Let the line $x + y = 1$ meet the axes of $x$ and $y$ at A and B , respectively. A right angled triangle AMN is inscribed in the triangle OAB , where O is the origin and the points M and N lie on the lines $O B$ and $A B$, respectively. If the area of the triangle $A M N$ is $\frac { 4 } { 9 }$ of the area of the triangle $O A B$ and $\mathrm { AN } : \mathrm { NB } = \lambda : 1$, then the sum of all possible value(s) of $\lambda$ is:
(1) 2
(2) $\frac { 5 } { 2 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 13 } { 6 }$

Let $m$ be a real number. On a plane with the coordinate system, in which the origin is denoted by O, consider the parabola $y = x^2$ and the two points on it,
$$\mathrm{A}(a,\, ma+1), \quad \mathrm{B}(b,\, mb+1) \quad (a < 0 < b)$$
(1) The $x$-coordinates $a$ and $b$ of the two points A and B can be expressed in terms of $m$ as
$$a = \frac{m - \sqrt{D}}{\mathbf{A}}, \quad b = \frac{m + \sqrt{D}}{\mathbf{B}},$$
where the expression $D$ is
$$D = m^2 + \mathbf{C}.$$
(2) Let the coordinates of the point of intersection of the segment AB and the $y$-axis be denoted by $(0, c)$. Then $c = \mathbf{D}$.
(3) Further, when the area $S$ of the triangle OAB with the three vertices O, A and B is expressed in terms of $a$ and $b$, we have
$$S = \frac{1}{2}\mathbf{E},$$
where $E$ is the appropriate choice from among (0) $\sim$ (5). (0) $a + b$
(1) $a - b$
(2) $b - a$
(3) $a^2 + b^2$
(4) $a^2 - b^2$
(5) $b^2 - a^2$
Also, when $S$ is represented in terms of $m$, we have
$$S = \frac{\mathbf{F}}{\mathbf{G}} \sqrt{m^2 + \mathbf{H}}.$$
Hence the value of $S$ is minimalized when $m = \mathbf{I}$, and its minimum value is $S = \mathbf{J}$.

Consider a figure made by cutting two corners from a rectangle, as in the diagram to the right. The lengths of the sides are
$$\begin{array}{lll} \mathrm{AB} = 11, & \mathrm{BC} = 4, & \mathrm{CD} = 2\sqrt{13} \\ \mathrm{DE} = 5, & \mathrm{EF} = 2\sqrt{5}, & \mathrm{FA} = 6 \end{array}$$
We are to find the area of this figure.
First, extend the sides of the figure as in the diagram and denote the sides forming the right angles by $x, y, u$ and $v$. Then
$$u = \mathbf{A} - y, \quad v = x + \mathbf{B}.$$
Substituting these expressions in the equation $u^2 + v^2 = \mathbf{CD}$ and also using the equation $x^2 + y^2 = \mathbf{EF}$, we obtain
$$x = \mathbf{G}.\, y - \mathbf{H}.$$
Then, since
$$\mathbf{I}y^2 - \mathbf{J} = \mathbf{J}y - \mathbf{K} = 0,$$
we obtain $y = \mathbf{L}$.
From this we have $x = \mathbf{M}$, and further $u = \mathbf{N}$ and $v = \mathbf{O}$. Finally we conclude that the area of this figure is $\mathbf{PQ}$.

4. On a coordinate plane, two points $A(1, 0)$ and $B(0, 1)$ are given. Consider three additional points $P(\pi, 1)$, $Q(-\sqrt{3}, 6)$, and $R\left(2, \log_{4} 32\right)$. Let the area of $\triangle PAB$ be $p$, the area of $\triangle QAB$ be $q$, and the area of $\triangle RAB$ be $r$. Which of the following options is correct?
(1) $p < q < r$
(2) $p < r < q$
(3) $q < p < r$
(4) $q < r < p$
(5) $r < q < p$

On the coordinate plane, there is a polygonal region $\Gamma$ (including boundary) as shown in the figure. If $k > 0$ , the line $7 x + 2 y = k$ and the two coordinate axes form a triangular region such that the polygonal region $\Gamma$ lies within this triangular region (including boundary), then the minimum positive real number $k =$ (7)(8).

