In the figure to the right, let $\angle \mathrm { XOY } = 60 ^ { \circ }$, and let OZ be the half-line (ray) which bisects $\angle \mathrm { XOY }$. In addition, the points A and B on the half-lines OX and OY satisfy $\mathrm { OA } = \mathrm { OB } = 1$.
Let $\mathrm { P } , \mathrm { Q }$ and R be moving points on $\mathrm { OX } , \mathrm { OZ }$ and OY that start simultaneously from $\mathrm { A } , \mathrm { O }$ and B, moving in the direction away from point O at the speeds of $1 , \sqrt { 3 }$ and 2 units per second.
We are to find the moment at which the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line by considering the area of the triangle PQR.
First, the lengths of $\mathrm { OP } , \mathrm { OQ }$ and OR at $t$ seconds after the start are
$$\mathrm { OP } = t + \mathbf { A } , \quad \mathrm { OQ } = \sqrt { \mathbf { B } } t , \quad \mathrm { OR } = \mathbf { C } t + \mathbf { D } .$$
At this time the areas of the triangles are
$$\begin{aligned} \triangle \mathrm { OPQ } & = \frac { \sqrt { \mathbf { E } } t ( t + \mathbf { F } ) } { 4 } , \\ \triangle \mathrm { ORQ } & = \frac { \sqrt { \mathbf { G } } t ( \mathbf { H } t + \mathbf { I } ) } { 4 } , \\ \triangle \mathrm { OPR } & = \frac { \sqrt { \mathbf { G } } ( t + \mathbf { K } ) ( \mathbf { L } t + \mathbf { M } ) } { 4 } \end{aligned}$$
Hence we obtain
$$\triangle \mathrm { PQR } = \frac { \sqrt { \mathbf { N } } } { 4 } \left| - t ^ { 2 } + t + \mathbf { O } \right| .$$
So, to find the moment such that the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line, we should find the case where
$$t ^ { 2 } - t - \mathbf { O } = \mathbf { P } .$$
Thus the time required is
$$t = \frac { \mathbf { Q } + \sqrt { \mathbf { R } } } { \mathbf { S } } \text{ (seconds).}$$

The triangle $ABC$ satisfies
$$\mathrm { AB } = 2 , \quad \mathrm { BC } = 3 , \quad \mathrm { CA } = 4 .$$
(1) When we set $\angle \mathrm { ABC } = \theta$, the inner product $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } }$ of the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { BC } }$ is
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \mathbf { AB } \cos \theta .$$
Finding the value of $\cos \theta$ from the law of cosines, we obtain
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \frac { \mathbf { C } } { \mathbf { D } } .$$
(2) We divide the side BC into $n$ equal parts by the points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \cdots , \mathrm { P } _ { n - 1 }$ which are arranged in ascending order of the distance from B, and set $\mathrm { B } = \mathrm { P } _ { 0 } , \mathrm { C } = \mathrm { P } _ { n }$. We are to find the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } }$.
When we calculate the inner product of $\overrightarrow { \mathrm { AP } _ { k - 1 } }$ and $\overrightarrow { \mathrm { AP } _ { k } }$ using (1), we have
$$\overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \mathbf { E } + \frac { \mathbf { F } } { 2 n } + \mathbf { G }$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \frac { \mathbf { IJ } } { \mathbf { K } }$$

A triangle OAB and a triangle OAC share one side OA. Further, they satisfy the following two conditions:
(i) $\overrightarrow { \mathrm { OC } } = x \overrightarrow { \mathrm { OA } } + \frac { 1 } { 2 } \overrightarrow { \mathrm { OB } }$;
(ii) the center of gravity G of the triangle OAC is on the segment AB.
We are to find the value of $x$, and express $\overrightarrow { \mathrm { OG } }$ in terms of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$.
Let us denote the point of intersection of the segment OC and the segment AB by D. Then we have
$$\overrightarrow { \mathrm { OD } } = \frac { x } { \mathbf { A } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { B } } { \mathbf { A } } \overrightarrow { \mathrm { OB } } .$$
Since D is on the segment AB, we have $x = \frac { \mathbf { D } } { \mathbf { E } }$. Hence we obtain
$$\overrightarrow { \mathrm { OG } } = \frac { \mathbf { F } } { \mathbf { G } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { H } } { \mathbf { I } } \overrightarrow { \mathrm { OB } } .$$
In particular, when $\mathrm { OA } = 1 , \mathrm { OB } = 2$ and $\angle \mathrm { AOB } = 60 ^ { \circ }$, we have $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.

