Two cones A and B have equal slant heights. The sum of the central angles of their lateral surface development diagrams is $2 \pi$. Let their lateral surface areas be $S_{\text{甲}}$ and $S_{\text{乙}}$, and their volumes be $V_{\text{甲}}$ and $V_{\text{乙}}$. If $\frac { S_{\text{甲}} } { S_{\text{乙}} } = 2$ , then $\frac { V_{\text{甲}} } { V_{\text{乙}} } =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 2 }$
C. $\sqrt { 10 }$
D. $\frac { 5 \sqrt { 10 } } { 4 }$

19. (12 points) As shown in the figure, a right triangular prism $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$ has volume 4, and the area of $\triangle A _ { 1 } B C$ is $2 \sqrt { 2 }$.
(1) Find the distance from $A$ to plane $A _ { 1 } B C$;
(2) Let $D$ be the midpoint of $A _ { 1 } C$, with $A A _ { 1 } = A B$ and plane $A _ { 1 } B C \perp$ plane [Figure] $A B B _ { 1 } A _ { 1 }$. Find the sine of the dihedral angle $A - B D - C$.

Execute the following flowchart, the output $B =$
A. $21$
B. $34$
C. $55$
D. $89$

As shown in the figure, the three-view drawing of a part drawn on grid paper, where the side length of each small square is 1. The volume of this part is
A. 24
B. 26
C. 28
D. 30

A cone with apex $P$ and base center $O$ has base radius $\sqrt { 3 }$. $PA$ and $PB$ are slant heights of the cone, $\angle A O B = 120 ^ { \circ }$. If the area of $\triangle P A B$ equals $\frac { 9 \sqrt { 3 } } { 4 }$, then the volume of the cone is
A. $\pi$
B. $\sqrt { 6 } \pi$
C. $3 \pi$
D. $3 \sqrt { 6 } \pi$

Given that $\triangle A B C$ is an isosceles right triangle with $AB$ as the hypotenuse, $\triangle A B D$ is an equilateral triangle, and the dihedral angle $C - A B - D$ is $150 ^ { \circ }$, then the tangent of the angle between line $C D$ and plane $A B C$ is
A. $\frac { 1 } { 5 }$
B. $\frac { \sqrt { 2 } } { 5 }$
C. $\frac { \sqrt { 3 } } { 5 }$
D. $\frac { 2 } { 5 }$

A cone has vertex $P$ and base center $O$. $AB$ is a diameter of the base, $\angle APB=120°$, $AP=2$. Point $C$ is on the base circle, and the dihedral angle $P$-$AC$-$O=45°$. Then
A. the volume of the cone is $\pi$
B. the lateral surface area of the cone is $4\sqrt{3}\pi$
C. $AC=2\sqrt{2}$
D. the area of $\triangle PAC$ is $\sqrt{3}$

In the cube $ABCD - A_{1}B_{1}C_{1}D_{1}$ , let $E , F$ be the midpoints of $CD , A_{1}B_{1}$ respectively. The total number of intersection points of the sphere with diameter $EF$ and each edge of the cube is $\_\_\_\_$ .

In the triangular prism $ABC - A_{1}B_{1}C_{1}$ , $AA_{1} = 2$ , $A_{1}C \perp$ base $ABC$ , $\angle ACB = 90^{\circ}$ , the distance from $A_{1}$ to plane $BCC_{1}B_{1}$ is 1 .
(1) Prove: $AC = A_{1}C$ ;
(2) If the distance between lines $AA_{1}$ and $BB_{1}$ is 2 , find the sine of the angle between $AB_{1}$ and plane $BCC_{1}B_{1}$ .

As shown in the figure, in the triangular pyramid $P - A B C$, $A B \perp B C$, $A B = 2$, $B C = 2 \sqrt { 2 }$, $P B = P C = \sqrt { 6 }$. The midpoints of $BP$, $AP$, and $BC$ are $D$, $E$, and $O$ respectively. $A D = \sqrt { 5 } D O$. Point $F$ is on $AC$ such that $B F \perp A O$.
(1) Prove that $EF \parallel$ plane $BEF$.
(2) Prove that plane $A D O \perp$ plane $B E F$.
(3) Find the sine of the dihedral angle $D - A O - C$.

[Elective 4-4: Coordinate Systems and Parametric Equations]
Given $P(2,1)$ and line $l : \left\{ \begin{array}{l} x = 2 + t\cos\alpha \\ y = 1 + t\sin\alpha \end{array} \right.$ ($t$ is a parameter). Line $l$ intersects the positive $x$-axis and positive $y$-axis at points $A , B$ respectively, with $|PA| \cdot |PB| = 4$ .
(1) Find the value of $\alpha$ ;
(2) With the origin as the pole and the positive $x$-axis as the polar axis, find the polar equation of line $l$ .

