The manufacturer of padlocks of the brand ``K'' wishes to print a logo for his company. This logo has the shape of a stylized capital letter K, inscribed in a square ABCD, with side length one unit of length, and respecting the following conditions C1 and C2:

• Condition C1: the letter K must consist of three lines:
• one of the lines is the segment $[AD]$;
• a second line has endpoints at point A and a point E on segment $[DC]$;
• the third line has endpoint at point B and a point G located on the second line.
• Condition C2: the area of each of the three surfaces delimited by the three lines drawn in the square must be between 0.3 and 0.4, with the unit of area being that of the square. These areas are denoted $r$, $s$, $t$.

We place ourselves in the orthonormal coordinate system $(A; \overrightarrow{AB}, \overrightarrow{AD})$.
Part A: study of Proposal A
In this proposal, the three lines are segments and the three areas are equal: $r = s = t = \frac{1}{3}$. Determine the coordinates of points E and G.

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

QUESTION 139
A rectangular piece of land has dimensions 30 m by 20 m. The owner wants to build a fence around the entire perimeter of the land. If the fence costs R\$ 15.00 per meter, the total cost of the fence will be
(A) R\$ 900.00
(B) R\$ 1,200.00
(C) R\$ 1,500.00
(D) R\$ 1,800.00
(E) R\$ 2,100.00

QUESTION 162
The lateral surface area of a cone with base radius 3 cm and slant height 5 cm is
(A) $10\pi$ cm$^2$
(B) $12\pi$ cm$^2$
(C) $15\pi$ cm$^2$
(D) $18\pi$ cm$^2$
(E) $20\pi$ cm$^2$

a) Let $\mathrm { E } , \mathrm { F } , \mathrm { G }$ and H respectively be the midpoints of the sides $\mathrm { AB } , \mathrm { BC } , \mathrm { CD }$ and DA of a convex quadrilateral ABCD. Show that EFGH is a parallelogram whose area is half that of ABCD. b) Let $\mathrm { E } = ( 0,0 ) , \mathrm { F } = ( 0 , - 1 ) , \mathrm { G } = ( 1 , - 1 ) , \mathrm { H } = ( 1,0 )$. Find all points $\mathrm { A } = ( p , q )$ in the first quadrant such that $\mathrm { E } , \mathrm { F } , \mathrm { G }$ and H respectively are the midpoints of the sides $\mathrm { AB } , \mathrm { BC } , \mathrm { CD }$ and DA of a convex quadrilateral ABCD.

Let $A = (h, k)$, $B = (2, 6)$, $C = (5, 2)$ be vertices of a triangle with area 12. Find the minimum distance from $A$ to the origin.
(A) $\dfrac{16}{\sqrt{5}}$ (B) $\dfrac{8}{\sqrt{5}}$ (C) $\dfrac{32}{\sqrt{5}}$ (D) $\dfrac{16}{\sqrt{5}}$

A unit square has its corners chopped off to form a regular polygon with eight sides. What is the area of this polygon?
(A) $2(\sqrt{3} - \sqrt{2})$
(B) $2\sqrt{2} - 2$
(C) $\frac{\sqrt{2}}{2}$
(D) $\frac{7}{9}$.

The sides of a regular hexagon $A B C D E F$ are extended by doubling them (for example, $B A$ extends to $B A ^ { \prime }$ with $B A ^ { \prime } = 2 B A$) to form a bigger regular hexagon $A ^ { \prime } B ^ { \prime } C ^ { \prime } D ^ { \prime } E ^ { \prime } F ^ { \prime }$. Then, the ratio of the areas of the bigger to the smaller hexagon is:
(A) 2
(B) 3
(C) $2 \sqrt { 3 }$
(D) $\pi$.

