For a quadrilateral ABCD inscribed in a circle of radius 1, let $\mathrm { AB } : \mathrm { AD } = 1 : 2$ and $\angle \mathrm { BAD } = 120 ^ { \circ }$. Also, when the point of intersection of diagonals BD and AC is denoted by E, let $\mathrm { BE } : \mathrm { ED } = 3 : 4$.
We are to find the area of quadrilateral $\mathrm{ABCD}$.
In order to find the area of quadrilateral ABCD, we are to find the area of triangle ABD, denoted by $\triangle \mathrm { ABD }$, and the area of triangle BCD, denoted by $\triangle \mathrm { BCD }$.
First, let us find $\triangle \mathrm { ABD }$. Since
$$\mathrm { BD } = \sqrt { \mathbf { A } } , \quad \mathrm { AB } = \frac { \sqrt { \mathbf { BC } } } { \mathbf { D } } ,$$
we have
$$\triangle \mathrm { ABD } = \frac { \mathbf { E } \sqrt { \mathbf { F } } } { \mathbf { GH } } .$$
Next, let us find $\triangle \mathrm { BCD }$. Since
$$\triangle \mathrm { ABC } : \triangle \mathrm { ACD } = \mathbf{I} : \mathbf { J } ,$$
we see that $\mathrm { BC } : \mathrm { CD } = \mathbf { K } : \mathbf { L }$. (Give the answers using the simplest integer ratios.)
Hence we have $\mathrm { BC } = \frac { \mathbf{M} \sqrt { \mathbf { N } } } { \mathbf{O} }$ and
$$\triangle \mathrm { BCD } = \frac { \mathbf { PQ } \sqrt { \mathbf { R } } } { \mathbf { ST } }$$
Thus, from (1) and (2) we obtain the result that the area of quadrilateral ABCD is $\frac{\mathbf{U}\sqrt{\mathbf{V}}}{\mathbf{WX}}$.

Consider a triangle ABC and its circumscribed circle O, where the lengths of the three sides of the triangle are $$\mathrm{AB} = 2, \quad \mathrm{BC} = 3, \quad \mathrm{CA} = 4.$$ Below, the area of a triangle such as PQR is expressed as $\triangle\mathrm{PQR}$.
(1) We see that $\cos\angle\mathrm{ABC} = \frac{\mathbf{AB}}{\mathbf{C}}$.
(2) Let us take a point D on the circumference of circle O such that it is on the opposite side of the circle from point B with respect to AC and $$\frac{\triangle\mathrm{ABD}}{\triangle\mathrm{BCD}} = \frac{8}{15}.$$ We are to find the lengths of line segments AD and CD.
First, since $$\angle\mathrm{BAD} = \mathbf{DEF}^\circ - \angle\mathrm{BCD},$$ we have $\sin\angle\mathrm{BAD} = \sin\angle\mathrm{BCD}$. Hence from (1) we have $$\frac{\mathrm{AD}}{\mathrm{CD}} = \frac{\mathbf{G}}{\mathbf{H}},$$ so we set $\mathrm{AD} = \mathbf{G}k$ and $\mathrm{CD} = \mathbf{H}k$, where $k$ is a positive number. Furthermore, since $$\angle\mathrm{ADC} = \mathbf{IJK}^\circ - \angle\mathrm{ABC},$$ we have $\cos\angle\mathrm{ADC} = \frac{\mathbf{L}}{\mathbf{L}}$. Hence, we obtain $k = \frac{\mathbf{N}}{\sqrt{\mathbf{OP}}}$, and then $$\mathrm{AD} = \frac{\mathbf{QR}\sqrt{\mathbf{OP}}}{\mathbf{OP}},$$ $$\mathrm{CD} = \frac{\mathbf{ST}\sqrt{\mathbf{OP}}}{\mathbf{OP}}.$$
(3) When we denote the point of intersection of the straight line DA and the straight line CB by E, we have $$\frac{\triangle\mathrm{ABE}}{\triangle\mathrm{CDE}} = \frac{\mathbf{UV}}{\mathbf{WXY}}.$$

