Geometry – Live with passion

460

Asif NaeemApril 10, 2016

#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cstdio>

using namespace std;

struct point{
	int x,y;
};
point LL1,LL2,LR1,LR2,UL1,UL2,UR1,UR2;
point New_LL,New_LR,New_UL,New_UR;

bool isOverlap()
{
	if(LL1.x>=LR2.x)
	return 0;
	if(LR1.x<=LL2.x)
	return 0;

	if(LL1.y>=UL2.y)
	return 0;
	if(UL1.y<=LL2.y)
	return 0;
	return 1;
}

bool isTotalInside()
{
	if((LL1.x>=LL2.x && LL1.x <= UR2.x) && (LL1.y>=LL2.y && LL1.y <= UR2.y))
	{
		if((LR1.x>=LL2.x && LR1.x <= UR2.x) && (LR1.y>=LL2.y && LR1.y <= UR2.y))
		{
			if((UR1.x>=LL2.x && UR1.x <= UR2.x) && (UR1.y>=LL2.y && UR1.y <= UR2.y))
			{
				if((UL1.x>=LL2.x && UL1.x <= UR2.x) && (UL1.y>=LL2.y && UL1.y <= UR2.y))
				{
					New_LL.x=LL1.x;
					New_LL.y=LL1.y;

					New_UR.x=UR1.x;
					New_UR.y=UR1.y;
					return 1;
				}
			}
		}
	}
	if((LL2.x>=LL1.x && LL2.x <= UR1.x) && (LL2.y>=LL1.y && LL2.y <= UR1.y))
	{
		if((LR2.x>=LL1.x && LR2.x <= UR1.x) && (LR2.y>=LL1.y && LR2.y <= UR1.y))
		{
			if((UR2.x>=LL1.x && UR2.x <= UR1.x) && (UR2.y>=LL1.y && UR2.y <= UR1.y))
			{
				if((UL2.x>=LL1.x && UL2.x <= UR1.x) && (UL2.y>=LL1.y && UL2.y <= UR1.y))
				{
					New_LL.x=LL2.x;
					New_LL.y=LL2.y;

					New_UR.x=UR2.x;
					New_UR.y=UR2.y;
					return 1;
				}
			}
		}
	}
	return 0;

}
int InsidePoints()
{
	if(isTotalInside())
	return 0;
	if(LL1.x>=LL2.x && LL1.x <= UR2.x)
	{
		if(LL1.y>=LL2.y && LL1.y <= UR2.y)
		{
			//vb.push_back(LL1);
			New_LL.x=LL1.x;
			New_LL.y=LL1.y;

			New_LR.x= (LR2.x<=LR1.x?LR2.x:LR1.x);
			New_LR.y= New_LL.y;

			New_UL.x=New_LL.x;
			New_UL.y=(UL2.y<=UL1.y?UL2.y:UL1.y);

			New_UR.x=New_LR.x;
			New_UR.y=New_UL.y;
			return 0;
		}

		New_LL.x=LL1.x;
		New_LL.y=LL2.y;

		New_UR.x=(LR1.x<=LR2.x?LR1.x:LR2.x);
		New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);
		return 0;
	}
	if(LR1.x>=LL2.x && LR1.x <= UR2.x)
	{
		if(LR1.y>=LL2.y && LR1.y <= UR2.y)
		{
			//vb.push_back(LL1);
			New_LR.x=LR1.x;
			New_LR.y=LR1.y;

			New_LL.x= (LL2.x<=LL1.x?LL1.x:LL2.x);
			New_LL.y= New_LR.y;

			New_UL.x=New_LL.x;
			New_UL.y=(UL2.y<=UL1.y?UL2.y:UL1.y);

			New_UR.x=New_LR.x;
			New_UR.y=New_UL.y;
			return 0;
		}
		New_LL.x=LL2.x;
		New_LL.y=LL2.y;

