The plane is equipped with an orthonormal coordinate system ( $\mathrm { O } , \vec { u } , \vec { v }$ ). For all integer $n \geqslant 4$, we consider $P _ { n }$ a regular polygon with $n$ sides, with centre $O$ and whose area is equal to 1. We admit that such a polygon is made up of $n$ triangles superimposable to a given triangle $\mathrm { OA } _ { n } \mathrm {~B} _ { n }$, isosceles at O. We denote $r _ { n } = \mathrm { OA } _ { n }$ the distance between the centre O and the vertex $\mathrm { A } _ { n }$ of such a polygon.
Part A: study of the particular case $n = 6$

1. Justify the fact that the triangle $\mathrm { OA } _ { 6 } \mathrm {~B} _ { 6 }$ is equilateral, and that its area is equal to $\frac { 1 } { 6 }$.
2. Express as a function of $r _ { 6 }$ the height of the triangle $\mathrm { OA } _ { 6 } \mathrm {~B} _ { 6 }$ from the vertex $\mathrm { B } _ { 6 }$.
3. Deduce that $r _ { 6 } = \sqrt { \frac { 2 } { 3 \sqrt { 3 } } }$.

Part B: general case with $n \geqslant 4$
In the method considered, we take as initial matrix the matrix $I = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)$.

1. Determine the two missing matrices $A$ and $B$, in the third row of the Stern-Brocot tree.
2. We associate to a matrix $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ of the Stern-Brocot tree the fraction $\frac { a + c } { b + d }$. Show that, in this association, the path ``left-right-left'' starting from the initial matrix in the tree, leads to a matrix corresponding to the fraction $\frac { 3 } { 5 }$.
3. Let $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ be a matrix of the tree. We recall that $a , b , c , d$ are integers. We denote $\Delta _ { M } = a d - b c$, the difference of the diagonal products of this matrix. a. Show that if $a d - b c = 1$, then $d ( a + c ) - c ( b + d ) = 1$. b. Deduce that if $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ is a matrix of the Stern-Brocot tree such that $\Delta _ { M } = a d - b c = 1$, then $\Delta _ { M \times G } = 1$, that is, the difference of the diagonal products of the matrix $M \times G$ is also equal to 1. We similarly admit that $\Delta _ { M \times D } = 1$, and that all other matrices $N$ of the Stern-Brocot tree satisfy the equality $\Delta _ { N } = 1$.
4. Deduce from the previous question that every fraction associated with a matrix of the Stern-Brocot tree is in lowest terms.
5. Let $m$ and $n$ be two non-zero natural integers that are coprime. Thus the fraction $\frac { m } { n }$ is in lowest terms. We consider the following algorithm: \begin{verbatim} VARIABLES : m and n are non-zero natural integers and coprime PROCESSING : While m = do If m

   Display		$\ldots$	$\ldots$	$\ldots$	$\ldots$
   $m$	4	$\ldots$	$\ldots$	$\ldots$	$\ldots$
   $n$	7	$\ldots$	$\ldots$	$\ldots$	$\ldots$

   
   b. Conjecture the role of this algorithm. Verify by a matrix calculation the result provided with the values $m = 4$ and $n = 7$.

Question 172
A figura mostra um setor circular com raio $r = 6$ cm e ângulo central de $60^\circ$.
[Figure]
A área desse setor circular, em cm², é
(A) $3\pi$ (B) $6\pi$ (C) $9\pi$ (D) $12\pi$ (E) $36\pi$

A figura mostra um setor circular com raio $r = 6$ cm e ângulo central de $60^\circ$. A área desse setor circular é
(A) $3\pi$ cm$^2$ (B) $6\pi$ cm$^2$ (C) $9\pi$ cm$^2$ (D) $12\pi$ cm$^2$ (E) $18\pi$ cm$^2$

Let $A _ { n } =$ the area of a regular $n$-sided polygon inscribed in a circle of radius 1 (i.e., vertices of this regular $n$-sided polygon lie on a circle of radius 1). (i) Find $A _ { 12 }$. (ii) Find $\left\lfloor A _ { 2014 } \right\rfloor$, i.e., the greatest integer $\leq A _ { 2014 }$.