In a triangle ABC, let $\angle \mathrm { B } = 45 ^ { \circ }$ and $\angle \mathrm { C } = 75 ^ { \circ }$, and let D be the intersection of the bisector of $\angle \mathrm { A }$ and side BC.
(1) From the law of sines we have
$$\mathrm { AC } = \frac { \sqrt { \mathbf { A } } } { \sqrt { \mathbf { B } } } \mathrm { BC } , \quad \mathrm { AD } = \sqrt { \mathbf { C } } \mathrm { BD } .$$
In particular, from $\angle \mathrm { ADC } = \mathbf { D E } ^ { \circ }$ we see that
$$\mathrm { BD } : \mathrm { BC } = \mathbf { F } : \sqrt { \mathbf { G } }$$
and hence we have
$$\mathrm { AB } : \mathrm { AC } = \mathbf { H } : \left( \sqrt { \mathbf { I } } - \frac { \mathbf { J } }{\mathbf{J} } \right) .$$
(2) Let $\mathrm { O } _ { 1 }$ be the center of the circumscribed circle of triangle ABD, and let $\mathrm { O } _ { 2 }$ be the center of the circumscribed circle of triangle ADC. Let us find the ratio of the areas of triangle ABC and triangle $\mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$, $\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$.
Since $\angle \mathrm { AO } _ { 1 } \mathrm { D } = \mathbf { K } \mathbf { L } ^ { \circ }$ and $\angle \mathrm { AO } _ { 2 } \mathrm { O } _ { 1 } = \mathbf { M N } ^ { \circ }$, by the same reasoning as (1), we have
$$\mathrm { AC } = \sqrt { \mathbf { O } } \mathrm { AO } _ { 1 } , \quad \mathrm { AO } _ { 2 } = ( \sqrt { \mathbf { P } } - \mathbf { Q } ) \mathrm { AO } _ { 1 } .$$
Hence we obtain
$$\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 } = \mathbf { R } : ( \mathbf { S } - \sqrt { \mathbf { T } } ) .$$

There is a triangular park with vertices at $O$, $A$, $B$. At vertex $O$ there is an observation tower 150 meters high. A person standing on the observation tower observes the other two vertices $A$, $B$ on the ground and the midpoint $C$ of $\overline{AB}$, measuring angles of depression of $30^{\circ}$, $60^{\circ}$, $45^{\circ}$ respectively. The area of this triangular park is （15）（16）（17）（18）$\sqrt{(19)}$ square meters. (Express as a simplified radical)

From 6, 8, 10, 12, select any three distinct numbers as the three sides of a triangle, and let $\theta$ be the largest interior angle of this triangle. Among all possible triangles formed, the minimum value of $\cos \theta$ is (Express as a fraction in lowest terms)

As shown in the figure, $\triangle A B C$ is an acute triangle, $P$ is a point outside the circumcircle $\Gamma$ of $\triangle A B C$, and both $\overline { P B }$ and $\overline { P C }$ are tangent to circle $\Gamma$. Let $\angle B P C = \theta$. What is the value of $\cos A$?
(1) $\sin 2 \theta$
(2) $\frac { \sin \theta } { 2 }$
(3) $\sin \frac { \theta } { 2 }$
(4) $\frac { \cos \theta } { 2 }$
(5) $\cos \frac { \theta } { 2 }$

In $\triangle A B C$, it is known that $\overline { A B } = 4$ and $\overline { A C } = 6$, which is insufficient to determine the shape and size of $\triangle A B C$. However, knowing certain additional conditions (for example, knowing the length of $\overline { B C }$) would uniquely determine the shape and size of $\triangle A B C$. Select the correct options.
(1) If we additionally know the value of $\cos A$, then $\triangle A B C$ can be uniquely determined
(2) If we additionally know the value of $\cos B$, then $\triangle A B C$ can be uniquely determined
(3) If we additionally know the value of $\cos C$, then $\triangle A B C$ can be uniquely determined
(4) If we additionally know the area of $\triangle A B C$, then $\triangle A B C$ can be uniquely determined
(5) If we additionally know the circumradius of $\triangle A B C$, then $\triangle A B C$ can be uniquely determined

On the coordinate plane, the three vertices of $\triangle A B C$ have coordinates $A ( 0,2 ) , B ( 1,0 ) , C ( 4,1 )$ respectively. Select the correct options.
(1) Among the three sides of $\triangle A B C$, $\overline { A C }$ is the longest
(2) $\sin A < \sin C$
(3) $\triangle A B C$ is an acute triangle
(4) $\sin B = \frac { 7 \sqrt { 2 } } { 10 }$
(5) The circumradius of $\triangle A B C$ is less than 2

As shown in the figure, the Wang family owns a triangular piece of land $\triangle A B C$ , where $\overline { B C } = 16$ meters. The government plans to requisition the trapezoid $D B C E$ portion to develop a road with straight lines $D E , B C$ as edges, with road width $h$ meters, leaving the Wang family with only $\frac { 9 } { 16 }$ of the original land area. After negotiation, the plan is changed to develop a road with parallel lines $B E , F C$ as edges, with the same road width, where $\angle E B C = 30 ^ { \circ }$ . Only the $\triangle B C E$ region needs to be requisitioned. According to this agreement, the Wang family's remaining land $\triangle A B E$ has (15-1)(15-2)(15-3) square meters.