		New_UR.x=LR1.x;
		New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);
		return 0;
	}

	if(UR1.x>=LL2.x && UR1.x <= UR2.x)
	{
		if(UR1.y>=LL2.y && UR1.y <= UR2.y)
		{
			//vb.push_back(LL1);
			New_UR.x=UR1.x;
			New_UR.y=UR1.y;

			New_UL.x=(UL1.x<=UL2.x?UL2.x:UL1.x);
			New_UL.y=New_UR.y;

			New_LL.x= New_UL.x;
			New_LL.y= (LL1.y<=LL2.y?LL2.y:LL1.y);

			New_LR.x=New_UR.x;
			New_LR.y=New_LL.y;
			return 0;
		}
		New_UR.x=UR1.x;
		New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);

		New_LL.x=(LL1.x>=LL2.x?LL1.x:LL2.x);
		New_LL.y=(LL1.y>=LL2.y?LL1.y:LL2.y);
		return 0;
	}

	if(UL1.x>=LL2.x && UL1.x <= UR2.x)
	{
		if(UL1.y>=LL2.y && UL1.y <= UR2.y)
		{
			//vb.push_back(LL1);
			New_UL.x=UL1.x;
			New_UL.y=UL1.y;

			New_UR.x=(UR1.x<=UR2.x?UR1.x:UR2.x);
			New_UR.y=New_UL.y;

			New_LL.x= New_UL.x;
			New_LL.y= (LL1.y<=LL2.y?LL2.y:LL1.y);

			New_LR.x=New_UR.x;
			New_LR.y=New_LL.y;
			return 0;
		}
		New_LL.x=UL1.x;
		New_LL.y=(LL1.y>=LL2.y?LL1.y:LL2.y);

		New_UR.x=(UR1.x<=UR2.x?UR1.x:UR2.x);
		New_UR.y=(UR1.y<=UR2.y?UR2.y:UR1.y);
		return 0;
	}
	return -1;
}
int main()
{
	int t;
	int length1,length2,width1,width2;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d%d",&LL1.x,&LL1.y,&UR1.x,&UR1.y);
		scanf("%d%d%d%d",&LL2.x,&LL2.y,&UR2.x,&UR2.y);

		length1 = abs(LL1.x - UR1.x);
		width1 = abs(LL1.y - UR1.y);
		length2 = abs(LL2.x - UR2.x);
		width2 = abs(LL2.y - UR2.y);

		LR1.x = UR1.x; 	LR1.y = LL1.y;
		UL1.x = LL1.x;	UL1.y = UR1.y;
		LR2.x = UR2.x;	LR2.y = LL2.y;
		UL2.x = LL2.x;	UL2.y = UR2.y;

		if(!isOverlap())
		{
			printf("No Overlap\n");
			if(t)
			printf("\n");
			continue;
		}
		if(InsidePoints()==0)
		printf("%d %d %d %d\n",New_LL.x,New_LL.y,New_UR.x,New_UR.y);
		if(t)
			printf("\n");
	}
	return 0;
}

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#include <iostream>

#include <cstdlib>

#include <cmath>

#include <cstdio>

using namespace std;

struct point{

int x,y;

};

point LL1,LL2,LR1,LR2,UL1,UL2,UR1,UR2;

point New_LL,New_LR,New_UL,New_UR;

bool isOverlap()

{

if(LL1.x>=LR2.x)

return 0;

if(LR1.x<=LL2.x)

return 0;

if(LL1.y>=UL2.y)

return 0;

if(UL1.y<=LL2.y)

return 0;

return 1;

}

bool isTotalInside()

{

if((LL1.x>=LL2.x && LL1.x <= UR2.x) && (LL1.y>=LL2.y && LL1.y <= UR2.y))

{

if((LR1.x>=LL2.x && LR1.x <= UR2.x) && (LR1.y>=LL2.y && LR1.y <= UR2.y))

{

if((UR1.x>=LL2.x && UR1.x <= UR2.x) && (UR1.y>=LL2.y && UR1.y <= UR2.y))

{

if((UL1.x>=LL2.x && UL1.x <= UR2.x) && (UL1.y>=LL2.y && UL1.y <= UR2.y))

