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 166
Um triângulo tem lados medindo 7 cm, 24 cm e 25 cm. A área desse triângulo, em cm², é
(A) 42 (B) 84 (C) 87,5 (D) 168 (E) 175

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

The owner of a boat must depart from point $P$ and arrive at point $R$ by means of two linear displacements and navigating at a constant speed. This trip will be made during the night, and since he has only a compass and a clock, he planned his route as follows: $1^{\text{st}}$ - depart from point $P$ in direction 110 and navigate for 4 hours, reaching a point $Q$; $2^{\text{nd}}$ - depart from point $Q$ in direction 90 and navigate for 2 hours, reaching the destination point $R$.
However, when directing the boat for the first displacement, he did so in direction 340, instead of 110. With this, he made the following displacements: $1^{\text{st}}$ - departed from point $P$ in direction 340 and navigated for 4 hours, reaching a point $S$; $2^{\text{nd}}$ - departed from point $S$ in direction 90 and navigated for 2 hours, reaching point $T$.
The boat owner only realized the mistake upon arriving at point $T$. With this, he now needs to define the direction and navigation time that will allow him, departing from point $T$, to reach the destination point $R$ through a straight route.
Consider 0.64 as an approximation for $\cos 50°$. The direction and approximate navigation time that the boat owner should use are, respectively,
(A) 135 and 7 hours and 15 minutes.
(B) 45 and 7 hours and 15 minutes.
(C) 135 and 12 hours.
(D) 135 and 6 hours.
(E) 45 and 6 hours.

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.

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) $\angle A = 30 ^ { \circ } \quad AB = 95 \quad BC = 55$

Consider the following construction in a circle. Choose points $A, B, C$ on the given circle such that $\angle ABC$ is $60^\circ$. Draw another circle that is tangential to the chords $AB$, $BC$ and to the original circle. Do the above construction in the unit circle to obtain a circle $S_1$. Repeat the process in $S_1$ to obtain another circle $S_2$. What is the radius of $S_2$?

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 $2xy + xz + \sqrt{3}\,yz$.

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 $N ( x ) = 3$.

[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.

The three sides of triangle $a < b < c$ are in arithmetic progression (AP) with common difference $d = b - a = c - b$. Denote the angles opposite to sides $a , b , c$ respectively by $A , B , C$.
Statements
(1) $d$ must be less than $a$.
(2) $d$ can be any positive number less than $a$.
(3) The numbers $\sin A , \sin B , \sin C$ are in AP.
(4) The numbers $\cos A , \cos B , \cos C$ are in AP.

[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 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]

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, 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) $42 + 35\sqrt{3}$

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) $\frac { 8 \sqrt { 10 } } { 5 }$

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]

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 $\mathrm { c } =$ $\_\_\_\_$ .