On the coordinate plane, there is an annular region formed by the intersection of the exterior of the circle $x ^ { 2 } + y ^ { 2 } = 3$ and the interior of the circle $x ^ { 2 } + y ^ { 2 } = 4$ . A person wants to use a straight scanning rod of length 1 to scan a certain region $R$ above the $x$-axis of this annular region. He designs the scanning rod with black and white ends moving respectively on the semicircles $C _ { 1 } : x ^ { 2 } + y ^ { 2 } = 3 ( y \geq 0 )$ and $C _ { 2 } : x ^ { 2 } + y ^ { 2 } = 4 ( y \geq 0 )$ . Initially, the black end of the scanning rod is at point $A ( \sqrt { 3 } , 0 )$ and the white end is at point $B$ on $C _ { 2 }$ . Then the black and white ends move counterclockwise along $C _ { 1 }$ and $C _ { 2 }$ respectively until the white end reaches point $B ^ { \prime } ( - 2,0 )$ on $C _ { 2 }$ , at which point scanning stops.
Let $O$ be the origin. When the scanning rod stops, the positions of the black and white ends are $A ^ { \prime }$ and $B ^ { \prime }$ respectively. In the diagram area of the answer sheet, use hatching to indicate the region $R$ swept by the scanning rod; and in the solution area, find $\cos \angle O A ^ { \prime } B ^ { \prime }$ and the polar coordinates of point $A ^ { \prime }$ . (Non-multiple choice question, 6 points)

China's Tiger Hill Tower, Pearl Tower, and Italy's Leaning Tower of Pisa are three famous leaning towers with tower heights of 48, 19, and 57 meters respectively, and offset distances of 2.3, 2.3, and 4 meters respectively. Their tilt angles are denoted as $\theta_1{}^{\circ}, \theta_2{}^{\circ}$, and $\theta_3{}^{\circ}$ respectively. Compare the size relationship of $\theta_1, \theta_2$, and $\theta_3$. (Non-multiple choice, 4 points)
Note: The tilt angle $\theta^{\circ}$ is the angle between the tower body and a vertical dashed line ($0 \leq \theta < 90$), and the offset distance is the distance from the tower top to the vertical dashed line.

Suppose there are two iron towers with equal tower heights. Their tilt angles $\alpha^{\circ}, \beta^{\circ}$ satisfy $\sin\alpha^{\circ} = \frac{1}{5}$ and $\sin\beta^{\circ} = \frac{7}{25}$ respectively. It is known that the offset distances of the two towers differ by 20 meters. Find the difference in the distance from the tower tops to the ground. (Non-multiple choice, 6 points)
Note: The tilt angle $\theta^{\circ}$ is the angle between the tower body and a vertical dashed line ($0 \leq \theta < 90$), the offset distance is the distance from the tower top to the vertical dashed line, and the distance from the tower top to the ground is the vertical height.

There are two tall buildings on the ground, Building A and Building B. It is known that Building A is taller than Building B, and the horizontal distance between the two buildings is 150 meters. A person pulls a rope from the top of Building A to the top of Building B, and measures the angle of depression to the top of Building B from the top of Building A as $22^{\circ}$. Assuming the rope is pulled straight, which of the following options is closest to the length of the rope (unit: meters)? (Note: The angle of depression is the angle between the line of sight and the horizontal line when looking down at an object)
(1) $150$ (2) $150 \sin 22^{\circ}$ (3) $150 \cos 22^{\circ}$ (4) $\frac{150}{\cos 22^{\circ}}$ (5) $\frac{150}{\sin 22^{\circ}}$

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 $\angle AOP = \theta$.
If $\sin\theta = \frac{3}{5}$, find the coordinates of point $Q$ and explain that $\overrightarrow{BQ} = 2\overrightarrow{AP}$. (Non-multiple choice question, 6 points)

As shown in the diagram on the right, there is a $\triangle ABC$. It is known that the altitude $\overline{AD} = 12$ on side $\overline{BC}$, and $\tan \angle B = \frac{3}{2}$, $\tan \angle C = \frac{2}{3}$. What is the length of $\overline{BC}$?
(1) 20
(2) 21
(3) 24
(4) 25
(5) 26

On the same plane, two artillery batteries $A$ and $B$ are 7 kilometers apart, with $A$ directly east of $B$. During an exercise, $A$ fires a projectile west-northwest at angle $\theta$, and $B$ fires a projectile east-northwest at angle $\theta$, where $\theta$ is an acute angle. Both projectiles hit the same target $P$ 9 kilometers away. Then $A$ fires another projectile west-northwest at angle $\frac{\theta}{2}$, landing at point $Q$ 9 kilometers away. What is the distance $\overline{BQ}$ between artillery battery $B$ and landing point $Q$?
(1) 4 kilometers
(2) 4.5 kilometers
(3) 5 kilometers
(4) 5.5 kilometers
(5) 6 kilometers