As shown in the figure, a sector $\mathrm { OA } _ { 1 } \mathrm {~B} _ { 1 }$ with radius equal to the line segment $\mathrm { OA } _ { 1 } ( 0,8 )$ connecting the origin $O$ and point $\mathrm { A } _ { 1 } ( 0,8 )$ and central angle $\theta$ is drawn. The foot of the perpendicular from point $\mathrm { B } _ { 1 }$ to the $x$-axis is $\mathrm { A } _ { 2 }$, and a sector $\mathrm { OA } _ { 2 } \mathrm {~B} _ { 2 }$ with radius equal to segment $\mathrm { OA } _ { 2 }$ and central angle $\theta$ is drawn. The foot of the perpendicular from point $\mathrm { B } _ { 2 }$ to the $y$-axis is $\mathrm { A } _ { 3 }$, and a sector $\mathrm { OA } _ { 3 } \mathrm {~B} _ { 3 }$ with radius equal to segment $\mathrm { OA } _ { 3 }$ and central angle $\theta$ is drawn. Continuing this process of alternately dropping perpendiculars to the $x$-axis and $y$-axis in the clockwise direction, let $l _ { n }$ be the length of arc $\mathrm { A } _ { n } \mathrm {~B} _ { n }$ of sector $\mathrm { OA } _ { n } \mathrm {~B} _ { n }$. When $\sum _ { n = 1 } ^ { \infty } l _ { n } = 12 \theta$, what is the value of $\sin \theta$? (Here, $0 < \theta < \frac { \pi } { 2 }$.) [4 points]
(1) $\frac { 1 } { 7 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 5 }$
(4) $\frac { 1 } { 4 }$
(5) $\frac { 1 } { 3 }$

As shown in the figure, a sector $\mathrm { OA } _ { 1 } \mathrm { B } _ { 1 }$ with radius equal to the line segment $\mathrm { OA } _ { 1 }$ connecting the origin O and the point $\mathrm { A } _ { 1 } ( 0,8 )$, and with central angle $\theta$, is drawn. The foot of the perpendicular from point $\mathrm { B } _ { 1 }$ to the $x$-axis is $\mathrm { A } _ { 2 }$, and a sector $\mathrm { OA } _ { 2 } \mathrm { B } _ { 2 }$ with radius equal to the line segment $\mathrm { OA } _ { 2 }$ and central angle $\theta$ is drawn. The foot of the perpendicular from point $\mathrm { B } _ { 2 }$ to the $y$-axis is $\mathrm { A } _ { 3 }$, and a sector $\mathrm { OA } _ { 3 } \mathrm { B } _ { 3 }$ with radius equal to the line segment $\mathrm { OA } _ { 3 }$ and central angle $\theta$ is drawn. Continuing this process of alternately dropping perpendiculars to the $x$-axis and $y$-axis in the clockwise direction, let $l _ { n }$ be the length of the arc $\mathrm { A } _ { n } \mathrm { B } _ { n }$ of the sector $\mathrm { OA } _ { n } \mathrm { B } _ { n }$. When $\sum _ { n = 1 } ^ { \infty } l _ { n } = 12 \theta$, what is the value of $\sin \theta$? (Here, $0 < \theta < \frac { \pi } { 2 }$.) [4 points]
(1) $\frac { 1 } { 7 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 5 }$
(4) $\frac { 1 } { 4 }$
(5) $\frac { 1 } { 3 }$