The parallelepiped satisfies
$$\begin{aligned} & \mathrm { AB } = 2 , \quad \mathrm { AD } = 3 , \quad \mathrm { AE } = 1 \\ & \angle \mathrm { BAD } = 60 ^ { \circ } , \quad \angle \mathrm { BAE } = 90 ^ { \circ } , \quad \angle \mathrm { DAE } = 120 ^ { \circ } \end{aligned}$$
and M is the midpoint of edge GH. Let us take points P and Q on edges BF and DH, respectively, such that the four points A, P, M, and Q are on the same plane. We are to find the points P and Q which maximize the length of the line segment PQ.
(1) Setting $\vec { a } = \overrightarrow { \mathrm { AB } } , \vec { b } = \overrightarrow { \mathrm { AD } }$ and $\vec { c } = \overrightarrow { \mathrm { AE } }$, the inner products of these vectors are
$$\vec { a } \cdot \vec { b } = \mathbf { A } , \quad \vec { b } \cdot \vec { c } = - \frac { \mathbf { B } } { \mathbf{C} } , \quad \vec { c } \cdot \vec { a } = \mathbf { D } .$$
(2) Let $s$ and $t$ satisfy $0 \leqq s \leqq 1,0 \leqq t \leqq 1$, and set $\mathrm { BP } : \mathrm { PF } = s : ( 1 - s )$, $\mathrm { DQ } : \mathrm { QH } = t : ( 1 - t )$. Since the four points A, P, M and Q are on the same plane, there exist two real numbers $\alpha$ and $\beta$ such that
$$\overrightarrow { \mathrm { AM } } = \alpha \overrightarrow { \mathrm { AP } } + \beta \overrightarrow { \mathrm { AQ } } .$$
Hence $s$ and $t$ satisfy
$$s = \mathbf { E } ( \mathbf { F } - t ) .$$
Then $| \overrightarrow { \mathrm { PQ } } |$ may be expressed in terms of $t$ as
$$| \overrightarrow { \mathrm { PQ } } | ^ { 2 } = \mathbf { G } t ^ { 2 } - \mathbf { H I } t + \mathbf { J K } .$$
Hence the length of segment PQ is maximized when $\square$ L. Here, for $\square$ L choose the correct answer from among choices (0) $\sim$ (5) below. (0) $s = 0 , \quad t = 1$
(1) $s = 0 , \quad t = \frac { 1 } { 2 }$
(2) $s = \frac { 1 } { 2 } , t = \frac { 3 } { 4 }$
(3) $s = \frac { 2 } { 3 } , t = \frac { 2 } { 3 }$
(4) $s = 1 , \quad t = \frac { 1 } { 2 }$
(5) $s = 1 , \quad t = \frac { 2 } { 3 }$

