In the coordinate plane, for a triangle ABC with area 9, let P, Q, R be points moving on the three sides AB, BC, CA respectively. When $$\overrightarrow { \mathrm { AX } } = \frac { 1 } { 4 } ( \overrightarrow { \mathrm { AP } } + \overrightarrow { \mathrm { AR } } ) + \frac { 1 } { 2 } \overrightarrow { \mathrm { AQ } }$$ is satisfied, the area of the region represented by point X is $\frac { q } { p }$. Find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

Suppose $ABCD$ is a parallelogram and $P, Q$ are points on the sides $BC$ and $CD$ respectively, such that $PB = \alpha BC$ and $DQ = \beta DC$. If the area of the triangles $ABP$, $ADQ$, $PCQ$ are 15, 15 and 4 respectively, then the area of $APQ$ is
(a) 14
(b) 15
(c) 16
(d) 18.

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat { \mathrm { i } } - 6 \hat { \mathrm { j } }$ and $3 \hat { \mathrm { i } } + 4 \hat { \mathrm { j } } - 12 \widehat { \mathrm { k } }$, is:
(1) 20
(2) 65
(3) 52
(4) 26

Let a vector $\alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } }$ be obtained by rotating the vector $\sqrt { 3 } \hat { \mathrm { i } } + \hat { \mathrm { j } }$ by an angle $45 ^ { \circ }$ about the origin in counterclockwise direction in the first quadrant. Then the area (in sq. units) of triangle having vertices $( \alpha , \beta ) , ( 0 , \beta )$ and $( 0,0 )$ is equal to
(1) $\frac { 1 } { 2 }$
(2) 1
(3) $\frac { 1 } { \sqrt { 2 } }$
(4) $2 \sqrt { 2 }$

$A(2, 6, 2)$, $B(-4, 0, \lambda)$, $C(2, 3, -1)$ and $D(4, 5, 0)$, $\lambda \leq 5$ are the vertices of a quadrilateral $ABCD$. If its area is 18 square units, then $5 - 6\lambda$ is equal to $\_\_\_\_$.

Let $\overrightarrow { \mathrm { OA } } = \overrightarrow { \mathrm { a } } , \overrightarrow { \mathrm { OB } } = 12 \overrightarrow { \mathrm { a } } + 4 \overrightarrow { \mathrm {~b} }$ and $\overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { b } }$, where O is the origin. If $S$ is the parallelogram with adjacent sides OA and OC, then $\frac { \text { area of the quadrilateral } \mathrm { OABC } } { \text { area of } \mathrm { S } }$ is equal to
(1) 6
(2) 10
(3) 7
(4) 8

Let $\vec { c }$ be the projection vector of $\vec { b } = \lambda \hat { i } + 4 \hat { k } , \lambda > 0$, on the vector $\vec { a } = \hat { i } + 2 \hat { j } + 2 \hat { k }$. If $| \vec { a } + \vec { c } | = 7$, then the area of the parallelogram formed by the vectors $\vec { b }$ and $\vec { c }$ is $\_\_\_\_$

Let D, E and F be the three points which divide internally the three sides AB, BC and CA, respectively, of a triangle ABC in the ratio of $k:(1-k)$, where $0 < k \leq \frac{1}{2}$.
(1) When $k = \frac{1}{3}$, we are to find how many times greater the area of triangle ABC is than the area of triangle DEF. Since
$$\triangle\mathrm{ADF} = \triangle\mathrm{BED} = \triangle\mathrm{CFE} = \frac{\mathbf{M}}{\mathbf{N}} \triangle\mathrm{ABC},$$
it follows that
$$\triangle\mathrm{ABC} = \mathbf{O}\, \triangle\mathrm{DEF}.$$
(2) The area of the triangle DEF is half of the area of the triangle ABC when
$$k(1-k) = \frac{\mathbf{P}}{\mathbf{Q}},$$
that is, when
$$k = \frac{\mathbf{R} - \sqrt{\mathbf{S}}}{\mathbf{T}}.$$

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
$$t = \frac { \mathbf { Q } + \sqrt { \mathbf { R } } } { \mathbf { S } } \text{ (seconds).}$$

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

A corner of a classroom is formed by two walls and the floor, which are mutually perpendicular. Let the corner be point $O$. There is a triangular baffle $A B C$ with vertices $A , B , C$ located on the intersection lines between walls or between walls and the floor, at distances of 20, 20, and 10 centimeters from corner $O$ respectively. The three sides $\overline { A B }$ , $\overline { B C }$ , $\overline { C A }$ are flush with the walls or floor, as shown in the figure. Find the area of the triangular baffle $ABC$.