[Elective 4-4] (10 points) In the rectangular coordinate system $x O y$, with the origin $O$ as the pole and the positive $x$-axis as the polar axis, establish a polar coordinate system. The polar equation of curve $C _ { 1 }$ is $\rho = 2 \sin \theta \left( \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 2 } \right)$. Curve $C _ { 2 }$ is given (see original paper for full problem statement).

A closed cylindrical container with base radius 4 cm and height 9 cm (container wall thickness is negligible) contains two iron spheres of equal radius. The maximum radius of the iron spheres is \_\_\_\_ cm.

For every integer $n \in \mathbb{N}$, we set $F_n(x) = \cos(n \arccos x)$.
Write a function \texttt{chebyshev} that takes as argument an integer $n$ and returns the display of the expression $F_n(x)$. Use the programming language associated with the usual computer algebra software.

We propose to prove Witt's theorem, whose statement is: ``let $F$ and $F'$ be two vector subspaces of $E$ such that there exists an isometry $f$ from $(F, q_{/F})$ to $(F', q_{/F'})$. Then there exists $g \in O(E,q)$ such that $g_{/F} = f$.''
Show that we can reduce to the case where $F$ and $F'$ are non-singular.

Let $\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}, \varepsilon_{4}$ be four strictly positive real numbers pairwise distinct and two strictly positive real numbers $E$ and $N$. Let $\Omega$ be the part, assumed to be non-empty, formed of the quadruplets $x = (x_{1}, x_{2}, x_{3}, x_{4})$ of $\mathbb{R}_{+}^{4}$ satisfying: $$\left\{\begin{array}{l} x_{1} + x_{2} + x_{3} + x_{4} = N \\ \varepsilon_{1} x_{1} + \varepsilon_{2} x_{2} + \varepsilon_{3} x_{3} + \varepsilon_{4} x_{4} = E \end{array}\right.$$ We define the function $F$ for all $x = (x_{1}, x_{2}, x_{3}, x_{4})$ of $\mathbb{R}_{+}^{4}$ by $$F(x_{1}, x_{2}, x_{3}, x_{4}) = -\sum_{i=1}^{4} \ln \Gamma(1 + x_{i})$$ We suppose that there exists $\bar{N} = (N_{1}, N_{2}, N_{3}, N_{4}) \in \Omega$, the numbers $N_{1}, N_{2}, N_{3}, N_{4}$ all being non-zero, such that $$\max_{x \in \Omega} F(x) = F(\bar{N})$$ Show the existence of two real numbers $\lambda$ and $\mu$ satisfying for all $i \in \{1,2,3,4\}$: $$\ln N_{i} + \frac{1}{2N_{i}} + \int_{0}^{+\infty} \frac{h(u)}{(u+N_{i})^{2}} du = \lambda + \mu \varepsilon_{i}$$

For $n \in \mathbb{N}$ and $k \in \llbracket 0; n \rrbracket$, we denote by $\binom{n}{k}$ the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Let $p \in \mathbb{N}^*$. Show that $\binom{2p}{p}$ is an even integer.
Deduce that, if $n \in \mathbb{N}^*$ and $p \in \llbracket 1; n \rrbracket$, then $\binom{n+p}{p}\binom{n}{p}$ is an even integer.

For $n \in \mathbb{N}$ and $k \in \llbracket 0; n \rrbracket$, we denote by $\binom{n}{k}$ the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. The family $\left(K_n\right)_{n \in \mathbb{N}}$ is the orthonormal family defined in question II.E.
For all $n \in \mathbb{N}$, show that we can write: $$K_n = \sqrt{2n+1} \, \Lambda_n$$ where $\Lambda_n$ is a polynomial with integer coefficients that we will make explicit.
Among the coefficients of $\Lambda_n$, which ones are even?

For $n \in \mathbb{N}^*$ and $(i,j) \in \llbracket 1, n \rrbracket^2$, we denote by $h_{i,j}^{(-1,n)}$ the coefficient at position $(i,j)$ of the matrix $H_n^{-1}$. For $n \in \mathbb{N}$ and $k \in \llbracket 0; n \rrbracket$, we denote by $\binom{n}{k}$ the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. The family $\left(K_p\right)_{p \in \mathbb{N}}$ is the orthonormal family defined in question II.E, and $K_n = \sqrt{2n+1}\,\Lambda_n$ where $\Lambda_n$ is a polynomial with integer coefficients.
Let $n \in \mathbb{N}^*$.
a) Calculate $h_{i,i}^{(-1,n)}$ for all $i \in \llbracket 1; n \rrbracket$; we will give in particular a very simple expression of $h_{1,1}^{(-1,n)}$ and $h_{n,n}^{(-1,n)}$ as a function of $n$.
b) Calculate $h_{i,j}^{(-1,n)}$ for all pairs $(i,j) \in \llbracket 1; n \rrbracket^2$; deduce that the coefficients of $H_n^{-1}$ are integers.
c) Show that $h_{i,j}^{(-1,n)}$ is divisible by 4 for all pairs $(i,j) \in \llbracket 2; n \rrbracket^2$.

Let $\pi$ be the endomorphism of $\mathbb{R}^n$ whose representation in the canonical basis is the matrix $P = I_n - \frac{1}{n}J$, where $J = Z^t Z$ and $Z = \left(\begin{array}{c}1\\ \vdots \\ 1\end{array}\right) \in \mathcal{M}_{n,1}(\mathbb{R})$.
Show that $\pi$ is an orthogonal projector and specify its characteristic elements.

We consider the endomorphism $\Phi$ of $\mathcal{M}_n(\mathbb{R})$ defined by: $$\forall M \in \mathcal{M}_n(\mathbb{R}), \quad \Phi(M) = PMP$$ where $P = I_n - \frac{1}{n}J$.
1) Show that $\Phi$ is an orthogonal projector in the Euclidean space $(\mathcal{M}_n(\mathbb{R}), (\cdot \mid \cdot))$.
2) Show that $\operatorname{Im}\Phi = \left\{M \in \mathcal{M}_n(\mathbb{R}) \mid MZ = 0 \text{ and } {}^t MZ = 0\right\}$.

Let $M = (m_{ij}) \in \mathcal{S}_n(\mathbb{R})$. We set $$S(M) = MZ = \left(\begin{array}{c}\sum_{i=1}^n m_{1i}\\ \vdots \\ \sum_{i=1}^n m_{ni}\end{array}\right) = \left(\begin{array}{c}S(M)_1\\ \vdots \\ S(M)_n\end{array}\right) \quad \text{and} \quad \sigma(M) = \langle Z, S(M)\rangle$$
Show that $$\Phi(M) = M - \frac{1}{n}\left(S(M)^t Z + Z^t S(M)\right) + \frac{\sigma(M)}{n^2}J$$

Let $U_1, U_2, \cdots, U_n$, $n$ elements of $\mathbb{R}^p$ satisfying $\sum_{i=1}^n U_i = 0$. We define the matrix of squared mutual distances $M = \left(\|U_i - U_j\|^2\right)_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{S}_n(\mathbb{R})$. We denote $U$ the matrix of $\mathcal{M}_{p,n}(\mathbb{R})$ having as column vectors the elements $U_1, U_2, \cdots, U_n$.
Show that ${}^t UU = -\frac{1}{2}\Phi(M)$.

Let $U_1, U_2, \cdots, U_n$, $n$ elements of $\mathbb{R}^p$ satisfying $\sum_{i=1}^n U_i = 0$. We define the matrix of squared mutual distances $M = \left(\|U_i - U_j\|^2\right)_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{S}_n(\mathbb{R})$. It has been shown that ${}^t UU = -\frac{1}{2}\Phi(M)$.
Deduce, for every pair $(i,j) \in \llbracket 1,n\rrbracket^2$, an expression for the inner product $\langle U_i, U_j\rangle = {}^t U_i U_j$ as a function of $$\alpha_{ij} = -\frac{1}{n}\left(S(M)_i + S(M)_j\right) + \frac{1}{n^2}\sigma(M)$$ and of $m_{ij}$ (Torgerson relation).

Let $M = (m_{ij})_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{S}_n(\mathbb{R})$ such that for every pair $(i,j) \in \llbracket 1,n\rrbracket^2$, $m_{ij} \geqslant 0$ and $m_{ii} = 0$. We assume in this question that there exist $U_1, U_2, \cdots, U_n$ elements of $\mathbb{R}^p$ such that for every $(i,j) \in \llbracket 1,n\rrbracket^2$, $m_{ij} = \|U_i - U_j\|^2$.
1) Show that the eigenvalues of $\Phi(M)$ are all real and non-positive.
2) We further assume (if necessary by performing a translation) that the $(U_i)_{i \in \llbracket 1,n\rrbracket}$ are centered, that is, $\sum_{i=1}^n U_i = 0$. Show that $\operatorname{rg}(U) = \operatorname{rg}(U_1 | U_2 | \cdots | U_n) = \operatorname{rg}(\Phi(M))$ and that $p \geqslant \operatorname{rg}(\Phi(M))$.