(Course 2) Q1 For $\mathbf { A } , \mathbf { B } , \mathbf { D } , \mathbf { E }$ and $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below, and for the other $\square$, enter the correct number.
Given a sphere of radius 2 with the center at point O, we have a tetrahedron ABCD whose four vertices are on the sphere. Let $\mathrm { AB } = \mathrm { BC } = \mathrm { CA } = 2$ and side BD be a diameter of the sphere.
Set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$.
(1) Let M and N denote the midpoints of segments DA and BC, respectively. Then we have
$$\overrightarrow { \mathrm { DA } } = \mathbf { A } , \quad \overrightarrow { \mathrm { MN } } = \frac { \mathbf { B } } { \mathbf { C } } + \mathbf { D }$$
(2) When the midpoint of segment MN is denoted by P and the center of gravity of triangle BCD is denoted by G, we see that
$$\overrightarrow { \mathrm { OP } } = \frac { \mathbf { E } } { \mathbf { F } } , \quad \overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { F } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$$
Also, since $\overrightarrow { \mathrm { AG } } = \frac { \mathbf { K } } { \mathbf { L } } \overrightarrow { \mathrm { AP } }$, we see that the three points A, P and G are on a straight line. (0) $\vec { a }$
(1) $\vec { b }$
(2) $\vec { c }$
(3) $\vec { a } - \vec { b }$
(4) $\vec { b } - \vec { c }$
(5) $\vec { c } - \vec { a }$ (6) $\vec { a } + \vec { b }$ (7) $\vec { b } + \vec { c }$ (8) $\vec { c } + \vec { a }$ (9) $\vec { a } + \vec { b } + \vec { c }$

On the coordinate plane, two parallel lines $L _ { 1 } , L _ { 2 }$ both have slope 2 and are at distance 5 apart. Point $A ( 2 , - 1 )$ is a point on $L _ { 1 }$ in the fourth quadrant. Point $B$ is a point on $L _ { 2 }$ in the second quadrant with $\overline { A B } = 5$ . Line $L _ { 3 }$ has slope 3, passes through point $A$, and intersects $L _ { 2 }$ at point $C$. Answer the following questions:
(1) Find the slope of line $AB$ . (2 points)
(2) Find the vector $\overrightarrow { A B }$ . (4 points)
(3) Find the dot product $\overrightarrow { A B } \cdot \overrightarrow { A C }$ . (3 points)
(4) Find the vector $\overrightarrow { A C }$ . (4 points)

In the coordinate plane, let $O$ be the origin, and let $A$ and $B$ be two distinct points different from $O$. Let $C_{1}$, $C_{2}$, $C_{3}$ be three points in the plane satisfying $\overrightarrow{OC}_{n} = \overrightarrow{OA} + n\overrightarrow{OB}$, $n = 1, 2, 3$. Select the correct options.
(1) $\overrightarrow{OC}_{1} \neq \overrightarrow{0}$
(2) $\overline{OC_{1}} < \overline{OC_{2}} < \overline{OC_{3}}$
(3) $\overrightarrow{OC}_{1} \cdot \overrightarrow{OA} < \overrightarrow{OC}_{2} \cdot \overrightarrow{OA} < \overrightarrow{OC}_{3} \cdot \overrightarrow{OA}$
(4) $\overrightarrow{OC_{1}} \cdot \overrightarrow{OB} < \overrightarrow{OC_{2}} \cdot \overrightarrow{OB} < \overrightarrow{OC_{3}} \cdot \overrightarrow{OB}$
(5) $C_{1}$, $C_{2}$, $C_{3}$ are collinear

Consider a trapezoid $ABCD$ where $\overline { AB }$ is parallel to $\overline { DC }$ . It is known that points $E$ and $F$ lie on diagonals $\overline { AC }$ and $\overline { BD }$ respectively, and $\overline { AB } = \frac { 2 } { 5 } \overline { DC }$ , $\overline { AE } = \frac { 3 } { 2 } \overline { EC }$ , $\overline { BF } = \frac { 2 } { 3 } \overline { FD }$ . If vector $\overrightarrow { FE }$ is expressed as $\alpha \overrightarrow { AC } + \beta \overrightarrow { AD }$ , then the real numbers $\alpha = \frac { \text{(16)} } { \text{(17)(18)} } , \beta = \frac { \text{(19)(20)} } { \text{(21)(22)} }$ . (Express as fractions in lowest terms)

