In triangle $ABC$, the bisector of angle $A$ meets side $BC$ in point $D$ and the bisector of angle $B$ meets side $AC$ in point $E$. Given that $DE$ is parallel to $AB$, show that $\mathrm{AE} = \mathrm{BD}$ and that the triangle $ABC$ is isosceles.

Let $ABC$ be an equilateral triangle with side length 2. Point $A'$ is chosen on side $BC$ such that the length of $A'B$ is $k < 1$. Likewise points $B'$ and $C'$ are chosen on sides $CA$ and $AB$ with $CB' = AC' = k$. Line segments are drawn from points $A', B', C'$ to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled $PQR$. Show that $PQR$ is an equilateral triangle with side length $\dfrac{4(1-k)}{\sqrt{k^{2}-2k+4}}$.

Three positive real numbers $x, y, z$ satisfy $$\begin{aligned} x^{2} + y^{2} &= 3^{2} \\ y^{2} + yz + z^{2} &= 4^{2} \\ x^{2} + \sqrt{3}\,xz + z^{2} &= 5^{2} \end{aligned}$$ Find the value of $2xy + xz + \sqrt{3}\,yz$.

[11 points] Given $\triangle XYZ$, the following constructions are made: mark point $W$ on segment $XZ$, point $P$ on segment $XW$ and point $Q$ on segment $YZ$ such that
$$\frac{WZ}{YX} = \frac{PW}{XP} = \frac{QZ}{YQ} = k$$
Extend segments $QP$ and $YX$ to meet at the point $R$ as shown. Prove that $XR = XP$.
Hint (use this or your own method): A suitable construction may help in calculations.

[14 points] In $\triangle A B C , \angle B A C = 2 \angle A C B$ and $0 ^ { \circ } < \angle B A C < 120 ^ { \circ }$. A point $M$ is chosen in the interior of $\triangle A B C$ such that $B A = B M$ and $M A = M C$. Prove that $\angle M C B = 30 ^ { \circ }$.
Hint (use this or your own method): Draw a suitable segment $C D$ of appropriate length making an appropriate angle with $C A$.

In triangle ABC, $\overline { \mathrm { AB } } = 1$, $\angle \mathrm { A } = \theta$, and $\angle \mathrm { B } = 2 \theta$. Point D on side AB is chosen so that $\angle \mathrm { ACD } = 2 \angle \mathrm { BCD }$. When $\lim _ { \theta \rightarrow + 0 } \frac { \overline { \mathrm { CD } } } { \theta } = a$, find the value of $27 a ^ { 2 }$. (Given that $0 < \theta < \frac { \pi } { 4 }$.) [4 points]