{

New_LL.x=LL1.x;

New_LL.y=LL1.y;

New_UR.x=UR1.x;

New_UR.y=UR1.y;

return 1;

}

if((LL2.x>=LL1.x && LL2.x <= UR1.x) && (LL2.y>=LL1.y && LL2.y <= UR1.y))

{

if((LR2.x>=LL1.x && LR2.x <= UR1.x) && (LR2.y>=LL1.y && LR2.y <= UR1.y))

{

if((UR2.x>=LL1.x && UR2.x <= UR1.x) && (UR2.y>=LL1.y && UR2.y <= UR1.y))

{

if((UL2.x>=LL1.x && UL2.x <= UR1.x) && (UL2.y>=LL1.y && UL2.y <= UR1.y))

{

New_LL.x=LL2.x;

New_LL.y=LL2.y;

New_UR.x=UR2.x;

New_UR.y=UR2.y;

return 1;

}

return 0;

}

int InsidePoints()

{

if(isTotalInside())

return 0;

if(LL1.x>=LL2.x && LL1.x <= UR2.x)

{

if(LL1.y>=LL2.y && LL1.y <= UR2.y)

{

//vb.push_back(LL1);

New_LL.x=LL1.x;

New_LL.y=LL1.y;

New_LR.x= (LR2.x<=LR1.x?LR2.x:LR1.x);

New_LR.y= New_LL.y;

New_UL.x=New_LL.x;

New_UL.y=(UL2.y<=UL1.y?UL2.y:UL1.y);

New_UR.x=New_LR.x;

New_UR.y=New_UL.y;

return 0;

}

New_LL.x=LL1.x;

New_LL.y=LL2.y;

New_UR.x=(LR1.x<=LR2.x?LR1.x:LR2.x);

New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);

return 0;

}

if(LR1.x>=LL2.x && LR1.x <= UR2.x)

{

if(LR1.y>=LL2.y && LR1.y <= UR2.y)

{

//vb.push_back(LL1);

New_LR.x=LR1.x;

New_LR.y=LR1.y;

New_LL.x= (LL2.x<=LL1.x?LL1.x:LL2.x);

New_LL.y= New_LR.y;

New_UL.x=New_LL.x;

New_UL.y=(UL2.y<=UL1.y?UL2.y:UL1.y);

New_UR.x=New_LR.x;

New_UR.y=New_UL.y;

return 0;

}

New_LL.x=LL2.x;

New_LL.y=LL2.y;

New_UR.x=LR1.x;

New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);

return 0;

}

if(UR1.x>=LL2.x && UR1.x <= UR2.x)

{

if(UR1.y>=LL2.y && UR1.y <= UR2.y)

{

//vb.push_back(LL1);

New_UR.x=UR1.x;

New_UR.y=UR1.y;

New_UL.x=(UL1.x<=UL2.x?UL2.x:UL1.x);

New_UL.y=New_UR.y;

New_LL.x= New_UL.x;

New_LL.y= (LL1.y<=LL2.y?LL2.y:LL1.y);

New_LR.x=New_UR.x;

New_LR.y=New_LL.y;

return 0;

}

New_UR.x=UR1.x;

New_UR.y=(UR1.y<=UR2.y?UR1.y:UR2.y);

New_LL.x=(LL1.x>=LL2.x?LL1.x:LL2.x);

New_LL.y=(LL1.y>=LL2.y?LL1.y:LL2.y);

return 0;

}

if(UL1.x>=LL2.x && UL1.x <= UR2.x)

{

if(UL1.y>=LL2.y && UL1.y <= UR2.y)

{

//vb.push_back(LL1);

New_UL.x=UL1.x;

New_UL.y=UL1.y;

New_UR.x=(UR1.x<=UR2.x?UR1.x:UR2.x);

New_UR.y=New_UL.y;

New_LL.x= New_UL.x;

New_LL.y= (LL1.y<=LL2.y?LL2.y:LL1.y);

New_LR.x=New_UR.x;

