Suppose $A B C D$ is a quadrilateral such that $\angle B A C = 50 ^ { \circ } , \angle C A D = 60 ^ { \circ } , \angle C B D = 30 ^ { \circ }$ and $\angle B D C = 25 ^ { \circ }$. If $E$ is the point of intersection of $A C$ and $B D$, then the value of $\angle A E B$ is
(a) $75 ^ { \circ }$.
(B) $85 ^ { \circ }$.
(C) $95 ^ { \circ }$.
(D) $110 ^ { \circ }$.

In a triangle $XYZ$, let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$, respectively, and $2s = x + y + z$. If $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2}$ and area of incircle of the triangle $XYZ$ is $\frac{8\pi}{3}$, then
(A) area of the triangle $XYZ$ is $6\sqrt{6}$
(B) the radius of circumcircle of the triangle $XYZ$ is $\frac{35}{6}\sqrt{6}$
(C) $\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{4}{35}$
(D) $\sin^2\left(\frac{X+Y}{2}\right) = \frac{3}{5}$

In a triangle $P Q R$, let $\angle P Q R = 30 ^ { \circ }$ and the sides $P Q$ and $Q R$ have lengths $10 \sqrt { 3 }$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE?
(A) $\angle Q P R = 45 ^ { \circ }$
(B) The area of the triangle $P Q R$ is $25 \sqrt { 3 }$ and $\angle Q R P = 120 ^ { \circ }$
(C) The radius of the incircle of the triangle $P Q R$ is $10 \sqrt { 3 } - 15$
(D) The area of the circumcircle of the triangle $P Q R$ is $100 \pi$

In a non-right-angled triangle $\triangle P Q R$, let $p , q , r$ denote the lengths of the sides opposite to the angles at $P , Q , R$ respectively. The median from $R$ meets the side $P Q$ at $S$, the perpendicular from $P$ meets the side $Q R$ at $E$, and $R S$ and $P E$ intersect at $O$. If $p = \sqrt { 3 } , q = 1$, and the radius of the circumcircle of the $\triangle P Q R$ equals 1, then which of the following options is/are correct?
(A) Length of $R S = \frac { \sqrt { 7 } } { 2 }$
(B) Area of $\triangle S O E = \frac { \sqrt { 3 } } { 12 }$
(C) Length of $O E = \frac { 1 } { 6 }$
(D) Radius of incircle of $\triangle P Q R = \frac { \sqrt { 3 } } { 2 } ( 2 - \sqrt { 3 } )$

In a triangle $ABC$, let $AB = \sqrt{23}$, $BC = 3$ and $CA = 4$. Then the value of $$\frac{\cot A + \cot C}{\cot B}$$ is ____.

Let $P Q R S$ be a quadrilateral in a plane, where $Q R = 1 , \angle P Q R = \angle Q R S = 70 ^ { \circ } , \angle P Q S = 15 ^ { \circ }$ and $\angle P R S = 40 ^ { \circ }$. If $\angle R P S = \theta ^ { \circ } , P Q = \alpha$ and $P S = \beta$, then the interval(s) that contain(s) the value of $4 \alpha \beta \sin \theta ^ { \circ }$ is/are
(A) $( 0 , \sqrt { 2 } )$
(B) $( 1,2 )$
(C) $( \sqrt { 2 } , 3 )$
(D) $( 2 \sqrt { 2 } , 3 \sqrt { 2 } )$

Consider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac { \pi } { 2 }$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Let $a$ be the area of the triangle $ABC$. Then the value of $( 64a ) ^ { 2 }$ is

ABCD is a trapezium such that AB and CD are parallel and $BC \perp CD$. If $\angle ADB = \theta$, $BC = p$ and $CD = q$, then $AB$ is equal to
(1) $\frac{(p^{2}+q^{2})\sin\theta}{p\cos\theta+q\sin\theta}$
(2) $\frac{p^{2}+q^{2}\cos\theta}{p\cos\theta+q\sin\theta}$
(3) $\frac{p^{2}+q^{2}}{p^{2}\cos\theta+q^{2}\sin\theta}$
(4) $\frac{(p^{2}+q^{2})\sin\theta}{(p\cos\theta+q\sin\theta)^{2}}$

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$, $BC = p$ and $CD = q$, then $AB$ is equal to
(1) $\frac{p^2 + q^2}{p^2 \cos\theta + q^2 \sin\theta}$
(2) $\frac{\left(p^2 + q^2\right)\sin\theta}{(p\cos\theta + q\sin\theta)^2}$
(3) $\frac{\left(p^2 + q^2\right)\sin\theta}{p\cos\theta + q\sin\theta}$
(4) $\frac{p^2 + q^2\cos\theta}{p\cos\theta + q\sin\theta}$

For a triangle $ABC$, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is 3 and incentre is $M$, then which of the following is NOT correct?
(1) Perimeter of $\triangle ABC$ is $18\sqrt{3}$
(2) $\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
(3) $\overrightarrow{MA} \cdot \overrightarrow{MB} = -18$
(4) area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$

Consider a triangle ABC where
$$\mathrm { AB } = 10 , \quad \angle \mathrm { B } = 30 ^ { \circ }$$
and the radius of its inscribed circle O is 1.
(1) Set $a = \mathrm { BC }$ and $b = \mathrm { CA }$. Finding the area $S$ of the triangle ABC by two different methods, we have
$$S = \frac { \mathbf { A } } { \mathbf{B} } \mathbf { B } ,$$
and
$$S = \frac { \mathbf { C } } { \mathbf{C} } \mathbf { D } ( a + b + \mathbf { E C } ) .$$
Hence we obtain
$$b = \mathbf { G } a - \mathbf { H I } .$$
Since the relationship between $a$ and $b$ can also be expressed by the equation
$$b ^ { 2 } = a ^ { 2 } - \mathbf { J K } \sqrt { \mathbf { L } } a + \mathbf { M N O } ,$$
we have
$$a = \frac { \mathbf { P Q } - \mathbf { R } } { 3 } \sqrt { \mathbf { S } } , \quad b = \frac { \mathbf { T U } } { \mathbf{P} } \cdot \frac { \mathbf { V } } { 3 } \sqrt { \mathbf { W } } .$$
(2) Let D denote the point of intersection of the segment BC and the straight line which passes through the two points A and O. We denote the area of the triangle OBC by $S ^ { \prime }$. Since
$$S : S ^ { \prime } = \mathbf { X } : 1 ,$$
it follows that
$$\mathrm { AO } : \mathrm { OD } = \mathbf { Y } : 1 .$$

Consider a triangle ABC where $\mathrm{AB} = 8$, $\mathrm{AC} = 5$ and $\angle\mathrm{BAC} = 60^\circ$. Take point D and point E on sides AB and AC, respectively, such that the segment DE divides triangle ABC into two parts with the equal areas. Set $x = \mathrm{AD}$.
(1) When we represent AE in terms of $x$, we have $\mathrm{AE} = \frac{\mathbf{JK}}{x}$.
(2) When E moves along side AC, the range of the values which $x$ can take is
$$\mathbf{L} \leq x \leqq \mathbf{M}.$$
(3) Since $\mathrm{DE}^2 = \left(x - \frac{\mathbf{NO}}{x}\right)^2 + \mathbf{PQ}$, the length of segment DE is minimized at $x = \mathbf{R}\sqrt{\mathbf{S}}$, and its length there is $\mathbf{T}$.$\mathbf{U}$.

Suppose that in the figure to the right
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 5 , \quad \cos \angle \mathrm { BAC } = \frac { 1 } { 8 }$$
and
$$\angle \mathrm { BAD } = \angle \mathrm { ACB } , \quad \angle \mathrm { CAE } = \angle \mathrm { ABC } .$$
(1) When we denote the area of $\triangle \mathrm { ABC }$ by $S$, we have
$$S = \frac { \square \mathbf { A B } \sqrt { \square \mathbf { C } } } { \square } .$$
Also $\mathrm { BC } = \mathbf { E }$.
(2) Furthermore, when we denote the areas of $\triangle \mathrm { ABD }$ and $\triangle \mathrm { ACE }$ by $S _ { 1 }$ and $S _ { 2 }$, respectively, we have
$$S : S _ { 1 } : S _ { 2 } = 1 : \frac { \mathbf { F } } { \mathbf{G} } : \frac { \mathbf { H I } } { \mathbf { J } } .$$
(3) When we denote the area of $\triangle \mathrm { ADE }$ by $T$, we have
$$T = \frac { \mathbf { L M } \sqrt { \mathbf { N } } } { \mathbf { O P } } .$$
Also $\mathrm { DE } = \dfrac { \mathbf { Q } } { \mathbf{R} }$.

In the figure to the right, let $$\mathrm{OA} = 6, \quad \mathrm{OB} = 3, \quad \angle\mathrm{AOB} = 120^\circ,$$ and let the point Q denote the point of intersection of the bisector of $\angle\mathrm{XAB}$ and the bisector of $\angle\mathrm{ABY}$. Let P denote the point of intersection of segment AB and segment OQ. We are to find the length of segment PQ.
(1) First of all, we see that $\mathrm{AB} = \square\sqrt{\mathbf{A}}$ and that the area of triangle OAB is $\mathbf{BC}\sqrt{\mathbf{D}}$.
(2) For $\mathbf{F}$ and $\mathbf{G}$ in the following, choose the correct answer from among choices (0) $\sim$ (4), just below. (0) AB (1) AP (2) AQ (3) BP (4) BQ
Since AQ is the bisector of the exterior angle of $\angle\mathrm{A}$ of triangle OAP and BQ is the bisector of the exterior angle of $\angle\mathrm{B}$ of triangle OBP, we have $$\begin{aligned} \mathrm{OQ} : \mathrm{PQ} &= \mathrm{OA} : \mathbf{F} \\ &= \mathrm{OB} : \mathbf{G} \end{aligned}$$
(3) Thus we see $\mathrm{AP} = \mathbf{H}\sqrt{\mathbf{I}}$. Since $\angle\mathrm{AOP} = \mathbf{JK}^\circ$, we have $\mathrm{OP} = \square\mathbf{L}$. Hence we have $\mathrm{PQ} = \mathbf{M} + \mathbf{N}\sqrt{\mathbf{O}}$.

Consider a triangle ABC where $\angle \mathrm{ BAC } = 60 ^ { \circ }$.
Let D be the point of intersection of the bisector of $\angle \mathrm{ BAC }$ and the side BC. Let DE and DF be the line segments perpendicular to sides AB and AC, respectively. Let us set
$$x = \frac { \mathrm{ AB } } { \mathrm{ AC } } , \quad k = \frac { \triangle \mathrm{ DEF } } { \triangle \mathrm{ ABC } } .$$
Note that $\triangle \mathrm{ ABC }$ denotes the area of the triangle ABC, and similarly for other triangles.
(1) We are to represent $k$ in terms of $x$. Since $\triangle \mathrm{ ABD } + \triangle \mathrm{ ACD } = \triangle \mathrm{ ABC }$, when we set $b = \mathrm{ AB }$, $c = \mathrm{ AC }$ and $d = \mathrm{ AD }$, we have
$$d = \frac { \sqrt { \mathbf { A } } \, b c } { b + c } .$$
Next, since $\mathrm{ DE } = \mathrm{ DF } = \dfrac { \mathbf { B } } { \mathbf { C } } d$, we have
$$\triangle \mathrm{ DEF } = \frac { \sqrt { \mathbf { D } } } { \mathbf { EF } } d ^ { 2 } .$$
From (1) and (2), we see that
$$k = \frac { d ^ { 2 } } { \mathbf { G } \, b c } = \frac { \mathbf { H } \, b c } { \mathbf { I } ( b + c ) ^ { 2 } } .$$
Since $x = \dfrac { b } { c }$, we have
$$k = \frac { \mathbf { J } \, x } { \mathbf { K } ( x + \mathbf { L } ) ^ { 2 } } .$$
(2) If $\mathrm{ BD } = 8$ and $\mathrm{ BC } = 10$, then $x = \mathbf { M }$ and $k = \dfrac { \mathbf { N } } { \mathbf { O P } }$.

The triangle ABC satisfies
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 3 \quad \text { and } \quad \angle \mathrm { B } = 30 ^ { \circ } .$$
D is the point on side BC such that $\mathrm { AC } = \mathrm { AD }$. Let us consider the circumscribed circle O of triangle ACD.
(1) Since $\sin B = \frac { \mathbf { A } } { \mathbf { B } }$, we have $\sin C = \frac { \mathbf { C } } { \mathbf{D} }$.
Hence the radius of circle O is $\frac { \mathbf { E } } { \mathbf{F} }$.
(2) We have
$$\mathrm { BC } = \mathrm { G } \sqrt { \mathrm { H } } + \sqrt { \mathrm{H} }$$
and
$$\mathrm { BD } = \mathrm { J } \sqrt { \mathrm {~K} } - \sqrt { \mathrm { K } } .$$
Let us denote the intersection of side AB and circle O by E . Then
$$\mathrm { BE } = \frac { \mathbf { M } } { \mathbf{N} } .$$
Hence the relationships between the areas of triangles $\mathrm { BDE } , \mathrm { ADE }$ and ACD are
$$\begin{aligned} & \triangle \mathrm { BDE } : \triangle \mathrm { ADE } = \mathbf { O } : \mathbf { P } , \\ & \triangle \mathrm { BDE } : \triangle \mathrm { ACD } = \mathbf { Q } ( \mathbf{J} \sqrt { \mathbf { K } } - \sqrt { \mathrm { K } } ) : \mathbf { R S } \sqrt { \mathbf { T } } . \end{aligned}$$

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
$$\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 } = \mathbf { R } : ( \mathbf { S } - \sqrt { \mathbf { T } } ) .$$

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

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)

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)

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

ABC isosceles triangle\ $AD \cap BC = \{ E \}$\ $AD \perp BC$\ $| \mathrm { AB } | = | \mathrm { BD } | = 6$ units\ $| AC | = | BC | = 9$ units\ $| CE | = \mathrm { x }$
Accordingly, what is x in units?\ A) 4\ B) 5\ C) 6\ D) 7\ E) 8

3 identical isosceles trapezoids are joined together such that any two of them share a vertex as shown below.
One side of the large triangle formed is 6 units, and one side of the small triangle is 3 units.
Accordingly, what is the perimeter of one of these isosceles trapezoids in units?
A) 10
B) 10.5
C) 11
D) 11.5
E) 12

In the square-shaped paper ABCD given in Figure 1, $|\mathrm{DE}| = 6$ and $|\mathrm{BF}| = 9$ units. When this paper is folded along the line segments $[\mathrm{CE}]$ and $[\mathrm{CF}]$ as shown in the figure, the BC side and DC side of the square coincide as shown in Figure 2.
Accordingly, what is the perimeter of square ABCD in units?
A) 64
B) 68
C) 72
D) 76
E) 80