Question 141
A figura representa um mapa com a localização de três cidades: A, B e C. As distâncias entre as cidades, em linha reta, são: A a B = 80 km, B a C = 60 km e A a C = 100 km.
[Figure]
O ângulo formado pelas estradas AB e AC, em graus, é
(A) $30^\circ$ (B) $37^\circ$ (C) $45^\circ$ (D) $53^\circ$ (E) $60^\circ$

Question 169
A figura mostra dois triângulos semelhantes $ABC$ e $DEF$.
[Figure]
Se $AB = 6$ cm, $BC = 8$ cm, $AC = 10$ cm e $DE = 9$ cm, o perímetro do triângulo $DEF$, em cm, é
(A) 24 (B) 30 (C) 36 (D) 40 (E) 45

Um triângulo tem lados medindo 5 cm, 12 cm e 13 cm. A área desse triângulo é
(A) 20 cm$^2$ (B) 25 cm$^2$ (C) 30 cm$^2$ (D) 35 cm$^2$ (E) 40 cm$^2$

A triangle has sides measuring 5 cm, 12 cm, and 13 cm. What is the area, in square centimeters, of this triangle?
(A) 20
(B) 25
(C) 30
(D) 35
(E) 40

In each of the following independent situations we want to construct a triangle $ABC$ satisfying the given conditions. In each case state how many such triangles $ABC$ exist up to congruence.
(i) $AB = 30 \quad BC = 95 \quad AC = 55$
(ii) $\angle A = 30 ^ { \circ } \quad \angle B = 95 ^ { \circ } \quad \angle C = 55 ^ { \circ }$
(iii) $\angle A = 30 ^ { \circ } \quad \angle B = 95 ^ { \circ } \quad BC = 55$
(iv) $\angle A = 30 ^ { \circ } \quad AB = 95 \quad BC = 55$

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 $\dfrac{4(1-k)}{\sqrt{k^{2}-2k+4}}$.

We want to construct a triangle ABC such that angle A is $20.21 ^ { \circ }$, side AB has length 1 and side BC has length $x$ where $x$ is a positive real number. Let $N ( x ) =$ the number of pairwise noncongruent triangles with the required properties.
(a) There exists a value of $x$ such that $N ( x ) = 0$.
(b) There exists a value of $x$ such that $N ( x ) = 1$.
(c) There exists a value of $x$ such that $N ( x ) = 2$.
(d) There exists a value of $x$ such that $N ( x ) = 3$.

(Calculus) As shown in the figure, for a positive angle $\theta$, there is an isosceles triangle ABC with $\angle \mathrm { ABC } = \angle \mathrm { ACB } = \theta$ and $\overline { \mathrm { BC } } = 2$. Let O be the center of the inscribed circle of triangle ABC, D be the point where segment AB meets the inscribed circle, and E be the point where segment AC meets the inscribed circle. [3 points]
(1) $\frac { \pi } { 4 } - 1$
(2) $\frac { \pi } { 4 }$
(3) $\frac { \pi } { 4 } + \frac { 1 } { 3 }$
(4) $\frac { \pi } { 4 } + \frac { 1 } { 2 }$
(5) $\frac { \pi } { 4 } + 1$

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]

As shown in the figure, there is an isosceles triangle ABC with AB as one side of length 4, $\overline { \mathrm { AC } } = \overline { \mathrm { BC } }$, and $\angle \mathrm { ACB } = \theta$. On the extension of segment AB, a point D is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AD } }$, and a point P is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AP } }$ and $\angle \mathrm { PAB } = 2 \theta$. Let $S ( \theta )$ be the area of triangle BDP. Find the value of $\lim _ { \theta \rightarrow + 0 } ( \theta \times S ( \theta ) )$. (Here, $0 < \theta < \frac { \pi } { 6 }$) [4 points]