New_LR.y=New_LL.y;

return 0;

}

New_LL.x=UL1.x;

New_LL.y=(LL1.y>=LL2.y?LL1.y:LL2.y);

New_UR.x=(UR1.x<=UR2.x?UR1.x:UR2.x);

New_UR.y=(UR1.y<=UR2.y?UR2.y:UR1.y);

return 0;

}

return -1;

}

int main()

{

int t;

int length1,length2,width1,width2;

scanf("%d",&t);

while(t--)

{

scanf("%d%d%d%d",&LL1.x,&LL1.y,&UR1.x,&UR1.y);

scanf("%d%d%d%d",&LL2.x,&LL2.y,&UR2.x,&UR2.y);

length1 = abs(LL1.x - UR1.x);

width1 = abs(LL1.y - UR1.y);

length2 = abs(LL2.x - UR2.x);

width2 = abs(LL2.y - UR2.y);

LR1.x = UR1.x; LR1.y = LL1.y;

UL1.x = LL1.x; UL1.y = UR1.y;

LR2.x = UR2.x; LR2.y = LL2.y;

UL2.x = LL2.x; UL2.y = UR2.y;

if(!isOverlap())

{

printf("No Overlap\n");

if(t)

printf("\n");

continue;

}

if(InsidePoints()==0)

printf("%d %d %d %d\n",New_LL.x,New_LL.y,New_UR.x,New_UR.y);

if(t)

printf("\n");

}

return 0;

}

11909

Asif NaeemAugust 23, 2015

#include <iostream>
#include <cmath>
#include <cstdio>

#define PI acos(-1)
using namespace std;
//using the formulae a/sin A = b/sin B = c/sin C
int main()
{
	double L,W,H,angle;
	double changed_hight,res,changed_base,empty_area;
	while(scanf("%lf%lf%lf%lf",&L,&W,&H,&angle)!=EOF)
	{
		changed_hight= sin(angle*PI/180)*L/sin((90-angle)*PI/180);

		if(angle==0)
		{
			printf("%.3lf mL\n",L*W*H);
		}
		else if(angle==90)
		{
			printf("%.3lf mL\n",0);
		}
		else if(changed_hight<=H)
		{
			empty_area=changed_hight*L*W/2;
			res = L*W*H-empty_area;
			printf("%.3lf mL\n",res);
		}
		else
		{
			changed_base = H * sin((90-angle)*PI/180)/sin(angle*PI/180);
			res = H*changed_base*W/2;
			printf("%.3lf mL\n",res);
		}
	}
	return 0;
}

#include <iostream>

#include <cmath>

#include <cstdio>

#define PI acos(-1)

using namespace std;

//using the formulae a/sin A = b/sin B = c/sin C

int main()

{

double L,W,H,angle;

double changed_hight,res,changed_base,empty_area;

while(scanf("%lf%lf%lf%lf",&L,&W,&H,&angle)!=EOF)

{

changed_hight= sin(angle*PI/180)*L/sin((90-angle)*PI/180);

if(angle==0)

{

printf("%.3lf mL\n",L*W*H);

}

else if(angle==90)

{

printf("%.3lf mL\n",0);

}

else if(changed_hight<=H)

{

empty_area=changed_hight*L*W/2;

res = L*W*H-empty_area;

printf("%.3lf mL\n",res);

}

else

{

changed_base = H * sin((90-angle)*PI/180)/sin(angle*PI/180);

res = H*changed_base*W/2;

printf("%.3lf mL\n",res);

}

return 0;

}

To get accepted I had to use PI = acos(-1). Lets assume X = middle portion; Y = dotted region; Z = stripped region; Three independent equations are: X+ 4*Y + 4*Z = a*a……………………………………………….(1) Y + 2*Z = a*a – PI*a*a/4………………………………………..(2) X/2 + Y + Z/2 = (PI*a*a)/4 – ΔPAR………………………..(3)

#include <iostream>
#include <cstdio>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <cstdlib>

