In $\triangle \mathrm { ABC } , \mathrm { BE }$ is a median, and O the mid-point of BE. The line joining A and O meets BC at D. Find the ratio $\overline { \mathrm { AO } } : \overline { \mathrm { OD } }$ (Hint: Draw a line through E parallel to AD.)

In coordinate space, for two points $\mathrm { A } ( 1 , a , - 6 ) , \mathrm { B } ( - 3,2 , b )$, when the point that externally divides the line segment AB in the ratio $3 : 2$ lies on the $x$-axis, what is the value of $a + b$? [3 points]
(1) $-1$
(2) $-2$
(3) $-3$
(4) $-4$
(5) $-5$

For any two points $M$ and $N$ in the $XY$-plane, let $\overrightarrow { MN }$ denote the vector from $M$ to $N$, and $\overrightarrow { 0 }$ denote the zero vector. Let $P , Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\triangle PQR$ such that
$$\overrightarrow { SP } + 5 \overrightarrow { SQ } + 6 \overrightarrow { SR } = \overrightarrow { 0 }$$
Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$, respectively. Then the value of
$$\frac { \text { length of the line segment } EF } { \text { length of the line segment } ES }$$
is $\_\_\_\_$ .

Let $O$ be the origin and the position vector of $A$ and $B$ be $2 \hat { i } + 2 \hat { j } + \widehat { k }$ and $2 \hat { i } + 4 \hat { j } + 4 \widehat { k }$ respectively. If the internal bisector of $\angle A O B$ meets the line $A B$ at $C$, then the length of $O C$ is
(1) $\frac { 2 } { 3 } \sqrt { 31 }$
(2) $\frac { 2 } { 3 } \sqrt { 34 }$
(3) $\frac { 3 } { 4 } \sqrt { 34 }$
(4) $\frac { 3 } { 2 } \sqrt { 31 }$

Let the point A divide the line segment joining the points $P ( - 1 , - 1,2 )$ and $Q ( 5,5,10 )$ internally in the ratio $\mathrm { r } : 1 ( \mathrm { r } > 0 )$. If O is the origin and $( \overrightarrow { \mathrm { OQ } } \cdot \overrightarrow { \mathrm { OA } } ) - \frac { 1 } { 5 } | \overrightarrow { \mathrm { OP } } \times \overrightarrow { \mathrm { OA } } | ^ { 2 } = 10$, then the value of r is :
(1) $\sqrt { 7 }$
(2) 14
(3) 3
(4) 7

Consider a triangle OAB. We denote by M the point dividing the side OA internally in the ratio 3:1 and denote by N the point dividing the side OB internally in the ratio 1:2. Also we denote by P the point of intersection of the line segment AN and the line segment BM.
(1) When the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ are denoted by $\vec{a}$ and $\vec{b}$ respectively, we are to express the vector $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}$ and $\vec{b}$.
When we set
$$\begin{array}{ll} \mathrm{AP}:\mathrm{PN} = s:(1-s) & (0we have
$$\overrightarrow{\mathrm{OP}} = (\mathbf{A} - t)\vec{b} = \frac{\mathbf{D}}{\mathbf{E}}t\vec{a}+\frac{\mathbf{B}}{\mathbf{E}}s\vec{b},$$
from which we obtain
$$s=\frac{\mathbf{G}}{\mathbf{H}}, \quad t=\frac{\square}{\mathbf{I}}.$$
Hence, $\overrightarrow{\mathrm{OP}}$ can be expressed in terms of $\vec{a}$ and $\vec{b}$ as
$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{K}}{\mathbf{L}}\vec{a}+\frac{\mathbf{M}}{\mathbf{N}}\vec{b}.$$
(2) We are to look at the relation between the length of the line segment OP and the size of $\angle\mathrm{AOB}$, when $\mathrm{OA}=6$ and $\mathrm{OB}=9$.
Let us denote the length of OP by $\ell$. When we express $\ell^2$ in terms of $\vec{a}\cdot\vec{b}$, we have
$$\ell^2 = \frac{\mathbf{Q}}{\mathbf{PQ}}\vec{a}\cdot\vec{b}+\mathbf{RS}$$
where $\vec{a}\cdot\vec{b}$ represents the inner product of $\vec{a}$ and $\vec{b}$.
Hence, for instance, if $\ell=4$, then we have
$$\cos\angle\mathrm{AOB}=\frac{\mathbf{TU}}{\mathbf{V}}$$
When the size of $\angle\mathrm{AOB}$ varies, the range of the values which $\ell$ can take is
$$\mathbf{W}<\ell<\mathbf{X}.$$

Take four points
$$\mathrm { A } ( 1,0 ) , \quad \mathrm { B } ( 0,1 ) , \quad \mathrm { C } ( 3,0 ) , \quad \mathrm { D } ( 0,2 )$$
on a plane with the coordinate system having the origin O. Take two points P and Q on segments AB and CD respectively, such that
$$\mathrm { AP } : \mathrm { PB } = \mathrm { CQ } : \mathrm { QD } = k : 2 .$$
We are to find the minimum possible length of the segment PQ.
(1) First, let us find the value of $x + 2 y$, where $\overrightarrow { \mathrm { PQ } } = ( x , y )$.
Since
$$\overrightarrow { \mathrm { OP } } = \frac { \square \mathbf { A } \overrightarrow { \mathrm { OA } } + k \overrightarrow { \mathrm { OB } } } { k + \mathbf { B } } , \quad \overrightarrow { \mathrm { OQ } } = \frac { \square \mathbf { C } \overrightarrow { \mathrm { OC } } + k \overrightarrow { \mathrm { OD } } } { k + \mathbf { D } } ,$$
we have
$$( x , y ) = \frac { 1 } { k + \mathbf { E } } ( \mathbf { F } , k ) ,$$
which gives $x + 2 y = \mathbf { G }$.
(2) When we represent $\mathrm { PQ } ^ { 2 }$ in terms of $y$, we have
$$\mathrm { PQ } ^ { 2 } = \mathbf { H } y ^ { 2 } - \mathbf { I } y + \mathbf { J } .$$
Hence PQ takes the minimum at $y = \frac { \mathbf { K } } { \mathbf { L } }$ and its value there is $\mathrm { PQ } = \frac { \square \mathbf { M } \sqrt { \mathbf { N } } } { \mathbf { O } }$. In this case, the value of $k$ is $k = \square \mathbf { P }$.

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 $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.