Dados 3 puntos no colineales en el plano 2D P, Q y R con sus respectivas coordenadas x e y, encuentra el circuncentro del triángulo.
Nota: Circuncentro de un triángulo es el centro del círculo, formado por los tres vértices de un triángulo. Tenga en cuenta que tres puntos pueden determinar de forma única un círculo.
Ejemplos:
Input : P(6, 0)
Q(0, 0)
R(0, 8)
Output : The circumcenter of the triangle PQR
is: (3, 4)
Input : P(1, 1)
Q(0, 0)
R(2, 2)
Output : The two perpendicular bisectors found
come parallel. Thus, the given points
do not form a triangle and are collinear
Dados tres puntos del triángulo, podemos encontrar fácilmente los lados del triángulo. Ahora, tenemos las ecuaciones de las rectas de los tres lados del triángulo. Después de obtener estos, podemos encontrar el circuncentro del triángulo mediante una propiedad simple que se indica a continuación:
El circuncentro del triángulo es el punto donde se cortan todas las mediatrices de los lados del triángulo.
Esto está bien explicado en el siguiente diagrama.
Tenga en cuenta aquí que no hay necesidad de encontrar los tres lados del triángulo. Encontrar dos lados es suficiente ya que solo podemos encontrar el punto de intersección usando solo dos bisectrices perpendiculares. La tercera bisectriz perpendicular pasará ella misma por el circuncentro así hallado.
Las cosas que se deben hacer se pueden dividir de la siguiente manera:
- Encuentra 2 líneas (digamos PQ y QR) que forman los lados del triángulo.
- Encuentre las mediatrices de PQ y QR (digamos las líneas L y M respectivamente).
- Encuentre el punto de intersección de las líneas L y M como el circuncentro del triángulo dado.
PASO 1
Consulte este programa de publicación para encontrar la línea que pasa por 2 puntos
PASO 2
Sea PQ representado como ax + by = c
Una línea perpendicular a esta línea se representa como -bx + ay = d para alguna d.
Sin embargo, estamos interesados en la bisectriz perpendicular. Entonces, encontramos el punto medio de P y Q y poniendo este valor en la ecuación estándar, obtenemos el valor de d.
Del mismo modo, repetimos el proceso para QR.
d = -bx + ay where, x = (xp + xq)/2 AND y = (yp + yq)/2
PASO 3
Consulte esta publicación Programa para Punto de intersección de dos líneas
CPP
// C++ program to find the CIRCUMCENTER of a
// triangle
#include <iostream>
#include <cfloat>
using namespace std;
// This pair is used to store the X and Y
// coordinate of a point respectively
#define pdd pair<double, double>
// Function to find the line given two points
void lineFromPoints(pdd P, pdd Q, double &a,
double &b, double &c)
{
a = Q.second - P.second;
b = P.first - Q.first;
c = a*(P.first)+ b*(P.second);
}
// Function which converts the input line to its
// perpendicular bisector. It also inputs the points
// whose mid-point lies on the bisector
void perpendicularBisectorFromLine(pdd P, pdd Q,
double &a, double &b, double &c)
{
pdd mid_point = make_pair((P.first + Q.first)/2,
(P.second + Q.second)/2);
// c = -bx + ay
c = -b*(mid_point.first) + a*(mid_point.second);
double temp = a;
a = -b;
b = temp;
}
// Returns the intersection point of two lines
pdd lineLineIntersection(double a1, double b1, double c1,
double a2, double b2, double c2)
{
double determinant = a1*b2 - a2*b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return make_pair(FLT_MAX, FLT_MAX);
}
else
{
double x = (b2*c1 - b1*c2)/determinant;
double y = (a1*c2 - a2*c1)/determinant;
return make_pair(x, y);
}
}
void findCircumCenter(pdd P, pdd Q, pdd R)
{
// Line PQ is represented as ax + by = c
double a, b, c;
lineFromPoints(P, Q, a, b, c);
// Line QR is represented as ex + fy = g
double e, f, g;
lineFromPoints(Q, R, e, f, g);
// Converting lines PQ and QR to perpendicular
// vbisectors. After this, L = ax + by = c
// M = ex + fy = g
perpendicularBisectorFromLine(P, Q, a, b, c);
perpendicularBisectorFromLine(Q, R, e, f, g);
// The point of intersection of L and M gives
// the circumcenter
pdd circumcenter =
lineLineIntersection(a, b, c, e, f, g);
if (circumcenter.first == FLT_MAX &&
circumcenter.second == FLT_MAX)
{
cout << "The two perpendicular bisectors "
"found come parallel" << endl;
cout << "Thus, the given points do not form "
"a triangle and are collinear" << endl;
}
else
{
cout << "The circumcenter of the triangle PQR is: ";
cout << "(" << circumcenter.first << ", "
<< circumcenter.second << ")" << endl;
}
}
// Driver code.