#define PI acos(-1)
#define ll long long

#define THIRD (12-2*PI-3*sqrt(3))/12 //0.0433886732
#define SECOND (PI+6*sqrt(3)-12)/12 //0.12782449020255

using namespace std;

int main()
{
	double a,b,c,d;
	//cout<<THIRD<<"\n";
	while(scanf("%lf",&a)!=EOF)
	{
		b = 4*SECOND*a*a;
		c = 4*THIRD*a*a;
		d = a*a - b - c;
		printf("%.3lf %.3lf %.3lf\n",d,b,c);
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <vector>

#include <cmath>

#include <cstdlib>

#define PI acos(-1)

#define ll long long

#define THIRD (12-2*PI-3*sqrt(3))/12 //0.0433886732

#define SECOND (PI+6*sqrt(3)-12)/12 //0.12782449020255

using namespace std;

int main()

{

double a,b,c,d;

//cout<<THIRD<<"\n";

while(scanf("%lf",&a)!=EOF)

{

b = 4*SECOND*a*a;

c = 4*THIRD*a*a;

d = a*a - b - c;

printf("%.3lf %.3lf %.3lf\n",d,b,c);

}

return 0;

}

12779

Asif NaeemAugust 20, 2015

#include <stdio.h>
#include <iostream>
#include <math.h>

using namespace std;

#define eps 1e-6
#define MX  1000000000
#define ll long long
struct POINT{int x,y;};

long long gcd(ll a,ll b)
{
	if(a%b==0)
	return b;
	else
	return gcd(b,a%b);
}

POINT pp[4];


bool find_perp_distance(POINT u,POINT v,POINT s,ll &B,ll &C,double &val)
{
	//form equation of st line uv (ax+by+c=0)
	double a,b,c,r;
	a = u.y-v.y;//y1-y2
	b = v.x-u.x;
	c = -(a*u.x+b*u.y);
	//distance from a point d = (a*x+b*y+c)/sqrt(a*a+b*b)
	r = fabs(a*s.x+b*s.y+c)/sqrt(a*a+b*b);

	//here if v can be considered as the edge of the square or diameter of the inscribed circle
	//so the area of the circle = d*d/4 * PI
	//cout<<r<<"\n";
	if(r<eps)
		return 0;
	if(r < val)
	{
		B=a*s.x+b*s.y+c;
		B=B*B;
		C=a*a+b*b;
		val=r;
	}
	return 1;
}
int main()
{
	int a,ans,b;
	double compare;
	ll B,C,g;
	bool f1,f2;
	while(cin>>a)
	{
		cin>>b;
		pp[0].x=a;
		pp[0].y=b;
		for(int i=1;i<=3;i++)
		{
			cin>>a>>b;
			pp[i].x=a;pp[i].y=b;
		}
		ans=0;
		for(int i=0;i<=3;i++)
		{
			ans+=(pp[i].x+pp[i].y);
		}

		if(ans==0)
		break;
		compare = MX;
		//find perpendicular distance from first point to the st line of 2nd and 3rd point
		f1 = find_perp_distance(pp[0],pp[1],pp[2],B,C,compare);
		f2 = find_perp_distance(pp[1],pp[2],pp[3],B,C,compare);

		if(f1*f2==0)
		{
			cout<<-1<<"\n";
			continue;
		}
		C*=4;
		g = gcd(B,C);
		B/=g;
		C/=g;
		cout<<"("<<B<<"/"<<C<<")*pi\n";
	}

	return 0;
}

#include <stdio.h>

#include <iostream>

#include <math.h>

using namespace std;

#define eps 1e-6

#define MX 1000000000

#define ll long long

struct POINT{int x,y;};

long long gcd(ll a,ll b)

{

if(a%b==0)

return b;

else

return gcd(b,a%b);

}

POINT pp[4];

bool find_perp_distance(POINT u,POINT v,POINT s,ll &B,ll &C,double &val)

{

//form equation of st line uv (ax+by+c=0)

double a,b,c,r;

a = u.y-v.y;//y1-y2

b = v.x-u.x;

c = -(a*u.x+b*u.y);