The sides of a regular hexagon $A B C D E F$ is extended by doubling them to form a bigger hexagon $A ^ { \prime } B ^ { \prime } C ^ { \prime } D ^ { \prime } E ^ { \prime } F ^ { \prime }$ as in the figure below. Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) $\sqrt { 3 }$
(B) 3
(C) $2 \sqrt { 3 }$
(D) 4

In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area $x$ of the remaining triangle. The four triangles have areas 4, 5, $x$, and 13 respectively.
(A) 12
(B) 13
(C) 14
(D) 15

In a triangle $A B C$, consider points $D$ and $E$ on $A C$ and $A B$, respectively, and assume that they do not coincide with any of the vertices $A , B , C$. If the segments $B D$ and $C E$ intersect at $F$, consider the areas $w , x , y , z$ of the quadrilateral $A E F D$ and the triangles $B E F , B F C , C D F$, respectively.
(a) Prove that $y ^ { 2 } > x z$.
(b) Determine $w$ in terms of $x , y , z$.

A line is drawn through the point $(1,2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle of area $\frac{9}{2}$ sq. units with the coordinate axes. The equation of the line $PQ$ is
(1) $x+2y=5$
(2) $3x+y=5$
(3) $x+2y=4$
(4) $2x+y=4$

If the extremities of the base of an isosceles triangle are the points $( 2 a , 0 )$ and $( 0 , a )$ and the equation of one of the sides is $x = 2 a$, then the area of the triangle, in square units, is :
(1) $\frac { 5 } { 4 } a ^ { 2 }$
(2) $\frac { 5 } { 2 } a ^ { 2 }$
(3) $\frac { 25 a ^ { 2 } } { 4 }$
(4) $5 a^2$

In a triangle $ABC$, coordinates of $A$ are $(1,2)$ and the equations of the medians through $B$ and $C$ are respectively, $x + y = 5$ and $x = 4$. Then area of $\triangle ABC$ (in sq. units) is :
(1) 12
(2) 4
(3) 9
(4) 5

In a triangle $A B C$, coordiantates of $A$ are $( 1,2 )$ and the equations of the medians through $B$ and $C$ are $x + y = 5$ and $x = 4$ respectively. Then area of $\triangle A B C$ (in sq. units) is
(1) 5
(2) 9
(3) 12
(4) 4

A triangle $ABC$ lying in the first quadrant has two vertices as $A ( 1,2 )$ and $B ( 3,1 )$. If $\angle BAC = 90 ^ { \circ }$, and $\operatorname { ar } ( \Delta \mathrm { ABC } ) = 5 \sqrt { 5 }$ sq. units, then the abscissa of the vertex C is :
(1) $1 + \sqrt { 5 }$
(2) $1 + 2 \sqrt { 5 }$
(3) $2 + \sqrt { 5 }$
(4) $2 \sqrt { 5 } - 1$

Let $A ( - 1,1 ) , B ( 3,4 )$ and $C ( 2,0 )$ be given three points. A line $y = m x , m > 0$, intersects lines $AC$ and $BC$ at point $P$ and $Q$ respectively. Let $A _ { 1 }$ and $A _ { 2 }$ be the areas of $\triangle ABC$ and $\triangle PQC$ respectively, such that $A _ { 1 } = 3 A _ { 2 }$, then the value of $m$ is equal to :
(1) $\frac { 4 } { 15 }$
(2) 1
(3) 2
(4) 3

Let $A ( a , 0 ) , B ( b , 2 b + 1 )$ and $C ( 0 , b ) , b \neq 0 , | b | \neq 1$, be points such that the area of triangle $A B C$ is 1 sq. unit, then the sum of all possible values of $a$ is: (1) $\frac { - 2 b } { b + 1 }$ (2) $\frac { 2 b ^ { 2 } } { b + 1 }$ (3) $\frac { - 2 b ^ { 2 } } { b + 1 }$ (4) $\frac { 2 b } { b + 1 }$

Let $R$ be the point $( 3,7 )$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is
(1) $\frac { 25 } { 4 \sqrt { 3 } }$
(2) $\frac { 25 \sqrt { 3 } } { 2 }$
(3) $\frac { 25 } { \sqrt { 3 } }$
(4) $\frac { 25 } { 2 \sqrt { 3 } }$

Let $A ( 1,1 ) , B ( - 4,3 ) , C ( - 2 , - 5 )$ be vertices of a triangle $A B C , P$ be a point on side $B C$, and $\Delta _ { 1 }$ and $\Delta _ { 2 }$ be the areas of triangle $A P B$ and $A B C$ respectively. If $\Delta _ { 1 } : \Delta _ { 2 } = 4 : 7$, then the area enclosed by the lines $A P , A C$ and the $x$-axis is
(1) $\frac { 1 } { 4 }$
(2) $\frac { 3 } { 4 }$
(3) $\frac { 1 } { 2 }$
(4) 1