For any positive integer $n \geq 2$, let $T_{n}$ denote a triangle with side lengths $n, n+1, n+2$. Select the correct options. (Note: If a triangle has side lengths $a, b, c$ respectively, let $s = \frac{a+b+c}{2}$, then the area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$)
(1) $T_{n}$ is always an acute triangle
(2) The perimeters of $T_{2}, T_{3}, T_{4}, \cdots, T_{10}$ form an arithmetic sequence
(3) The area of $T_{n}$ increases as $n$ increases
(4) The three altitudes of $T_{5}$ form an arithmetic sequence in order
(5) The largest angle of $T_{3}$ is greater than the largest angle of $T_{2}$

As shown in the figure, consider a rectangular stone block with a vertex $A$ and a face containing point $A$. Let the midpoints of the edges of this face be $B , E , F , D$ respectively. Another face of the rectangular block containing point $B$ has its edge midpoints as $B , C , H , G$ respectively. Given that $\overline { B C } = 8$ and $\overline { B D } = \overline { D C } = 9$. The stone block is now cut to remove eight corners, such that the cutting plane for each corner passes through the midpoints of the three adjacent edges of that corner.
Find the area of $\triangle B C D$. (Non-multiple choice question, 4 points)

On a plane, there is a triangle $A B C$ where $\angle A = 91 ^ { \circ }, \angle C = 29 ^ { \circ }$. Let $\overline { B C } = a, \overline { C A } = b, \overline { A B } = c$. Select the correct options.
(1) $a ^ { 2 } > b ^ { 2 } + c ^ { 2 }$
(2) $\frac { c } { a } > \sin 29 ^ { \circ }$
(3) $\frac { b } { a } > \cos 29 ^ { \circ }$
(4) $\frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { a b } < \sqrt { 3 }$
(5) The circumradius of triangle $A B C$ is less than $c$

In $\triangle A B C$, $\overline { A B } = \overline { B C } = 3$ and $\cos \angle A B C = - \frac { 1 } { 8 }$. On the circumcircle of $\triangle A B C$ there is a point $D$ satisfying $\overline { B D } = 4$ and $\overline { A D } \leq \overline { C D }$. Then $\overline { C D } = $ (17-1) $+$ $\sqrt{\text{(17-2)}}$. (Express as a simplified radical form.)

ABC is a triangle
$$\begin{aligned} & \mathrm { m } ( \widehat { \mathrm { ABC } } ) = 50 ^ { \circ } \\ & \mathrm { m } ( \widehat { \mathrm { CAB } } ) = 100 ^ { \circ } \end{aligned}$$
According to the given information, the expression $\frac { | a - b | + | b - c | + | c - a | } { 2 }$ is equal to which of the following?
A) a-c
B) $a - b$
C) $b - c$
D) $b - a$
E) $\mathrm { c } - \mathrm { b }$

The triangle ABC is drawn on unit squares as shown above. What is the tangent of angle $B$?
A) $\frac { 25 } { 4 }$
B) $\frac { 34 } { 5 }$
C) $\frac { 40 } { 9 }$
D) 4
E) 5

ABCD is a square, $|BE| = 5$ cm, $|EC| = 7$ cm, $m(\widehat{EAC}) = x$.
According to the given information, what is $\tan x$?
A) $\frac { 4 } { 13 }$
B) $\frac { 6 } { 13 }$
C) $\frac { 9 } { 13 }$
D) $\frac { 5 } { 17 }$
E) $\frac { 7 } { 17 }$

ABC and DEC are triangles
$$\begin{aligned} & \mathrm { m } ( \widehat { \mathrm { CAB } } ) = \mathrm { m } ( \widehat { \mathrm { DEC } } ) \\ & | \mathrm { AD } | = 5 \mathrm {~cm} \\ & | \mathrm { DC } | = 3 \mathrm {~cm} \\ & | \mathrm {~EB} | = 2 \mathrm {~cm} \\ & | \mathrm { BC } | = x \end{aligned}$$
According to the given information, what is x in cm?
A) 4
B) 5
C) $\frac { 9 } { 2 }$
D) $\frac { 10 } { 3 }$
E) $\frac { 13 } { 3 }$

$| \mathrm { AB } | = 6$ units $| \mathrm { BH } | = 2$ units $[ \mathrm { BC } ] \cap [ \mathrm { GF } ] = \mathrm { H }$ $\mathrm { m } ( \widehat { \mathrm { GHC } } ) = \mathrm { x }$
In the figure, $ABCD$ and $AEFG$ are congruent squares. Accordingly, what is the value of $\tan ( x )$?
A) $\frac { 1 } { 3 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 5 } { 3 }$
D) $\frac { 3 } { 4 }$
E) $\frac { 5 } { 4 }$