _ { 1 } S _ { 2 } } \right) = \frac { 1 } { 3 } \times \sqrt { 2 } \times ( 16 + 4 + \sqrt { 64 } ) = \frac { 28 } { 3 } \sqrt { 2 }$ . Therefore, the answer is: D.