Let $B$ and $C$ be the two points on the line $y + x = 0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $\mathrm { y } - 2 \mathrm { x } = 2$ such that $\triangle A B C$ is an equilateral triangle. Then, the area of the $\triangle A B C$ is
(1) $3 \sqrt { 3 }$
(2) $2 \sqrt { 3 }$
(3) $\frac { 8 } { \sqrt { 3 } }$
(4) $\frac { 10 } { \sqrt { 3 } }$

Let a variable line of slope $m > 0$ passing through the point $( 4 , - 9 )$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is
(1) 30
(2) 25
(3) 15
(4) 10

Let the line $x + y = 1$ meet the axes of $x$ and $y$ at A and B , respectively. A right angled triangle AMN is inscribed in the triangle OAB , where O is the origin and the points M and N lie on the lines $O B$ and $A B$, respectively. If the area of the triangle $A M N$ is $\frac { 4 } { 9 }$ of the area of the triangle $O A B$ and $\mathrm { AN } : \mathrm { NB } = \lambda : 1$, then the sum of all possible value(s) of $\lambda$ is:
(1) 2
(2) $\frac { 5 } { 2 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 13 } { 6 }$

Let $m$ be a real number. On a plane with the coordinate system, in which the origin is denoted by O, consider the parabola $y = x^2$ and the two points on it,
$$\mathrm{A}(a,\, ma+1), \quad \mathrm{B}(b,\, mb+1) \quad (a < 0 < b)$$
(1) The $x$-coordinates $a$ and $b$ of the two points A and B can be expressed in terms of $m$ as
$$a = \frac{m - \sqrt{D}}{\mathbf{A}}, \quad b = \frac{m + \sqrt{D}}{\mathbf{B}},$$
where the expression $D$ is
$$D = m^2 + \mathbf{C}.$$
(2) Let the coordinates of the point of intersection of the segment AB and the $y$-axis be denoted by $(0, c)$. Then $c = \mathbf{D}$.
(3) Further, when the area $S$ of the triangle OAB with the three vertices O, A and B is expressed in terms of $a$ and $b$, we have
$$S = \frac{1}{2}\mathbf{E},$$
where $E$ is the appropriate choice from among (0) $\sim$ (5). (0) $a + b$
(1) $a - b$
(2) $b - a$
(3) $a^2 + b^2$
(4) $a^2 - b^2$
(5) $b^2 - a^2$
Also, when $S$ is represented in terms of $m$, we have
$$S = \frac{\mathbf{F}}{\mathbf{G}} \sqrt{m^2 + \mathbf{H}}.$$
Hence the value of $S$ is minimalized when $m = \mathbf{I}$, and its minimum value is $S = \mathbf{J}$.

Consider a figure made by cutting two corners from a rectangle, as in the diagram to the right. The lengths of the sides are
$$\begin{array}{lll} \mathrm{AB} = 11, & \mathrm{BC} = 4, & \mathrm{CD} = 2\sqrt{13} \\ \mathrm{DE} = 5, & \mathrm{EF} = 2\sqrt{5}, & \mathrm{FA} = 6 \end{array}$$
We are to find the area of this figure.
First, extend the sides of the figure as in the diagram and denote the sides forming the right angles by $x, y, u$ and $v$. Then
$$u = \mathbf{A} - y, \quad v = x + \mathbf{B}.$$
Substituting these expressions in the equation $u^2 + v^2 = \mathbf{CD}$ and also using the equation $x^2 + y^2 = \mathbf{EF}$, we obtain
$$x = \mathbf{G}.\, y - \mathbf{H}.$$
Then, since
$$\mathbf{I}y^2 - \mathbf{J} = \mathbf{J}y - \mathbf{K} = 0,$$
we obtain $y = \mathbf{L}$.
From this we have $x = \mathbf{M}$, and further $u = \mathbf{N}$ and $v = \mathbf{O}$. Finally we conclude that the area of this figure is $\mathbf{PQ}$.