As shown in the figure, let $\mathrm { Q } _ { 1 }$ be the point where the tangent line to the circle $x ^ { 2 } + y ^ { 2 } = 1$ at point $\mathrm { P } _ { 1 }$ meets the $x$-axis. The area of triangle $\mathrm { P } _ { 1 } \mathrm { OQ } _ { 1 }$ is $\frac { 1 } { 4 }$. Let $\mathrm { P } _ { 2 }$ be the point obtained by rotating $\mathrm { P } _ { 1 }$ about the origin O by $\frac { \pi } { 4 }$, and let $\mathrm { Q } _ { 2 }$ be the point where the tangent line at $\mathrm { P } _ { 2 }$ meets the $x$-axis. What is the area of triangle $\mathrm { P } _ { 2 } \mathrm { OQ } _ { 2 }$? (Here, point $\mathrm { P } _ { 1 }$ is in the first quadrant.) [3 points]
(1) 1
(2) $\frac { 5 } { 4 }$
(3) $\frac { 3 } { 2 }$
(4) $\frac { 7 } { 4 }$
(5) 2

(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) $\frac { \pi } { 4 } + 1$

There is a rectangle $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$ with $\overline { \mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } } = 1$ and $\overline { \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } } = 2$. As shown in the figure, let $\mathrm { M } _ { 1 }$ be the midpoint of segment $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$, and on segment $\mathrm { A } _ { 1 } \mathrm { D } _ { 1 }$, two points $\mathrm { B } _ { 2 } , \mathrm { C } _ { 2 }$ are determined such that $\angle \mathrm { A } _ { 1 } \mathrm { M } _ { 1 } \mathrm {~B} _ { 2 } = \angle \mathrm { C } _ { 2 } \mathrm { M } _ { 1 } \mathrm { D } _ { 1 } = 15 ^ { \circ } , \angle \mathrm { B } _ { 2 } \mathrm { M } _ { 1 } \mathrm { C } _ { 2 } = 60 ^ { \circ }$. Let $S _ { 1 }$ be the sum of the area of triangle $\mathrm { A } _ { 1 } \mathrm { M } _ { 1 } \mathrm {~B} _ { 2 }$ and the area of triangle $\mathrm { C } _ { 2 } \mathrm { M } _ { 1 } \mathrm { D } _ { 1 }$.
Quadrilateral $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ is a rectangle with $\overline { \mathrm { B } _ { 2 } \mathrm { C } _ { 2 } } = 2 \overline { \mathrm {~A} _ { 2 } \mathrm {~B} _ { 2 } }$ such that two points $\mathrm { A } _ { 2 } , \mathrm { D } _ { 2 }$ are determined as shown in the figure. Let $\mathrm { M } _ { 2 }$ be the midpoint of segment $\mathrm { B } _ { 2 } \mathrm { C } _ { 2 }$, and on segment $\mathrm { A } _ { 2 } \mathrm { D } _ { 2 }$, two points $\mathrm { B } _ { 3 } , \mathrm { C } _ { 3 }$ are determined such that $\angle \mathrm { A } _ { 2 } \mathrm { M } _ { 2 } \mathrm {~B} _ { 3 } = \angle \mathrm { C } _ { 3 } \mathrm { M } _ { 2 } \mathrm { D } _ { 2 } = 15 ^ { \circ }$, $\angle \mathrm { B } _ { 3 } \mathrm { M } _ { 2 } \mathrm { C } _ { 3 } = 60 ^ { \circ }$. Let $S _ { 2 }$ be the sum of the area of triangle $\mathrm { A } _ { 2 } \mathrm { M } _ { 2 } \mathrm {~B} _ { 3 }$ and the area of triangle $\mathrm { C } _ { 3 } \mathrm { M } _ { 2 } \mathrm { D } _ { 2 }$. Continuing this process, what is the value of $\sum _ { n = 1 } ^ { \infty } S _ { n }$ for the obtained $S _ { n }$? [4 points]
(1) $\frac { 2 + \sqrt { 3 } } { 6 }$
(2) $\frac { 3 - \sqrt { 3 } } { 2 }$
(3) $\frac { 4 + \sqrt { 3 } } { 9 }$
(4) $\frac { 5 - \sqrt { 3 } } { 5 }$
(5) $\frac { 7 - \sqrt { 3 } } { 8 }$