Let $\vec { a }$ and $\vec { b }$ be non-zero vectors in the plane. If the area of the triangle formed by $2 \vec { a } + \vec { b }$ and $\vec { a } + 2 \vec { b }$ is 6, what is the area of the triangle formed by $3 \vec { a } + \vec { b }$ and $\vec { a } + 3 \vec { b }$?
(1) 8
(2) 9
(3) 12
(4) 13.5
(5) 16

On a plane, there is a trapezoid $A B C D$ with upper base $\overline { A B } = 10$, lower base $\overline { C D } = 15$, and leg length $\overline { A D } = \overline { B C } + 1$. Select the correct options.
(1) $\angle A > \angle B$
(2) $\angle B + \angle D < 180 ^ { \circ }$
(3) $\overrightarrow { B A } \cdot \overrightarrow { B C } < 0$
(4) The length of $\overline { B C }$ could be 2
(5) $\overrightarrow { C B } \cdot \overrightarrow { C D } < 30$

Let $P$ be a point inside $\triangle A B C$, and $\overrightarrow { A P } = a \overrightarrow { A B } + b \overrightarrow { A C }$ , where $a , b$ are distinct real numbers. Let $Q , R$ be on the same plane, with $\overrightarrow { A Q } = b \overrightarrow { A B } + a \overrightarrow { A C } , ~ \overrightarrow { A R } = a \overrightarrow { A B } + ( b - 0.05 ) \overrightarrow { A C }$ . Select the correct options.
(1) $Q , R$ are also both inside $\triangle A B C$
(2) $| \overrightarrow { A P } | = | \overrightarrow { A Q } |$
(3) Area of $\triangle A B P$ = Area of $\triangle A C Q$
(4) Area of $\triangle B C P$ = Area of $\triangle B C Q$
(5) Area of $\triangle A B P$ > Area of $\triangle A B R$

Consider points $O(0,0), A, B, C, D, E, F, G$ on a coordinate plane, where points $B$, $C$ and $D$, $E$ and $F$, $G$ and $A$ are located in the first, second, third, and fourth quadrants respectively. If $\vec{v}$ is a vector on the coordinate plane satisfying $\vec{v} \cdot \overrightarrow{OA} > 0$ and $\vec{v} \cdot \overrightarrow{OB} > 0$, then the dot product of $\vec{v}$ with which of the following vectors must be negative?
(1) $\overrightarrow{OC}$
(2) $\overrightarrow{OD}$
(3) $\overrightarrow{OE}$
(4) $\overrightarrow{OF}$
(5) $\overrightarrow{OG}$

On a coordinate plane, a particle starts from point $( - 3 , - 2 )$ and moves 5 units in the direction of vector $( a , 1 )$ and arrives exactly at the $x$-axis, where $a$ is a positive real number. What is the value of $a$?
(1) $\frac { \sqrt { 13 } } { 2 }$
(2) 2
(3) $\sqrt { 5 }$
(4) $\frac { \sqrt { 21 } } { 2 }$
(5) $2 \sqrt { 6 }$

In coordinate space, consider a unit cube with edge length 1, with one vertex $O$ fixed. From the seven vertices other than $O$, two distinct points are randomly selected, denoted as $P$ and $Q$. What is the expected value of the dot product $\overrightarrow{OP} \cdot \overrightarrow{OQ}$ among the following options?
(1) $\frac{4}{7}$
(2) $\frac{5}{7}$
(3) $\frac{6}{7}$
(4) 1
(5) $\frac{8}{7}$

It is known that a right triangle $\triangle A B C$ has side lengths $\overline { A B } = \sqrt { 7 }$, $\overline { A C } = \sqrt { 3 }$, $\overline { B C } = 2$. If isosceles triangles $\triangle M A B$ and $\triangle N A C$ with vertex angles equal to $120 ^ { \circ }$ are constructed outside $\triangle A B C$ using $\overline { A B }$ and $\overline { A C }$ as bases respectively, then $\overline { M N } ^ { 2 } =$ . (Express as a fraction in lowest terms)