int main()
{
pdd P = make_pair(6, 0);
pdd Q = make_pair(0, 0);
pdd R = make_pair(0, 8);
findCircumCenter(P, Q, R);
return 0;
}
Python3
# Python3 program to find the CIRCUMCENTER of a
# triangle
# This pair is used to store the X and Y
# coordinate of a point respectively
# define pair<double, double>
# Function to find the line given two points
def lineFromPoints(P, Q, a, b, c):
a = Q[1] - P[1]
b = P[0] - Q[0]
c = a * (P[0]) + b * (P[1])
return a, b, c
# Function which converts the input line to its
# perpendicular bisector. It also inputs the points
# whose mid-point lies on the bisector
def perpendicularBisectorFromLine(P, Q, a, b, c):
mid_point = [(P[0] + Q[0])//2, (P[1] + Q[1])//2]
# c = -bx + ay
c = -b * (mid_point[0]) + a * (mid_point[1])
temp = a
a = -b
b = temp
return a, b, c
# Returns the intersection point of two lines
def lineLineIntersection(a1, b1, c1, a2, b2, c2):
determinant = a1 * b2 - a2 * b1
if (determinant == 0):
# The lines are parallel. This is simplified
# by returning a pair of (10.0)**19
return [(10.0)**19, (10.0)**19]
else:
x = (b2 * c1 - b1 * c2)//determinant
y = (a1 * c2 - a2 * c1)//determinant
return [x, y]
def findCircumCenter(P, Q, R):
# Line PQ is represented as ax + by = c
a, b, c = 0.0, 0.0, 0.0
a, b, c = lineFromPoints(P, Q, a, b, c)
# Line QR is represented as ex + fy = g
e, f, g = 0.0, 0.0, 0.0
e, f, g = lineFromPoints(Q, R, e, f, g)
# Converting lines PQ and QR to perpendicular
# vbisectors. After this, L = ax + by = c
# M = ex + fy = g
a, b, c = perpendicularBisectorFromLine(P, Q, a, b, c)
e, f, g = perpendicularBisectorFromLine(Q, R, e, f, g)
# The point of intersection of L and M gives
# the circumcenter
circumcenter = lineLineIntersection(a, b, c, e, f, g)
if (circumcenter[0] == (10.0)**19 and circumcenter[1] == (10.0)**19):
print("The two perpendicular bisectors found come parallel")
print("Thus, the given points do not form a triangle and are collinear")
else:
print("The circumcenter of the triangle PQR is: ", end="")
print("(", circumcenter[0], ",", circumcenter[1], ")")
# Driver code.
if __name__ == '__main__':
P = [6, 0]
Q = [0, 0]
R = [0, 8]
findCircumCenter(P, Q, R)
# This code is contributed by mohit kumar 29
C#
using System;
using System.Collections.Generic;
public static class GFG {
// C# program to find the CIRCUMCENTER of a
// triangle
// This pair is used to store the X and Y
// coordinate of a point respectively
// Function to find the line given two points
public static void
lineFromPoints(List<double> P, List<double> Q,
ref double a, ref double b, ref double c)
{
a = Q[1] - P[1];
b = P[0] - Q[0];
c = a * (P[0]) + b * (P[1]);
}
// Function which converts the input line to its
// perpendicular bisector. It also inputs the points
// whose mid-point lies on the bisector
public static void perpendicularBisectorFromLine(
List<double> P, List<double> Q, ref double a,
ref double b, ref double c)
{
List<double> mid_point = new List<double>();
mid_point.Add((P[0] + Q[0]) / 2);
mid_point.Add((P[1] + Q[1]) / 2);
// c = -bx + ay
c = -b * (mid_point[0]) + a * (mid_point[1]);
double temp = a;
a = -b;
b = temp;
}
// Returns the intersection point of two lines
public static List<double>
lineLineIntersection(double a1, double b1, double c1,
double a2, double b2, double c2)
{
List<double> ans = new List<double>();
double determinant = a1 * b2 - a2 * b1;
if (determinant == 0) {
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
ans.Add(double.MaxValue);
ans.Add(double.MaxValue);
}
else {
double x = (b2 * c1 - b1 * c2) / determinant;
double y = (a1 * c2 - a2 * c1) / determinant;
ans.Add(x);
ans.Add(y);
}
return ans;
}
public static void findCircumCenter(List<double> P,
List<double> Q,
List<double> R)
{
// Line PQ is represented as ax + by = c
double a = 0;
double b = 0;
double c = 0;
lineFromPoints(P, Q, ref a, ref b, ref c);
// Line QR is represented as ex + fy = g
double e = 0;
double f = 0;
double g = 0;
lineFromPoints(Q, R, ref e, ref f, ref g);
// Converting lines PQ and QR to perpendicular
// vbisectors. After this, L = ax + by = c
// M = ex + fy = g
perpendicularBisectorFromLine(P, Q, ref a, ref b,
ref c);
perpendicularBisectorFromLine(Q, R, ref e, ref f,
ref g);
// The point of intersection of L and M gives
// the circumcenter
List<double> circumcenter
= lineLineIntersection(a, b, c, e, f, g);
if (circumcenter[0] == float.MaxValue
&& circumcenter[1] == float.MaxValue) {
Console.Write("The two perpendicular bisectors "
+ "found come parallel");
Console.Write("\n");
Console.Write(
"Thus, the given points do not form "
+ "a triangle and are collinear");
Console.Write("\n");
}
else {
Console.Write(
"The circumcenter of the triangle PQR is: ");
Console.Write("(");
Console.Write(circumcenter[0]);
Console.Write(", ");
Console.Write(circumcenter[1]);
Console.Write(")");
Console.Write("\n");
}
}
// Driver code.