In rectangle $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$, $\overline { \mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } } = 1$ and $\overline { \mathrm {~A} _ { 1 } \mathrm { D } _ { 1 } } = 2$. As shown in the figure, let $\mathrm { M } _ { 1 }$ and $\mathrm {~N} _ { 1 }$ be the midpoints of segments $\mathrm { A } _ { 1 } \mathrm { D } _ { 1 }$ and $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$, respectively.
Draw a circular sector $\mathrm { N } _ { 1 } \mathrm { M } _ { 1 } \mathrm {~B} _ { 1 }$ with center $\mathrm { N } _ { 1 }$, radius $\overline { \mathrm { B } _ { 1 } \mathrm {~N} _ { 1 } }$, and central angle $\frac { \pi } { 2 }$, and draw a circular sector $\mathrm { D } _ { 1 } \mathrm { M } _ { 1 } \mathrm { C } _ { 1 }$ with center $\mathrm { D } _ { 1 }$, radius $\overline { \mathrm { C } _ { 1 } \mathrm { D } _ { 1 } }$, and central angle $\frac { \pi } { 2 }$. The region bounded by the arc $\mathrm { M } _ { 1 } \mathrm {~B} _ { 1 }$ and segment $\mathrm { M } _ { 1 } \mathrm {~B} _ { 1 }$ of sector $\mathrm { N } _ { 1 } \mathrm { M } _ { 1 } \mathrm {~B} _ { 1 }$ and the region bounded by the arc $\mathrm { M } _ { 1 } \mathrm { C } _ { 1 }$ and segment $\mathrm { M } _ { 1 } \mathrm { C } _ { 1 }$ of sector $\mathrm { D } _ { 1 } \mathrm { M } _ { 1 } \mathrm { C } _ { 1 }$ form a checkmark shape. Color this shape to obtain figure $R _ { 1 }$.
In figure $R _ { 1 }$, construct a rectangle $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ with vertices at point $\mathrm { A } _ { 2 }$ on segment $\mathrm { M } _ { 1 } \mathrm {~B} _ { 1 }$, point $\mathrm { D } _ { 2 }$ on arc $\mathrm { M } _ { 1 } \mathrm { C } _ { 1 }$, and two points $\mathrm { B } _ { 2 } , \mathrm { C } _ { 2 }$ on side $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$, such that $\overline { \mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } } : \overline { \mathrm { A } _ { 2 } \mathrm { D } _ { 2 } } = 1 : 2$. Color the shape created in the same way as for figure $R _ { 1 }$ to obtain figure $R _ { 2 }$.
Continue this process. Let $S _ { n }$ be the area of the colored region in figure $R _ { n }$ obtained at the $n$-th step. What is the value of $\lim _ { n \rightarrow \infty } S _ { n }$? [4 points]
(1) $\frac { 25 } { 19 } \left( \frac { \pi } { 2 } - 1 \right)$
(2) $\frac { 5 } { 4 } \left( \frac { \pi } { 2 } - 1 \right)$
(3) $\frac { 25 } { 21 } \left( \frac { \pi } { 2 } - 1 \right)$
(4) $\frac { 25 } { 22 } \left( \frac { \pi } { 2 } - 1 \right)$
(5) $\frac { 25 } { 23 } \left( \frac { \pi } { 2 } - 1 \right)$