As shown in the figure, in triangle ABC with $\overline { \mathrm { AB } } = 5 , \overline { \mathrm { AC } } = 2 \sqrt { 5 }$, let D be the foot of the perpendicular from vertex A to segment BC.
For point E that divides segment AD internally in the ratio $3 : 1$, we have $\overline { \mathrm { EC } } = \sqrt { 5 }$. If $\angle \mathrm { ABD } = \alpha , \angle \mathrm { DCE } = \beta$, what is the value of $\cos ( \alpha - \beta )$? [4 points]
(1) $\frac { \sqrt { 5 } } { 5 }$
(2) $\frac { \sqrt { 5 } } { 4 }$
(3) $\frac { 3 \sqrt { 5 } } { 10 }$
(4) $\frac { 7 \sqrt { 5 } } { 20 }$
(5) $\frac { 2 \sqrt { 5 } } { 5 }$

In triangle ABC with $\angle \mathrm { A } = \frac { \pi } { 3 }$ and $\overline { \mathrm { AB } } : \overline { \mathrm { AC } } = 3 : 1$, the radius of the circumcircle of triangle ABC is 7. What is the length of segment AC? [3 points]
(1) $2 \sqrt { 5 }$
(2) $\sqrt { 21 }$
(3) $\sqrt { 22 }$
(4) $\sqrt { 23 }$
(5) $2 \sqrt { 6 }$

In triangle ABC, $\angle \mathrm { A } = \frac { \pi } { 3 }$ and $\overline { \mathrm { AB } } : \overline { \mathrm { AC } } = 3 : 1$. If the circumradius of triangle ABC is 7, let $k$ be the length of segment AC. Find the value of $k ^ { 2 }$. [4 points]

As shown in the figure, quadrilateral ABCD is inscribed in a circle and $$\overline { \mathrm { AB } } = 5 , \overline { \mathrm { AC } } = 3 \sqrt { 5 } , \overline { \mathrm { AD } } = 7 , \angle \mathrm { BAC } = \angle \mathrm { CAD }$$ What is the radius of this circle? [4 points]
(1) $\frac { 5 \sqrt { 2 } } { 2 }$
(2) $\frac { 8 \sqrt { 5 } } { 5 }$
(3) $\frac { 5 \sqrt { 5 } } { 3 }$
(4) $\frac { 8 \sqrt { 2 } } { 3 }$
(5) $\frac { 9 \sqrt { 3 } } { 4 }$

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) $42 + 35\sqrt{3}$

In triangle ABC, $\overline { \mathrm { AB } } = 5$, $\overline { \mathrm { AC } } = 6$, and $\cos ( \angle \mathrm { BAC } ) = - \frac { 3 } { 5 }$. Find the area of triangle ABC. [3 points]

18. If the three interior angles of $\triangle A B C$ satisfy $\sin A : \sin B : \sin C = 5 : 11 : 13$ , then $\triangle A B C$
A. must be an acute triangle
B. must be a right triangle
C. must be an obtuse triangle
D. could be either an acute triangle or an obtuse triangle

III. Solution Problems (Total Score: 74 points)

8. At two points $A$ and $B$ that are 2 kilometers apart, target $C$ is measured. If $\angle CAB = 75°, \angle CBA = 60°$, then the distance between points $A$ and $C$ is $\_\_\_\_$ kilometers.

In $\triangle ABC$, $a = 3$, $b = \sqrt { 6 }$, $\angle A = \frac { 2 \pi } { 3 }$, then $\angle B =$

12. In $\triangle A B C$, $a = 4 , b = 5 , c = 6$, then $\frac { \sin 2 A } { \sin C } =$ $\_\_\_\_$.

12. If the area of acute triangle $A B C$ is $10 \sqrt { 3 }$, and $A B = 5$, $A C = 8$, then $B C$ equals $\_\_\_\_$.

In $\triangle A B C$, let the sides opposite to angles $\mathrm { A } , \mathrm { B } , \mathrm { C }$ be $a , b , c$ respectively. Given $a = 2 , \cos C = - \frac { 1 } { 4 } , 3 \sin A = 2 \sin B$, then $\mathrm { c } =$ $\_\_\_\_$ .

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.

13. As shown in the figure, a car is traveling due west on a horizontal road. At point $A$, the mountain peak $D$ on the north side of the road is measured to be in the direction $30°$ west of north. After traveling 600 m to reach point $B$, the peak is measured to be in the direction $75°$ west of north with an elevation angle of $30°$. The height of the mountain $CD = $ $\_\_\_\_$ m.
[Figure]
Figure for Question 13
[Figure]
Figure for Question 14