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, 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, there is a right triangle $\mathrm { OA } _ { 1 } \mathrm {~B} _ { 1 }$ with $\overline { \mathrm { OA } _ { 1 } } = 4$ and $\overline { \mathrm { OB } _ { 1 } } = 4 \sqrt { 3 }$. Let $\mathrm { B } _ { 2 }$ be the point where the circle with center O and radius $\overline { \mathrm { OA } _ { 1 } }$ meets the line segment $\mathrm { OB } _ { 1 }$. The figure $R _ { 1 }$ is obtained by shading the region that is inside triangle $\mathrm { OA } _ { 1 } \mathrm {~B} _ { 1 }$ but outside the sector $\mathrm { OA } _ { 1 } \mathrm {~B} _ { 2 }$. In figure $R _ { 1 }$, let $\mathrm { A } _ { 2 }$ be the point where the line passing through $\mathrm { B } _ { 2 }$ and parallel to segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ meets segment $\mathrm { OA } _ { 1 }$, and let $\mathrm { B } _ { 3 }$ be the point where the circle with center O and radius $\overline { \mathrm { OA } _ { 2 } }$ meets segment $\mathrm { OB } _ { 2 }$. The figure $R _ { 2 }$ is obtained by shading the region that is inside triangle $\mathrm { OA } _ { 2 } \mathrm {~B} _ { 2 }$ but outside the sector $\mathrm { OA } _ { 2 } \mathrm {~B} _ { 3 }$. Continuing this process, let $S _ { n }$ be the area of the shaded region in the $n$-th figure $R _ { n }$. What is the value of $\lim _ { n \rightarrow \infty } S _ { n }$? [4 points] [Figure] [Figure]
(1) $\frac { 3 } { 2 } \pi$
(2) $\frac { 5 } { 3 } \pi$
(3) $\frac { 11 } { 6 } \pi$
(4) $2 \pi$
(5) $\frac { 13 } { 6 } \pi$

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) $\frac { 5 } { 4 }$

As shown in the figure, a sector ABD with center A and central angle $90 ^ { \circ }$ is drawn in a square ABCD with side length 5. Let $\mathrm { A } _ { 1 }$ be the point that divides segment AD in the ratio $3 : 2$, and let $\mathrm { B } _ { 1 }$ be the point where the line passing through $\mathrm { A } _ { 1 }$ and parallel to segment AB meets arc BD. A square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$ is drawn with segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ as one side and meeting segment DC, and then a sector $\mathrm { D } _ { 1 } \mathrm {~A} _ { 1 } \mathrm { C } _ { 1 }$ with center $\mathrm { D } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn. Let $\mathrm { E } _ { 1 } , \mathrm {~F} _ { 1 }$ be the points where segment DC meets arc $\mathrm { A } _ { 1 } \mathrm { C } _ { 1 }$ and segment $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$ respectively. The region enclosed by two segments $\mathrm { DA } _ { 1 } , \mathrm { DE } _ { 1 }$ and arc $\mathrm { A } _ { 1 } \mathrm { E } _ { 1 }$, and the region enclosed by two segments $\mathrm { E } _ { 1 } \mathrm {~F} _ { 1 } , \mathrm {~F} _ { 1 } \mathrm { C } _ { 1 }$ and arc $\mathrm { E } _ { 1 } \mathrm { C } _ { 1 }$ are shaded to obtain the figure $R _ { 1 }$.
In figure $R _ { 1 }$, a sector $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { D } _ { 1 }$ with center $\mathrm { A } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn in square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$. Let $\mathrm { A } _ { 2 }$ be the point that divides segment $\mathrm { A } _ { 1 } \mathrm { D } _ { 1 }$ in the ratio $3 : 2$, and let $\mathrm { B } _ { 2 }$ be the point where the line passing through $\mathrm { A } _ { 2 }$ and parallel to segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ meets arc $\mathrm { B } _ { 1 } \mathrm { D } _ { 1 }$. A square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ is drawn with segment $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 }$ as one side and meeting segment $\mathrm { D } _ { 1 } \mathrm { C } _ { 1 }$, and then a shaded figure is drawn and colored in square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ in the same way as obtaining figure $R _ { 1 }$ to obtain the figure $R _ { 2 }$. Continuing this process, let $S _ { n }$ denote the area of the shaded part in the $n$-th obtained figure $R _ { n }$. What is the value of $\lim _ { n \rightarrow \infty } S _ { n }$? [4 points]
(1) $\frac { 50 } { 3 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(2) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(3) $\frac { 50 } { 3 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(4) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(5) $\frac { 100 } { 9 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$