As shown in the figure, there is an equilateral triangle $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 }$ with side length 1. Let $\mathrm { D } _ { 1 }$ be the midpoint of segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$, and let $\mathrm { B } _ { 2 }$ be a point on segment $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$ such that $\overline { \mathrm { C } _ { 1 } \mathrm { D } _ { 1 } } = \overline { \mathrm { C } _ { 1 } \mathrm {~B} _ { 2 } }$. Draw a sector $\mathrm { C } _ { 1 } \mathrm { D } _ { 1 } \mathrm {~B} _ { 2 }$ with center $\mathrm { C } _ { 1 }$. Let $\mathrm { A } _ { 2 }$ be the foot of the perpendicular from $\mathrm { B } _ { 2 }$ to segment $\mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$, and let $\mathrm { C } _ { 2 }$ be the midpoint of segment $\mathrm { C } _ { 1 } \mathrm {~B} _ { 2 }$. The figure $R _ { 1 }$ is obtained by shading the region enclosed by two segments $\mathrm { B } _ { 1 } \mathrm {~B} _ { 2 } , \mathrm {~B} _ { 1 } \mathrm { D } _ { 1 }$ and arc $\mathrm { D } _ { 1 } \mathrm {~B} _ { 2 }$, and the interior of triangle $\mathrm { C } _ { 1 } \mathrm {~A} _ { 2 } \mathrm { C } _ { 2 }$. In figure $R _ { 1 }$, let $\mathrm { D } _ { 2 }$ be the midpoint of segment $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 }$, and let $\mathrm { B } _ { 3 }$ be a point on segment $\mathrm { B } _ { 2 } \mathrm { C } _ { 2 }$ such that $\overline { \mathrm { C } _ { 2 } \mathrm { D } _ { 2 } } = \overline { \mathrm { C } _ { 2 } \mathrm {~B} _ { 3 } }$. Draw a sector $\mathrm { C } _ { 2 } \mathrm { D } _ { 2 } \mathrm {~B} _ { 3 }$ with center $\mathrm { C } _ { 2 }$. Let $\mathrm { A } _ { 3 }$ be the foot of the perpendicular from $\mathrm { B } _ { 3 }$ to segment $\mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$, and let $\mathrm { C } _ { 3 }$ be the midpoint of segment $\mathrm { C } _ { 2 } \mathrm {~B} _ { 3 }$. The figure $R _ { 2 }$ is obtained by shading the region enclosed by two segments $\mathrm { B } _ { 2 } \mathrm {~B} _ { 3 }$, $\mathrm { B } _ { 2 } \mathrm { D } _ { 2 }$ and arc $\mathrm { D } _ { 2 } \mathrm {~B} _ { 3 }$, and the interior of triangle $\mathrm { C } _ { 2 } \mathrm {~A} _ { 3 } \mathrm { C } _ { 3 }$. Continuing this process, let $S _ { n }$ be the area of the shaded part in the $n$-th figure $R _ { n }$. Find the value of $\lim _ { n \rightarrow \infty } S _ { n }$. [4 points]
(1) $\frac { 11 \sqrt { 3 } - 4 \pi } { 56 }$
(2) $\frac { 11 \sqrt { 3 } - 4 \pi } { 52 }$
(3) $\frac { 15 \sqrt { 3 } - 6 \pi } { 56 }$
(4) $\frac { 15 \sqrt { 3 } - 6 \pi } { 52 }$
(5) $\frac { 15 \sqrt { 3 } - 4 \pi } { 52 }$

As shown in the figure, in a right triangle ABC with $\overline { \mathrm { AB } } = 1 , \angle \mathrm { B } = \frac { \pi } { 2 }$, let D be the intersection of the angle bisector of $\angle \mathrm { C }$ and segment AB, and let E be the intersection of the circle with center A and radius $\overline { \mathrm { AD } }$ and segment AC. When $\angle \mathrm { A } = \theta$, let $S ( \theta )$ be the area of sector ADE and $T ( \theta )$ be the area of triangle BCE. What is the value of $\lim _ { \theta \rightarrow 0 + } \frac { \{ S ( \theta ) \} ^ { 2 } } { T ( \theta ) }$? [4 points] [Figure]
(1) $\frac { 1 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 3 } { 4 }$
(4) 1
(5) $\frac { 5 } { 4 }$

3. B

Solution: Let the slant height of the cone be $l$. According to the property that the arc length of a semicircle equals the circumference of the cone's base, we have $\pi l = 2 \sqrt { 2 } \pi$, so $l = 2 \sqrt { 2 }$.