public static void Main()
{
List<double> P = new List<double>();
P.Add(6);
P.Add(0);
List<double> Q = new List<double>();
Q.Add(0);
Q.Add(0);
List<double> R = new List<double>();
R.Add(0);
R.Add(8);
findCircumCenter(P, Q, R);
}
}
// The code is contributed by Aarti_Rathi
Javascript
// JavaScript program to find the CIRCUMCENTER of a
// triangle
// This pair is used to store the X and Y
// coordinate of a point respectively
// #define pdd pair<double, double>
// Function to find the line given two points
function lineFromPoints(P, Q)
{
let a = Q[1] - P[1];
let b = P[0] - Q[0];
let c = a*(P[0])+ b*(P[1]);
return [a, b, c];
}
// Function which converts the input line to its
// perpendicular bisector. It also inputs the points
// whose mid-point lies on the bisector
function perpendicularBisectorFromLine(P, Q, a, b, c)
{
let mid_point = [(P[0] + Q[0])/2, (P[1] + Q[1])/2];
// c = -bx + ay
c = -b*(mid_point[0]) + a*(mid_point[1]);
let temp = a;
a = -b;
b = temp;
return [a, b, c];
}
// Returns the intersection point of two lines
function lineLineIntersection(a1, b1, c1, a2, b2, c2)
{
let determinant = a1*b2 - a2*b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return [(10.0)**19, (10.0)**19];
}
else
{
let x = (b2*c1 - b1*c2)/determinant;
let y = (a1*c2 - a2*c1)/determinant;
return [x, y];
}
}
function findCircumCenter(P, Q, R)
{
// Line PQ is represented as ax + by = c
let PQ_line = lineFromPoints(P, Q);
let a = PQ_line[0];
let b = PQ_line[1];
let c = PQ_line[2];
// Line QR is represented as ex + fy = g
let QR_line = lineFromPoints(Q, R);
let e = QR_line[0];
let f = QR_line[1];
let g = QR_line[2];
// Converting lines PQ and QR to perpendicular
// vbisectors. After this, L = ax + by = c
// M = ex + fy = g
let PQ_perpendicular = perpendicularBisectorFromLine(P, Q, a, b, c);
a = PQ_perpendicular[0];
b = PQ_perpendicular[1];
c = PQ_perpendicular[2];
let QR_perpendicular = perpendicularBisectorFromLine(Q, R, e, f, g);
e = QR_perpendicular[0];
f = QR_perpendicular[1];
g = QR_perpendicular[2];
// The point of intersection of L and M gives
// the circumcenter
let circumcenter = lineLineIntersection(a, b, c, e, f, g);
if (circumcenter[0] == (10.0)**19 && circumcenter[1] == (10.0)**19){
console.log("The two perpendicular bisectors found come parallel" )
console.log("Thus, the given points do not form a triangle and are collinear");
}
else{
console.log("The circumcenter of the triangle PQR is: (", circumcenter[0], ",", circumcenter[1], ")");
}
}
// Driver code.
let P = [6, 0];
let Q = [0, 0];
let R = [0, 8];
findCircumCenter(P, Q, R);
// The code is contributed by Gautam goel (gautamgoel962)
Producción:
The circumcenter of the triangle PQR is: (3, 4)
Complejidad temporal : O(1) desde la realización de operaciones constantes
Espacio Auxiliar: O(1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA