How to find the perimeter of a triangle on the coordinate plane

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Preliminary information

The perimeter of any flat geometric figure in the plane is defined as the sum of the lengths of all its sides. The triangle is no exception to this. First, we give the concept of a triangle, as well as the types of triangles depending on the sides.

Definition 1

We'll call it a triangle. geometric figure, which is composed of three points connected by segments (Fig. 1).

Definition 2

The points within Definition 1 will be called the vertices of the triangle.

Definition 3

The segments within the framework of Definition 1 will be called the sides of the triangle.

Obviously any triangle will have 3 vertices as well as 3 sides.

Depending on the ratio of the sides to each other, triangles are divided into scalene, isosceles and equilateral.

Definition 4

A triangle is said to be scalene if none of its sides is equal to any other.

Definition 5

We will call a triangle isosceles if two of its sides are equal to each other, but not equal to the third side.

Definition 6

A triangle is called equilateral if all its sides are equal to each other.

You can see all types of these triangles in Figure 2.

How to find the perimeter of a scalene triangle?

Let us be given a scalene triangle with side lengths equal to $α$, $β$ and $γ$.

Conclusion: To find the perimeter scalene triangle all the lengths of its sides must be added together.

Example 1

Find the perimeter of a scalene triangle equal to $34$ cm, $12$ cm and $11$ cm.

$P=34+12+11=57$ cm

Answer: $57 see.

Example 2

Find the perimeter right triangle, whose legs are $6$ and $8$ cm.

First, we find the length of the hypotenuses of this triangle using the Pythagorean theorem. Denote it by $α$, then

$α=10$ According to the rule for calculating the perimeter of a scalene triangle, we get

$P=10+8+6=24$ cm

Answer: $24 see.

How to find the perimeter of an isosceles triangle?

Let us be given an isosceles triangle whose side lengths will be equal to $α$, and the length of the base will be equal to $β$.

By definition of the perimeter of a flat geometric figure, we get that

$P=α+α+β=2α+β$

Conclusion: To find the perimeter of an isosceles triangle, add twice the length of its sides to the length of its base.

Example 3

Find the perimeter of an isosceles triangle if its sides are $12$ cm and its base is $11$ cm.

From the example above, we see that

$P=2\cdot 12+11=35$ cm

Answer: $35 see.

Example 4

Find the perimeter of an isosceles triangle if its height drawn to the base is $8$ cm and the base is $12$ cm.

Consider the figure according to the condition of the problem:

Since the triangle is isosceles, $BD$ is also a median, hence $AD=6$ cm.

By the Pythagorean theorem, from the triangle $ADB$, we find the side. Denote it by $α$, then

According to the rule for calculating the perimeter of an isosceles triangle, we get

$P=2\cdot 10+12=32$ cm

Answer: $32 see.

How to find the perimeter of an equilateral triangle?

Let us be given an equilateral triangle with lengths of all sides equal to $α$.

By definition of the perimeter of a flat geometric figure, we get that

$P=α+α+α=3α$

Conclusion: To find the perimeter of an equilateral triangle, multiply the side length of the triangle by $3$.

Example 5

Find the perimeter of an equilateral triangle if its side is $12$ cm.

From the example above, we see that

$P=3\cdot 12=36$ cm

Petya and Vasya were preparing for control work on the topic "Perimeter and area of ​​\u200b\u200bfigures." Petya drew a geometric figure, painting some cells in blue on a sheet of paper, and Vasya calculated the perimeter educated figure and draw the maximum number of squares in red so that the perimeter of the newly formed figure remains the same.
Write a program that, given the coordinates of the filled blue squares, will find the maximum number of red squares that can be drawn so that the perimeter of the newly formed figure does not change.

Input data

The first line contains the number of blue squares $n$ ($0< n < 40404$). Далее идут $n$ строк, каждая из которых содержит координаты $x$, $y$ ($-101 \leq x, y \leq 101$) левых нижних углов синих квадратов.

Every blue square has at least one common point with at least one other blue square. The figure formed by the blue squares is connected.

Output

Output the number of red squares.

Tests

Input data

 Output
$3$
$1$ $2$
$2$ $1$
$3$ $1$ 		$3$
$3$
$1$ $1$
$2$ $2$
$1$ $3$ 		$6$
$10$
$1$ $1$
$2$ $2$
$1$ $3$
$2$ $4$
$1$ $5$
$2$ $6$
$1$ $7$
$2$ $8$
$1$ $9$
$2$ $10$ 		$90$

Program code

e-olymp 2817 Solution

#include

using namespace std ;

#define MAX_PAGE_SIZE 210

int squares [ MAX_PAGE_SIZE ] [ MAX_PAGE_SIZE ] ;

int main()(

int n ;

cin >> n ;

for (int i = 0 ; i< n ; ++ i ) { int x , y ; cin >> x >> y ;

squares [ x + MAX_PAGE_SIZE / 2 ] [ y + MAX_PAGE_SIZE / 2 ] = 1 ;

int perimeter = 0 ;

for (int i = 0 ; i< MAX_PAGE_SIZE ; ++ i ) {

for (int j = 0 ; j< MAX_PAGE_SIZE ; ++ j ) {

if (squares [ i ] [ j ] ) (

perimeter += ! squares [ i + 1 ] [ j ] + ! squares [ i - 1 ] [ j ] + ! squares [ i ] [ j + 1 ] + ! squares [ i ] [ j - 1 ] ;

int max = 0 ;

for (int j = 1 ; (perimiter - 2 * j ) / 2 > 0 ; ++ j ) (

int i = (perimiter - 2 * j ) / 2 ;

<< max ;

return 0 ;

The solution of the problem

First, you need to understand that for each connected figure made up of identical squares, there is at least one rectangle with the same perimeter as the figure. Then each figure can be completed to a rectangle, keeping the perimeter.

To prove this, let the side of the square be $1$. Then the perimeter of the figure made up of these squares will always be divisible by $2$ (this is easy to understand when building such figures on a piece of paper: adding each new square to the figure can only change the perimeter by $-4, -2, 0, 2, 4 $). And since the perimeter of the rectangle is equal to $2 * (a + b)$, where $a, b$ are the sides of the rectangle, then for the existence of a rectangle with the same perimeter, the condition $\forall p \in \mathbb(N) , p > 2 \rightarrow \exists a,b \in \mathbb(N) : 2p = 2*(a + b)$. Obviously, the condition is indeed satisfied for all $p>2$.

Let's write our figure to the squares array. Then we calculate its perimeter: each non-empty square of the figure adds $1$ to the perimeter for each empty cell to the left, right, top or bottom of it. Next, we will search for all suitable rectangles, writing the maximum area to the variable max: sorting through the values ​​of the first side $j$, we calculate the second side $i = \displaystyle \frac(p)(2) - j$ through the perimeter. We will consider the area as the difference between the area of ​​the rectangle and the original figure (the number $n$ is equal to the area of ​​the figure, because the area of ​​each square is $1$).
At the end, we print the difference between the maximum area and the area of ​​the original figure (the area of ​​the original figure is $n$, because the area of ​​each square is $1$).

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