4. The BeiDou-3 Global Navigation Satellite System is an important achievement of China's space program. In satellite navigation systems, geostationary satellites orbit in the plane of Earth's equator at an orbital altitude of 36,000 km (orbital altitude is the distance from the satellite to Earth's surface). Consider Earth as a sphere with center $O$ and radius $r = 6400$ km. The latitude of a point $A$ on Earth's surface is defined as the angle that $OA$ makes with the equatorial plane. The maximum latitude at which a geostationary satellite can be directly observed from Earth's surface is $a$. The surface area covered by the satellite signal on Earth's surface is $S = 2 \pi r ^ { 2 } ( 1 - \cos a )$ (in $\mathrm { km } ^ { 2 }$). What percentage of Earth's total surface area does $S$ represent? ( )
A. $26 \%$
B. $34 \%$
C. $42 \%$
D. $50 \%$
【Answer】C 【Solution】 【Analysis】From the given information, use the provided surface area formula and the formula for the surface area of a sphere to calculate the result. 【Detailed Solution】From the given information, the percentage of $S$ relative to Earth's surface area is approximately: $\frac { 2 \pi r ^ { 2 } ( 1 - \cos a ) } { 4 \pi r ^ { 2 } } = \frac { 1 - \cos a } { 2 } = \frac { 1 - \frac { 6400 } { 6400 + 36000 } } { 2 } \approx 0.42 = 42 \%$ . Therefore, the answer is: C.

As shown in the figure, $O$ is the center of a circle, $OA$ is the radius, arc $AB$ is part of the circle with center $O$ and radius $OA$, $C$ is the midpoint of chord $AB$, $D$ is on arc $AB$, and $CD \perp AB$. The formula for calculating the chord value $s$ is: $s = AB + \frac { CD ^ { 2 } } { OA }$. When $OA = 2$ and $\angle AOB = 60 ^ { \circ }$, then $s =$
A. $\frac { 11 - 3 \sqrt { 3 } } { 2 }$
B. $\frac { 11 - 4 \sqrt { 3 } } { 2 }$
C. $\frac { 9 - 3 \sqrt { 3 } } { 2 }$
D. $\frac { 9 - 4 \sqrt { 3 } } { 2 }$

Let $n \geq 3$ be an integer. Assume that inside a big circle, exactly $n$ small circles of radius $r$ can be drawn so that each small circle touches the big circle and also touches both its adjacent small circles. Then, the radius of the big circle is
(A) $r \operatorname{cosec} \frac{\pi}{n}$
(B) $r \left( 1 + \operatorname{cosec} \frac{2\pi}{n} \right)$
(C) $r \left( 1 + \operatorname{cosec} \frac{\pi}{2n} \right)$
(D) $r \left( 1 + \operatorname{cosec} \frac{\pi}{n} \right)$

Two parallel chords of a circle of radius 2 are at a distance $\sqrt { 3 } + 1$ apart. If the chords subtend at the center, angles of $\frac { \pi } { k }$ and $\frac { 2 \pi } { k }$, where $k > 0$, then the value of $[ k ]$ is [Note : [k] denotes the largest integer less than or equal to k]

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B ( = a )$ subtends an angle of $60 ^ { \circ }$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $30 ^ { \circ }$. The height of the tower is
(1) $\frac { 2 a } { \sqrt { 3 } }$
(2) $2 a \sqrt { 3 }$
(3) $\frac { a } { \sqrt { 3 } }$
(4) $a \sqrt { 3 }$

A huge circular arc of length 4.4 ly subtends an angle 4 s at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second? Given: $1\mathrm{ly} = 9.46 \times 10^{15}\mathrm{~m}$ $1\mathrm{AU} = 1.5 \times 10^{11}\mathrm{~m}$
(1) $3.5 \times 10^{6}\mathrm{~s}$
(2) $4.5 \times 10^{10}\mathrm{~s}$
(3) $4.1 \times 10^{8}\mathrm{~s}$
(4) $7.2 \times 10^{8}$