//distance from a point d = (a*x+b*y+c)/sqrt(a*a+b*b)

r = fabs(a*s.x+b*s.y+c)/sqrt(a*a+b*b);

//here if v can be considered as the edge of the square or diameter of the inscribed circle

//so the area of the circle = d*d/4 * PI

//cout<<r<<"\n";

if(r<eps)

return 0;

if(r < val)

{

B=a*s.x+b*s.y+c;

B=B*B;

C=a*a+b*b;

val=r;

}

return 1;

}

int main()

{

int a,ans,b;

double compare;

ll B,C,g;

bool f1,f2;

while(cin>>a)

{

cin>>b;

pp[0].x=a;

pp[0].y=b;

for(int i=1;i<=3;i++)

{

cin>>a>>b;

pp[i].x=a;pp[i].y=b;

}

ans=0;

for(int i=0;i<=3;i++)

{

ans+=(pp[i].x+pp[i].y);

}

if(ans==0)

break;

compare = MX;

//find perpendicular distance from first point to the st line of 2nd and 3rd point

f1 = find_perp_distance(pp[0],pp[1],pp[2],B,C,compare);

f2 = find_perp_distance(pp[1],pp[2],pp[3],B,C,compare);

if(f1*f2==0)

{

cout<<-1<<"\n";

continue;

}

C*=4;

g = gcd(B,C);

B/=g;

C/=g;

cout<<"("<<B<<"/"<<C<<")*pi\n";

}

return 0;

}

10784

Asif NaeemJuly 24, 2015

A few things figuring out helped to solve this problem. for n+2 gon there are maximum n*(n+1)/2 -1 (summation of n natural numbers -1 = S) diagonals and minimum S-(n-1) diagonals, that means for S-(n-1) number of diagonals we need minimum n+2 polygon. Example : there is no diagonal for triangle , 2 diagonals for Quadrilaterals

Continue Reading “10784” →

273

Asif NaeemFebruary 5, 2015

Line Segment Intersection: 1. General Case: a) (p1, q1, p2) and (p1, q1, q2) have different orientations and b) (p2, q2, p1) and (p2, q2, q1) have different orientations 2. Special Case a) (p1, q1, p2), (p1, q1, q2), (p2, q2, p1), and (p2, q2, q1) are all collinear and b) the x-projections of (p1,

Continue Reading “273” →

12816

Asif NaeemOctober 20, 2014

#include <iostream>
#include <math.h>
//#include <string>

using namespace std;


struct point{
	double x;
	double y;
}p[101];

int isIsosceles(int a,int b,int c)
{
	//dist of p[a],p[b]

	double  d1 = sqrt((p[a].x-p[b].x)*(p[a].x-p[b].x) + (p[a].y-p[b].y)*(p[a].y-p[b].y));
	double  d2 = sqrt((p[c].x-p[b].x)*(p[c].x-p[b].x) + (p[c].y-p[b].y)*(p[c].y-p[b].y));
	double  d3 = sqrt((p[a].x-p[c].x)*(p[a].x-p[c].x) + (p[a].y-p[c].y)*(p[a].y-p[c].y));

	//cout<<d1<<" "<<d2<<" "<<d3<<"\n";
	if((d1+d2)<=d3 || (d3+d1)<=d2 || (d3+d2)<=d1)
	return 0;
	else if(d1==d2 && d2==d3 && d1==d3)
	return 0;
	else if(d1!=d2 && d2!=d3 && d1!=d3)
	return 0;

	return 1;
}


int main()
{
	int n;
	double a,b;
	int Isosceles;

	while(cin>>n)
	{
		for(int i=0;i<n;i++)
		{
			cin>>a>>b;
			p[i].x=a;
			p[i].y=b;
		}