On the coordinate plane, there is a trapezoid with four vertices at $A ( 0,0 ) , B ( 1,0 ) , P , Q$ , where the line passing through $P$ and $Q$ has equation $y = 2 x + 4$ . If the coordinates of point $Q$ are $( a , 2 a + 4 )$ , where $a \geq 0$ is a real number, then the area of trapezoid $A B P Q$ is (14)$a +$ (16). (Reduce to lowest terms)

On the coordinate plane, $O$ is the origin, and points $A(1,0)$ and $B(-2,0)$ are given. There are also two points $P$ and $Q$ in the upper half-plane satisfying $\overline{AP} = \overline{OA}$, $\overline{BQ} = \overline{OB}$, $\angle POQ$ is a right angle. Let $\angle AOP = \theta$.
(Continuing from question 19, where $\sin\theta = \frac{3}{5}$) Find the distance from point $A$ to line $BQ$, and find the area of quadrilateral $PABQ$. (Non-multiple choice question, 6 points)

On the coordinate plane, $P ( a , 0 )$ is a point on the $x$-axis, where $a > 0$. Let $L _ { 1 }$ and $L _ { 2 }$ be lines passing through point $P$ with slopes $- \frac { 4 } { 3 }$ and $- \frac { 3 } { 2 }$ respectively. Given that the difference in areas of the two right triangles formed by $L _ { 1 }$ and $L _ { 2 }$ with the two coordinate axes is 3, what is the value of $a$?
(1) $3 \sqrt { 2 }$
(2) 6
(3) $6 \sqrt { 2 }$
(4) 9
(5) $8 \sqrt { 2 }$

On the coordinate plane, given three points $A ( 0,2 )$ , $B ( - 1,0 )$ , $C ( 4,0 )$ . If the line $y = m x$ divides triangle $A B C$ into two equal areas, then $m = \frac { \text{(14--1)} } { \text{(14--2)} }$ . (Reduce to lowest terms)

ABCD is a parallelogram AECD is a trapezoid $| \mathrm { BE } | = 3 \mathrm {~cm}$ $| \mathrm { DC } | = 4 \mathrm {~cm}$
If the area of the parallelogram ABCD in the figure is $20 \mathrm {~cm} ^ { 2 }$, what is the area of triangle $CBE$ in $\mathbf { cm } ^ { \mathbf { 2 } }$?
A) 7
B) 7,5
C) 8
D) 8,5
E) 9

Below are given squares $\mathrm { ABCD }$, $\mathrm { BLPR }$, and KLMN with side lengths of 3, 2, and 1 units respectively.
In the figure, points $\mathrm { A }$, $\mathrm { B }$, $\mathrm { K }$, and L are collinear.\ Accordingly, what is the area of triangle DNP in square units?\ A) 3\ B) 4\ C) 5\ D) 6\ E) 8

In the Cartesian coordinate plane; a triangle with one vertex at the origin and the other vertices on the lines $y = x$ and $y = - x$ has its medians intersecting at point $(2,4)$.
Accordingly, what is the area of this triangle in square units?
A) 18 B) 24 C) 27 D) $9 \sqrt { 2 }$ E) $18 \sqrt { 2 }$

In the rectangular coordinate plane, points $A(2, 7)$ and $B(-1, 4)$ are translated 3 units in the positive direction along the x-axis to obtain points $D$ and $C$ respectively.
Accordingly, what is the area of the quadrilateral with vertices at points A, B, C, and D in square units?
A) 9
B) 10
C) 11
D) 12
E) 13

In the rectangular coordinate plane, two lines that intersect perpendicularly at point $A ( 3,4 )$ have slopes whose sum is $\frac { 3 } { 2 }$.
If the points where these two lines intersect the x-axis are points B and C, what is the area of triangle ABC in square units?
A) 24
B) 20
C) 16
D) 12
E) 8

In a rectangular coordinate plane, what is the area of the triangular region bounded by the lines $2x - y = 0$, $x + 2y = 0$ and $x - 8y + 30 = 0$ in square units?
A) 9 B) 12 C) 15 D) 18 E) 21

In the rectangular coordinate plane, a triangle $OAB$ with one vertex at the origin and the other two vertices on the axes, and the line segment $[PR]$ connecting the points $P(6, -3)$ and $R(-2, 9)$ are drawn. The line segment $[PR]$ passes through the midpoints of both $[OA]$ and $[OB]$.
According to this, what is the area of triangle $OAB$ in square units?
A) 36 B) 42 C) 48 D) 54 E) 60