Let $AB$ and $PQ$ be two vertical poles, 160 m apart from each other. Let $C$ be the middle point of $B$ and $Q$, which are feet of these two poles. Let $\frac{\pi}{8}$ and $\theta$ be the angles of elevation from $C$ to $P$ and $A$, respectively. If the height of pole $PQ$ is twice the height of pole $AB$, then $\tan^2\theta$ is equal to
(1) $\frac{3-2\sqrt{2}}{2}$
(2) $\frac{3+\sqrt{2}}{2}$
(3) $\frac{3-2\sqrt{2}}{4}$
(4) $\frac{3-\sqrt{2}}{4}$

Let $O$ be the vertex (apex) of a right circular cone such that the radius of the base is 1 and the slant height is 3.
(1) Consider the net of the cone, which consists of a sector and a circle. (The net of a solid is a 2-dimensional shape that can be folded to form that solid.) The central angle of the sector is $\square$ ABC , and the area of the sector is $\square$ D $\pi$.
(2) Take two points A and B on the circumference of the base such that the line segment AB is a diameter. Take a point P on the segment OB and consider a path on the side of this circular cone which starts from the point A , passes through the point P and returns to A . Denote the length of the path by $\ell$.
(i) If $\mathrm { OP } = 2$, then the smallest value of $\ell$ is $\mathbf { E }$. $\mathbf { F }$.
(ii) Let point P be any point on the line segment OB . When $\ell$ is minimized, then $\mathrm { OP } = \frac { \mathbf { G } } { \mathbf { G } }$, and the value of $\ell$ is $\square \sqrt { } \square$.

As shown in the figure, a robot starts from a point $P$ on the ground and moves according to the following rules: First, move forward 1 meter in a certain direction, then rotate counterclockwise $45 ^ { \circ }$ in the direction of movement; move forward 1 meter in the new direction, then rotate clockwise $90 ^ { \circ }$ in the direction of movement; move forward 1 meter in the new direction, then rotate counterclockwise $45 ^ { \circ }$ in the direction of movement; move forward 1 meter in the new direction, then rotate clockwise $90 ^ { \circ }$ in the direction of movement, and so on. The path traced by the robot forms a closed region. The area of this closed region is (28) + (29) $\sqrt { (30) }$ square meters. (Express as a fraction in simplest radical form)

A light show display uses color-changing flashing lights. After each activation, the flashing color changes periodically according to the following sequence: Blue–White–Red–White–Blue–White–Red–White–Blue–White–Red–White…, with one cycle every four flashes. Blue light lasts 5 seconds each time, white light lasts 2 seconds each time, and red light lasts 6 seconds each time. Assuming the time to change lights is negligible, select the light color(s) between the 99th and 101st seconds after activation.
(1) All blue lights
(2) All white lights
(3) All red lights
(4) Blue light first, then white light
(5) White light first, then red light

Assuming the Earth is a sphere with radius $r$, a point particle moves north from location A along the meridian passing through that location. When it reaches the North Pole, the arc length traveled is $\frac{7}{12}\pi r$. Which of the following options is most likely the location of A?
(1) East longitude $75^{\circ}$, North latitude $15^{\circ}$
(2) East longitude $30^{\circ}$, South latitude $75^{\circ}$
(3) East longitude $75^{\circ}$, South latitude $15^{\circ}$
(4) West longitude $30^{\circ}$, North latitude $75^{\circ}$
(5) West longitude $15^{\circ}$, South latitude $30^{\circ}$

It is known that the world's tallest tilted skyscraper is located in Abu Dhabi with a tilt angle of $18^{\circ}$. Converting this tilt angle to radians is which of the following options? (Single choice, 5 points)
(1) $\frac{\pi}{36}$
(2) $\frac{\pi}{18}$
(3) $\frac{\pi}{20}$
(4) $\frac{\pi}{10}$
(5) $\frac{\pi}{8}$