		Isosceles = 0;
		for(int i=0;(i-2)<n;i++)
		{
			for(int j=i+1;(j-1)<n;j++)
			{
				for(int k=j+1;k<n;k++)
				{
					//cout<<i<<","<<j<<","<<k<<"\n";
					if(isIsosceles(i,j,k))
					Isosceles++;
				}
			}
		}

		cout<<Isosceles<<"\n";
	}

	return 0;
}

#include <iostream>

#include <math.h>

//#include <string>

using namespace std;

struct point{

double x;

double y;

}p[101];

int isIsosceles(int a,int b,int c)

{

//dist of p[a],p[b]

double d1 = sqrt((p[a].x-p[b].x)*(p[a].x-p[b].x) + (p[a].y-p[b].y)*(p[a].y-p[b].y));

double d2 = sqrt((p[c].x-p[b].x)*(p[c].x-p[b].x) + (p[c].y-p[b].y)*(p[c].y-p[b].y));

double d3 = sqrt((p[a].x-p[c].x)*(p[a].x-p[c].x) + (p[a].y-p[c].y)*(p[a].y-p[c].y));

//cout<<d1<<" "<<d2<<" "<<d3<<"\n";

if((d1+d2)<=d3 || (d3+d1)<=d2 || (d3+d2)<=d1)

return 0;

else if(d1==d2 && d2==d3 && d1==d3)

return 0;

else if(d1!=d2 && d2!=d3 && d1!=d3)

return 0;

return 1;

}

int main()

{

int n;

double a,b;

int Isosceles;

while(cin>>n)

{

for(int i=0;i<n;i++)

{

cin>>a>>b;

p[i].x=a;

p[i].y=b;

}

Isosceles = 0;

for(int i=0;(i-2)<n;i++)

{

for(int j=i+1;(j-1)<n;j++)

{

for(int k=j+1;k<n;k++)

{

//cout<<i<<","<<j<<","<<k<<"\n";

if(isIsosceles(i,j,k))

Isosceles++;

}

cout<<Isosceles<<"\n";

}

return 0;

}

12302

Asif NaeemSeptember 21, 2014

Theory: Construction of the Nine-Point Circle 1. Draw a triangle ABC. 2. Construct the midpoints of the three sides. Label them as L, M, N. 3. Construct the feet of the altitudes of the triangle ABC. Label them as D, E, F. Label the point of intersection of the three altitudes as H. This

Continue Reading “12302” →

356

Asif NaeemSeptember 17, 2014

#include <iostream>
#include <math.h>
#include <stdio.h>

using namespace std;


int main()
{
	double n,m,d1,d2;
	int cnt,n_cnt,c=0;
	while(cin>>n)
	{
		m=(n-0.5)*(n-0.5);
		cnt=0;
		n_cnt=0;
		for(int i=0;i<=n;i++)
		{
			for(int j=0;j<=n;j++)
			{
				d1=i*i+j*j;
				d2=(i+1)*(i+1)+(j+1)*(j+1);

				if(d1<m && d2<=m)
				{
					//cout<<i<<" "<<j<<" "<<i+1<<" "<<j+1<<"\n";
					cnt++;
				}
				else if(d1<m && d2>m)
				n_cnt++;
			}
		}
		if(c)
		cout<<"\n";
		c=1;
		cout<<"In the case n = "<<n<<", "<<4*n_cnt<<" cells contain segments of the circle.\n";
		cout<<"There are "<<4*cnt<<" cells completely contained in the circle.\n";

	}
	return 0;
}

#include <iostream>

#include <math.h>

#include <stdio.h>

using namespace std;

int main()

{

double n,m,d1,d2;

int cnt,n_cnt,c=0;

while(cin>>n)

{

m=(n-0.5)*(n-0.5);

cnt=0;

n_cnt=0;

for(int i=0;i<=n;i++)

{

for(int j=0;j<=n;j++)

{

d1=i*i+j*j;

d2=(i+1)*(i+1)+(j+1)*(j+1);

if(d1<m && d2<=m)

{

//cout<<i<<" "<<j<<" "<<i+1<<" "<<j+1<<"\n";

cnt++;

}

else if(d1<m && d2>m)

n_cnt++;