As shown in the figure, in a right triangle ABC with $\overline { \mathrm { AB } } = 2$ and $\angle \mathrm {~B} = \frac { \pi } { 2 }$, let D and E be the points where the circle with center A and radius 1 meets the two segments $\mathrm { AB }$ and $\mathrm { AC }$ respectively. Let F be the trisection point of arc DE closer to point D, and let G be the point where line AF meets segment BC. Let $\angle \mathrm { BAG } = \theta$. Let $f ( \theta )$ be the area of the common part of the interior of triangle ABG and the exterior of sector ADF, and let $g ( \theta )$ be the area of sector AFE. Find the value of $40 \times \lim _ { \theta \rightarrow 0 + } \frac { f ( \theta ) } { g ( \theta ) }$. (where $0 < \theta < \frac { \pi } { 6 }$) [3 points]

127. If the area of a regular hexagon inscribed in a circle is $6\sqrt{3}$, then the area of a regular hexagon circumscribed about this circle is how many times $\sqrt{3}$?
(1) $7/2$ (2) $7/5$ (3) $8$ (4) $9$

For all natural numbers $n$, let $$A_{n} = \sqrt{2 - \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}} \quad (n \text{ many radicals})$$
(a) Show that for $n \geq 2$, $$A_{n} = 2\sin\frac{\pi}{2^{n+1}}$$
(b) Hence, or otherwise, evaluate the limit $$\lim_{n \rightarrow \infty} 2^{n} A_{n}$$

8. Carlo Emilio Gadda writes in one of the stories from L'Adalgisa—Milanese Sketches: ``The service rooms, the bathroom, the corridors, the antechamber and one of the two toilets were paved with small red tiles: hexagonal [. . .]. The apothem of those tiles measured 5.196 centimeters: whereas the radius of the circumscribed circle reached 60 millimeters''. Express the exact relationship between the radius of the circumscribed circle and the apothem (that is, the radius of the inscribed circle) for a regular hexagon. Verify the result obtained in light of the measurements indicated by the writer. Explain why, using regular hexagonal tiles all congruent to each other, it is possible to tile a plane. With which other regular polygons, congruent to each other, is it possible to tile a plane? Justify your answer.
\footnotetext{Maximum duration of the test: 6 hours. The use of scientific or graphical calculators is permitted provided they are not equipped with symbolic algebraic processing capability and do not have Internet connectivity. The use of a bilingual dictionary (Italian–language of the country of origin) is permitted for candidates whose native language is not Italian. It is not permitted to leave the Institute before 3 hours have elapsed from the distribution of the test. }

Let $P$ be a point on the parabola $y ^ { 2 } = 4 a x$, where $a > 0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $P F Q$, where $F$ is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $( a , m )$ is
(A) $( 2,3 )$
(B) $( 1,3 )$
(C) $( 2,4 )$
(D) $( 3,4 )$

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) $7.2 \times 10^{8}$

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) $\frac{\pi}{8}$

These sectors of circles are similar.
The arc length of the smaller sector is 6 .
The difference between the areas of the sectors is 21 .
Find the positive difference between the perimeters of the sectors.

ABCD is a rectangle $\wideparen { \mathrm { CE } }$ is a circular arc with center A $| \mathrm { DA } | = 4 \mathrm {~cm}$ $| \mathrm { AC } | = 8 \mathrm {~cm}$
According to the given information, what is the area of the shaded circular sector in $\mathbf { cm } ^ { \mathbf { 2 } }$?
A) $\frac { 16 \pi } { 3 }$
B) $\frac { 20 \pi } { 3 }$
C) $\frac { 25 \pi } { 3 }$
D) $\frac { 28 \pi } { 3 }$
E) $\frac { 32 \pi } { 3 }$

O is the center of the circle
AT is tangent to the circle at point T
$$\begin{aligned} & | A T | = 3 \mathrm {~cm} \\ & \mathrm {~m} ( \widehat { \mathrm { OAT } } ) = 45 ^ { \circ } \end{aligned}$$
According to the given information, what is the length of arc BT in cm?
A) $\frac { \pi } { 2 }$
B) $\frac { 2 \pi } { 3 }$
C) $\frac { 3 \pi } { 4 }$
D) $\frac { 4 \pi } { 5 }$
E) $\frac { 5 \pi } { 6 }$

$$|\widehat{AD}| = a \text{ units}$$ $$|\widehat{BC}| = b \text{ units}$$ $$|DC| = c \text{ units}$$
The OAD and OBC circular sectors with center O are given above.
Accordingly, the area of the shaded region is equal to which of the following in terms of $a, b$ and $c$?
A) $\frac{(a + b) \cdot c}{2}$ B) $\frac{(b - a) \cdot c}{2}$ C) $\frac{2(a + b)}{c}$ D) $\frac{2(b - a)}{c}$ E) $\frac{a \cdot b \cdot c}{2}$

O is the center of a semicircle, ABCD is a rectangle.
A, O and B are collinear.
$|AE| = |ED| = \dfrac{1}{2}$ unit, $m(\widehat{AOE}) = x$
Points C and D lie on the semicircle with center O.
What is the length of a diagonal of rectangle ABCD in terms of $x$?
A) $\tan x$ B) $\operatorname{cosec} x$ C) $\sec x$ D) $\sin x$ E) $\cos x$