O número de diagonais de um polígono convexo de $n$ lados é dado pela fórmula $D = \dfrac{n(n-3)}{2}$. Um polígono convexo tem 20 diagonais. O número de lados desse polígono é
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10

QUESTION 152
The number of diagonals of a polygon with 8 sides is
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24

The number of diagonals of a polygon with 8 sides is:
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24

Let $P$ be a regular twelve-sided polygon. The number of right-angled triangles formed by the vertices of $P$ is
(A) 60
(B) 120
(C) 160
(D) 220 .

Let $n \geq 2$ be an integer. Take $n$ distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of $n$ is

Let $T_n$ be the number of all possible triangles formed by joining vertices of an $n$-sided regular polygon. If $T_{n+1} - T_n = 10$, then the value of $n$ is:
(1) 10
(2) 8
(3) 5
(4) 7

Let $n > 2$ be an integer. Suppose that there are $n$ Metro stations in a city located around a circular path. Each pair of the nearest stations is connected by a straight track only. Further, each pair of the nearest station is connected by blue line, whereas all remaining pairs of stations are connected by red line. If number of red lines is 99 times the number of blue lines, then the value of $n$ is
(1) 201
(2) 200
(3) 101
(4) 199

Consider a rectangle $ABCD$ having $5,6,7,9$ points in the interior of the line segments $AB , BC , CD , DA$ respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta - \alpha)$ is equal to
(1) 795
(2) 1173
(3) 1890
(4) 717

Let $P _ { 1 } , \quad P _ { 2 } \ldots , \quad P _ { 15 }$ be 15 points on a circle. The number of distinct triangles formed by points $P _ { i } , \quad P _ { j } , \quad P _ { k }$ such that $i + j + k \neq 15$, is :
(1) 455
(2) 419
(3) 12
(4) 443

If the sides $A B , B C$ and $C A$ of a triangle $A B C$ have 3,5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to:
(1) 364
(2) 240
(3) 333
(4) 360

The lines $L_1, L_2, \ldots, L_{20}$ are distinct. For $n = 1, 2, 3, \ldots, 10$ all the lines $L_{2n-1}$ are parallel to each other and all the lines $L_{2n}$ pass through a given point $P$. The maximum number of points of intersection of pairs of lines from the set $\{L_1, L_2, \ldots, L_{20}\}$ is equal to:

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is
(1) 48
(2) 56
(3) 24
(4) 16

On a coordinate plane, 12 points are arranged as shown in the figure to the right. If we are to select three points as vertices of a triangle, how many triangles are possible in total?
First, there are $\mathbf { K L M }$ ways to select three points from the 12 points.
Next, let us find how many ways it is possible to select three or more points in a straight line.
Let us look at the two cases.
(i) There are $\mathbf { N }$ straight lines that pass through four points.
(ii) There are $\mathbf { O }$ straight lines that pass through three points.
Hence, among all combinations of three points that are in a straight line and so cannot be the vertices of a triangle, $\mathbf { P Q }$ combinations belong to case (i), and $\mathbf { Q }$ combinations belong to case (ii).
Thus, the total number of possible triangles is $\mathbf{STU}$.
In particular, if we set $( 1,1 )$ as point A and $( 4,1 )$ as point B , then $\mathbf { V W }$ triangles have two vertices on segment AB.

On a coordinate plane, 12 points are arranged as shown in the figure to the right. If we are to select three points as vertices of a triangle, how many triangles are possible in total?
First, there are $\mathbf { K L M }$ ways to select three points from the 12 points.
Next, let us find how many ways it is possible to select three or more points in a straight line.
Let us look at the two cases.
(i) There are $\mathbf { N }$ straight lines that pass through four points.
(ii) There are $\mathbf { O }$ straight lines that pass through three points.
Hence, among all combinations of three points that are in a straight line and so cannot be the vertices of a triangle, $\mathbf { P Q }$ combinations belong to case (i), and $\mathbf { Q }$ combinations belong to case (ii).
Thus, the total number of possible triangles is $\mathbf{STU}$.
In particular, if we set $( 1,1 )$ as point A and $( 4,1 )$ as point B , then $\mathbf { V W }$ triangles have two vertices on segment AB.

In the Elements of Geometry, it is stated: "Two distinct points determine a line." In general, three distinct points determine $C_{2}^{3} = 3$ lines; however, if these three points are collinear, only one line is determined. On the coordinate plane, circle $\Gamma_{1}: x^{2} + y^{2} = 4$ intersects the two coordinate axes at 4 points, circle $\Gamma_{2}: x^{2} + y^{2} = 2$ intersects the line $x - y = 0$ at 2 points, and circle $\Gamma_{2}$ intersects the line $x + y = 0$ at 2 points. How many different lines can these 8 points determine?
(1) 12
(2) 16
(3) 20
(4) 24
(5) 28