pi

A surface area calculation problem.
A polygon’s area in term of its radius is:
$A_p = n \cdot {1 \over 2} r^2 sin(\theta)$
$r^2 = {2 A \over n sin(\theta)}$
The area of the outer circle is:
$A_o = \pi r^2$
And the inner circle is:
$A_i = \pi (r cos(\theta / 2))^2$
Coded in C++:

#include <iostream>
#include <cmath>
using namespace std;

typedef long double gtype;
const gtype pi = 2 * acos(0);
#define polA(n) (((n) - 2) * pi / (n))

int main() {
	int caseI = 0;
	gtype Ap, n, rSqr, Ao, Ai, a;
	cout.precision(5);
	while (cin >> n >> Ap && n > 2) {
		a = pi - polA(n);
		rSqr = 2 * Ap / (n * sin(a));
		Ao = pi * rSqr;
		Ai = pi * rSqr * cos(a / 2) * cos(a / 2);
		cout << "Case " << ++caseI << ": ";
		cout << fixed << (Ao - Ap) << " " << (Ap - Ai) << endl;
	}
	return 0;
}

A circumcircle problem.
Given a triangle, find the circumference of its circumcircle. There’s an easy way and a hard way.
The hard way is to find the circle’s centre then calculate the radius. I used this solution for a previous problem. Code:

#include <iostream>
#include <complex>
#include <limits>
using namespace std;

typedef long double gtype;
#define equ(a, b, e) ( abs((a) - (b)) <= (e) )
const gtype len_before_point_digits = 1E6;
const gtype epsLen = numeric_limits<gtype>::epsilon() * len_before_point_digits;
const gtype pi = M_PI;
typedef complex<gtype> point;
#define x real()
#define y imag()
#define crs(a, b) ( (conj(a) * (b)).y )
#define perp(p) ( point((p).y, -(p).x) )
#define intrN(a, b, p, q) ( crs((p)-(a),(b)-(a)) * (q) - crs((q)-(a),(b)-(a)) * (p) )
#define intrD(a, b, p, q) ( crs((p)-(a),(b)-(a)) - crs((q)-(a),(b)-(a)) )
#define circC(r) (2 * pi * abs(r))
#define plus(x) ((x) > 0 ? " + " : " - ")

int main() {
	point p[3], v[2][2], c;
	gtype r;
	cout.precision(2);
	while (cin >> p[0].x >> p[0].y >> p[1].x >> p[1].y >> p[2].x >> p[2].y) {
		// perp 1
		v[0][0] = 0.5l * (p[0] + p[1]);
		v[0][1] = perp(p[1] - p[0]) + v[0][0];
		// perp 2
		v[1][0] = 0.5l * (p[1] + p[2]);
		v[1][1] = perp(p[2] - p[1]) + v[1][0];
		// intersect
		gtype d = intrD(v[0][0], v[0][1], v[1][0], v[1][1]);
		if (!equ(d, 0, epsLen)) {
			c = intrN(v[0][0], v[0][1], v[1][0], v[1][1]) / d;
			r = abs(c - p[0]);
		} else
			r = 0;
		// out
		cout << fixed << circC(r) << endl;
	}
	return 0;
}

The easy way is to find the circumcircle’s radius through:

$sin(\theta) = {a / 2 \over R} = {a \over 2 R}$
And since:
$area(ABC) = 1/2 \cdot b \cdot c \cdot sin(\theta)$
$area(ABC) = 1/2 \cdot b \cdot c \cdot {a \over 2 R}$
$R = a \cdot b \cdot c / (4 \cdot area(ABC))$
This method is less prone to error. Implementation:

#include <iostream>
#include <complex>
#include <limits>
using namespace std;

typedef long double gtype;
const gtype pi = M_PI;
typedef complex<gtype> point;
#define x real()
#define y imag()
#define crs(a, b) ( (conj(a) * (b)).y )
#define circC(r) (2 * pi * abs(r))
#define area(a, b, c) ( (gtype) 0.5 * crs((b) - (a), (c) - (a)) )
#define plus(x) ((x) > 0 ? " + " : " - ")

int main() {
	point p[3];
	gtype r;
	cout.precision(2);
	while (cin >> p[0].x >> p[0].y >> p[1].x >> p[1].y >> p[2].x >> p[2].y) {
		r = abs(p[0] - p[1]) * abs(p[1] - p[2]) * abs(p[2] - p[0]) / area(p[0], p[1], p[2]) * 0.25;
		cout << fixed << circC(r) << endl;
	}
	return 0;
}

Tag: pi

10451 – Ancient Village Sports

438 – The Circumference of the Circle

Follow Blog