There are two circles $C _ { 1 } , C _ { 2 }$ with centers $\mathrm { O } _ { 1 } , \mathrm { O } _ { 2 }$ respectively and radii equal to $\overline { \mathrm { O } _ { 1 } \mathrm { O } _ { 2 } }$. As shown in the figure, three distinct points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ on circle $C _ { 1 }$ and a point $\mathrm { D }$ on circle $C _ { 2 }$ are given, with three points $\mathrm { A } , \mathrm { O } _ { 1 } , \mathrm { O } _ { 2 }$ and three points $\mathrm { C } , \mathrm { O } _ { 2 } , \mathrm { D }$ each on a line.
Let $\angle \mathrm { BO } _ { 1 } \mathrm {~A} = \theta _ { 1 } , \angle \mathrm { O } _ { 2 } \mathrm { O } _ { 1 } \mathrm { C } = \theta _ { 2 } , \angle \mathrm { O } _ { 1 } \mathrm { O } _ { 2 } \mathrm { D } = \theta _ { 3 }$.
The following is the process of finding the ratio of the lengths of segments AB and CD when $\overline { \mathrm { AB } } : \overline { \mathrm { O } _ { 1 } \mathrm { D } } = 1 : 2 \sqrt { 2 }$ and $\theta _ { 3 } = \theta _ { 1 } + \theta _ { 2 }$.
Since $\angle \mathrm { CO } _ { 2 } \mathrm { O } _ { 1 } + \angle \mathrm { O } _ { 1 } \mathrm { O } _ { 2 } \mathrm { D } = \pi$, we have $\theta _ { 3 } = \frac { \pi } { 2 } + \frac { \theta _ { 2 } } { 2 }$, and from $\theta _ { 3 } = \theta _ { 1 } + \theta _ { 2 }$, we get $2 \theta _ { 1 } + \theta _ { 2 } = \pi$, so $\angle \mathrm { CO } _ { 1 } \mathrm {~B} = \theta _ { 1 }$. Since $\angle \mathrm { O } _ { 2 } \mathrm { O } _ { 1 } \mathrm {~B} = \theta _ { 1 } + \theta _ { 2 } = \theta _ { 3 }$, triangles $\mathrm { O } _ { 1 } \mathrm { O } _ { 2 } \mathrm {~B}$ and $\mathrm { O } _ { 2 } \mathrm { O } _ { 1 } \mathrm { D }$ are congruent. Let $\overline { \mathrm { AB } } = k$. Since $\overline { \mathrm { BO } _ { 2 } } = \overline { \mathrm { O } _ { 1 } \mathrm { D } } = 2 \sqrt { 2 } k$, we have $\overline { \mathrm { AO } _ { 2 } } =$ (a) and since $\angle \mathrm { BO } _ { 2 } \mathrm {~A} = \frac { \theta _ { 1 } } { 2 }$, we have $\cos \frac { \theta _ { 1 } } { 2 } =$ (b). In triangle $\mathrm { O } _ { 2 } \mathrm { BC }$, with $\overline { \mathrm { BC } } = k , \overline { \mathrm { BO } _ { 2 } } = 2 \sqrt { 2 } k , \angle \mathrm { CO } _ { 2 } \mathrm {~B} = \frac { \theta _ { 1 } } { 2 }$, by the law of cosines, $\overline { \mathrm { O } _ { 2 } \mathrm { C } } =$ (c). Since $\overline { \mathrm { CD } } = \overline { \mathrm { O } _ { 2 } \mathrm { D } } + \overline { \mathrm { O } _ { 2 } \mathrm { C } } = \overline { \mathrm { O } _ { 1 } \mathrm { O } _ { 2 } } + \overline { \mathrm { O } _ { 2 } \mathrm { C } }$, $\overline { \mathrm { AB } } : \overline { \mathrm { CD } } = k : \left( \frac { \text{(a)} } { 2 } + \text{(c)} \right)$.
Let the expressions for (a) and (c) be $f ( k )$ and $g ( k )$ respectively, and let the number for (b) be $p$. What is the value of $f ( p ) \times g ( p )$? [4 points]
(1) $\frac { 169 } { 27 }$
(2) $\frac { 56 } { 9 }$
(3) $\frac { 167 } { 27 }$
(4) $\frac { 166 } { 27 }$
(5) $\frac { 55 } { 9 }$

As shown in the figure, $$\overline{\mathrm{AB}} = 3, \quad \overline{\mathrm{BC}} = \sqrt{13}, \quad \overline{\mathrm{AD}} \times \overline{\mathrm{CD}} = 9, \quad \angle\mathrm{BAC} = \frac{\pi}{3}$$ For quadrilateral ABCD, let $S_1$ denote the area of triangle ABC, $S_2$ denote the area of triangle ACD, and $R$ denote the circumradius of triangle ACD. If $S_2 = \frac{5}{6}S_1$, find the value of $\frac{R}{\sin(\angle\mathrm{ADC})}$. [4 points]
(1) $\frac{54}{25}$
(2) $\frac{117}{50}$
(3) $\frac{63}{25}$
(4) $\frac{27}{10}$
(5) $\frac{72}{25}$

As shown in the figure, in triangle ABC, point D is taken on segment AB such that $\overline{\mathrm{AD}} : \overline{\mathrm{DB}} = 3 : 2$, and a circle $O$ centered at A passing through D intersects segment AC at point E. $\sin A : \sin C = 8 : 5$, and the ratio of the areas of triangles ADE and ABC is $9 : 35$. When the circumradius of triangle ABC is 7, what is the maximum area of triangle PBC for a point P on circle $O$? (Given: $\overline{\mathrm{AB}} < \overline{\mathrm{AC}}$) [4 points]
(1) $18 + 15\sqrt{3}$
(2) $24 + 20\sqrt{3}$
(3) $30 + 25\sqrt{3}$
(4) $36 + 30\sqrt{3}$
(5) $42 + 35\sqrt{3}$

As shown in the figure, there is a right triangle ABC with $\overline { \mathrm { AB } } = 3$, $\overline { \mathrm { BC } } = 4$, and $\angle \mathrm { B } = \frac { \pi } { 2 }$. Let D be the point that divides segment AB internally in the ratio $2 : 1$, let E be the point where the circle centered at A with radius $\overline { \mathrm { AD } }$ meets segment AC, let F be the point where the line AB meets this circle other than D, and let G be a point on arc EF such that $\overline { \mathrm { CG } } = 2 \sqrt { 6 }$. When point H on the circle passing through the three points C, E, G satisfies $\angle \mathrm { HCG } = \angle \mathrm { BAC }$, what is the length of segment GH? [4 points]
(1) $\frac { 6 \sqrt { 15 } } { 5 }$
(2) $\frac { 38 \sqrt { 10 } } { 25 }$
(3) $\frac { 14 \sqrt { 3 } } { 5 }$
(4) $\frac { 32 \sqrt { 15 } } { 25 }$
(5) $\frac { 8 \sqrt { 10 } } { 5 }$

13. In $\triangle \mathrm { ABC }$, $\mathrm { B } = 120 ^ { \circ } , \mathrm { AB } = \sqrt { 2 }$, and the angle bisector from $A$ is $\mathrm { AD } = \sqrt { 3 }$, then $\mathrm { AC } = $ $\_\_\_\_$ . Note for Candidates: Questions (14), (15), and (16) are optional. Please choose any two to answer. If all three are answered, only the first two will be graded.

In $\triangle \mathrm { ABC }$, $D$ is a point on $BC$, $AD$ bisects $\angle \mathrm { BAC }$, and the area of $\triangle \mathrm { ABD }$ is 2 times the area of $\triangle \mathrm { ADC }$.
(I) Find $\frac { \sin \angle B } { \sin \angle C }$ ;
(II) If $A D = 1 , D C = \frac { \sqrt { 2 } } { 2 }$, find the lengths of $B D$ and $A C$.

In planar quadrilateral $A B C D$, $\angle A D C = 90 ^ { \circ }$, $\angle A = 45 ^ { \circ }$, $A B = 2$, $B D = 5$.
(1) Find $\cos \angle A D B$;
(2) If $D C = 2 \sqrt { 2 }$, find $B C$.

In $\triangle A B C$, $\cos C = \frac { 2 } { 3 } , A C = 4 , B C = 3$. Then $\tan B =$
A. $\sqrt { 5 }$
B. $2 \sqrt { 5 }$
C. $4 \sqrt { 5 }$
D. $8 \sqrt { 5 }$

19.
(1)
By the sine rule:
$$B D = \frac { a \sin C } { \sin B } = \frac { a c } { b } = \frac { b ^ { 2 } } { b } = b$$
(2)
Note that $\angle B D A$ and $\angle B D C$ are supplementary, so their cosines are opposite in sign. Applying the cosine rule in triangles $\triangle B D A$ and $\triangle B D C$:
$$\cos \angle B D A + \cos \angle B D C = \frac { b ^ { 2 } + \left( \frac { 2 } { 3 } b \right) ^ { 2 } - c ^ { 2 } } { 2 \cdot \left( \frac { 2 } { 3 } b \right) \cdot b } + \frac { b ^ { 2 } + \left( \frac { 1 } { 3 } b \right) ^ { 2 } - a ^

18. In $\triangle A B C$ , the sides opposite to angles $A$ , $B$ , $C$ are $a$ , $b$ , $c$ respectively, with $b = a + 1$ , $c = a + 2$ .
(1) If $2 \sin C = 3 \sin A$ , find the area of $\triangle A B C$ ;
(2) Does there exist a positive integer $a$ such that $\triangle A B C$ is an obtuse triangle? If it exists, find the value of $a$ ; if not, explain the reason.
Answer: (1) $\frac { 15 \sqrt { 7 } } { 4 }$ ; (2) Yes, and $a = 2$ .

[Solution]

[Analysis] (1) By the law of sines, we can obtain $2 c = 3 a$ . Combined with the known conditions, find the value of $a$ , and further obtain the values of $b$ and $c$ . Use the law of cosines and the fundamental trigonometric identity to find $\sin B$ , then use the area formula of a triangle to obtain the result;
(2) Analyze that angle $C$ is obtuse. From $\cos C < 0$ combined with the triangle inequality, find the value of the integer $a$ . [Detailed Solution] (1) Since $2 \sin C = 3 \sin A$ , by the law of sines we have $2 c = 3 a$ . Thus $2 ( a + 2 ) = 3 a$ , so $a = 4$ . Therefore $b = 5$ , $c = 6$ . By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { 16 + 25 - 36 } { 2 \times 4 \times 5 } = \frac { 5 } { 40 } = \frac { 1 } { 8 }$ . Since $C$ is acute, $\sin C = \sqrt { 1 - \cos ^ { 2 } C } = \sqrt { 1 - \frac { 1 } { 64 } } = \frac { 3 \sqrt { 7 } } { 8 }$ .
Therefore, $S _ { \triangle A B C } = \frac { 1 } { 2 } a b \sin C = \frac { 1 } { 2 } \times 4 \times 5 \times \frac { 3 \sqrt { 7 } } { 8 } = \frac { 15 \sqrt { 7 } } { 4 }$ ;
(2) Clearly $c > b > a$ . If $\triangle A B C$ is an obtuse triangle, then $C$ is obtuse. By the law of cosines: $\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b } = \frac { a ^ { 2 } + ( a + 1 ) ^ { 2 } - ( a + 2 ) ^ { 2 } } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } + a ^ 2 + 2a + 1 - a ^ 2 - 4a - 4 } { 2 a ( a + 1 ) } = \frac { a ^ { 2 } - 2 a - 3 } { 2 a ( a + 1 ) } < 0$ , Solving: $- 1 < a < 3$ , thus $0 < a < 3$ . By the triangle inequality: $a + a + 1 > a + 2$ , we get $a > 1$ . Since $a \in \mathbb{Z}$ , we have $a = 2$ .

In $\triangle A B C$, point $D$ is on side $B C$, $\angle A D B = 120 ^ { \circ } , A D = 2 , C D = 2 B D$ . If $S_{\triangle ABD} : S_{\triangle ACD} = 1 : 2$, then $B D =$ $\_\_\_\_$ .

In $\triangle ABC$, point $D$ is on side $BC$, $\angle ADB = 120 ^ { \circ }$, $AD = 2$, and $CD = 2 BD$. Then $BD =$ $\_\_\_\_$.

In $\triangle ABC$ , $\angle BAC = 60^{\circ} , AB = 2 , BC = \sqrt{6}$ . $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$ . Then $AD =$ $\_\_\_\_$ .

In $\triangle A B C$, it is given that $\angle B A C = 120 ^ { \circ } , A B = 2 , A C = 1$.
(1) Find $\sin \angle A B C$.
(2) If $D$ is a point on $BC$ such that $\angle B A D = 90 ^ { \circ }$, find the area of $\triangle A D C$.

128. In triangle $ABC$, the side lengths are $BC = 9$, $AC = 8$, and $AB = 2$. The angle bisectors of angle $A$ intersect side $BC$ at $M$ and $N$. What is the length of $MN$?
(1) $4/2$ (2) $4/5$ (3) $4/8$ (4) $5/1$

28. In triangle $ABC$, the medians drawn from vertices $B$ and $C$ are perpendicular to each other. If the length of the median drawn from vertex $C$ is $4.5$ and the area of this triangle is $18$, what is the ratio of the lengths of the medians drawn from vertices $B$ and $C$?
(1) $\dfrac{17}{9}$ (2) $\dfrac{19}{9}$ (3) $\dfrac{5}{3}$ (4) $\dfrac{4}{3}$

32. In the figure below, if $\widehat{DAC} = 3\widehat{BAD}$, what is the length of side $AC$?
[Figure: Triangle with vertices $A$, $B$, $C$, point $D$ on $BC$ with $BD = 4$, $AB = 8$, $AD = 6$]

• [(1)] $19/2$
• [(2)] $16/8$
• [(3)] $18/6$
• [(4)] $15/4$

In a triangle, $E$ is the midpoint of $AC$. Let $\angle BCE = \angle ABE$. Prove that $AB + BD = CD$ (where $D$ is the midpoint of $BC$), i.e., $AB + BD = l_1 + l_2$.

Let $ABC$ be a triangle $A \neq B$ and let $P \in (AB)$ be a point for which denote $m(\widehat{ACP}) = x$ and $m(\widehat{BCP}) = y$. Prove that $\frac{\sin A \sin B}{\sin(A-B)} = \frac{\sin x \sin y}{\sin(x-y)}$ if and only if $PA = PB$.