}

if(c)

cout<<"\n";

c=1;

cout<<"In the case n = "<<n<<", "<<4*n_cnt<<" cells contain segments of the circle.\n";

cout<<"There are "<<4*cnt<<" cells completely contained in the circle.\n";

}

return 0;

}

378

Asif NaeemSeptember 17, 2014

#include <cmath>
#include <iostream>
#include <stdio.h>

#define INF   1000000000

using namespace std;
int n;
int c=1;
double m1,m2,d,d1,d2,x1,x2,x3,x4,y_1,y2,y3,y4,c1,c2;
double px,py;

void intersect()
{
	if(x1!=x2)
		{
			m1=((y_1-y2)/(x1-x2));

		}

		else
		{
			m1=INF;

		}


		if(x3!=x4)
			m2=(y3-y4)/(x3-x4);
		else
			m2=INF;


		if(m1==INF)
		c1=0;
		else
		c1=y_1-m1*x1;

		if(m2==INF)
		c2=0;
		else
		c2=y3-m2*x3;


		if(m1==INF)
		{
			px=x1;
			if(m2==0)
			{
				py=y3;

				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
			else
			{
				py=m2*px+c2;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
		}
		if(m2==INF)
		{
			px=x3;
			if(m1==0)
			{
				py=y_1;

				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
			else
			{
				py=m1*px+c1;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
		}

		if(m1==0)
		{
			py=y_1;
			if(m2==INF)
			{
				px=x3;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
			else
			{
				px=(py-c2)/m2;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
		}
		if(m2==0)
		{
			py=y3;
			if(m1==INF)
			{
				px=x1;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
			else
			{
				px=(py-c1)/m1;
				printf("POINT %.2lf %.2lf\n",px,py);
				return;
			}
		}

		if(m1!=m2)
		{
			px=-(c1-c2)/(m1-m2);
			py=m1*px+c1;
			printf("POINT %.2lf %.2lf\n",px,py);
		}
}

int main()
{
	cin>>n;
	cout<<"INTERSECTING LINES OUTPUT\n";
	while(n--)
	{
		cin>>x1>>y_1>>x2>>y2>>x3>>y3>>x4>>y4;
		/*
		comparing the slopes of the st. line x1y1-x2y2, x3y3-x4y4,
		st. line x1y1-x3y3, x2y2-x4y4,
		st. line x1y1-x4y4, x2y2-x3y3,
		*/
		d=(y_1-y2)*(x3-x4)-(y3-y4)*(x1-x2);
		d1=(y_1-y3)*(x2-x4)-(y2-y4)*(x1-x3);
		d2=(y_1-y4)*(x2-x3)-(y2-y3)*(x1-x4);

		if(d==0 && d1==0 && d2==0)
		{
			cout<<"LINE\n";
			continue;
		}
		else if(d==0)
		{
			cout<<"NONE\n";
			continue;
		}
		else
		{
			intersect();
		}

	}
	cout<<"END OF OUTPUT\n";
	return 0;
}

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#include <cmath>

#include <iostream>

#include <stdio.h>

#define INF 1000000000

using namespace std;

int n;

int c=1;

double m1,m2,d,d1,d2,x1,x2,x3,x4,y_1,y2,y3,y4,c1,c2;

double px,py;

void intersect()

{

if(x1!=x2)

{

m1=((y_1-y2)/(x1-x2));

}

else

{

m1=INF;

}

if(x3!=x4)

m2=(y3-y4)/(x3-x4);

else

m2=INF;

if(m1==INF)

c1=0;

else

c1=y_1-m1*x1;

if(m2==INF)

c2=0;

else

c2=y3-m2*x3;

if(m1==INF)

{

px=x1;

if(m2==0)

{

py=y3;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

else

{

py=m2*px+c2;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

if(m2==INF)

{

px=x3;

if(m1==0)

{

py=y_1;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

else

{

py=m1*px+c1;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

if(m1==0)

{

py=y_1;

if(m2==INF)

{

px=x3;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

else

{

px=(py-c2)/m2;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

if(m2==0)

{

py=y3;

if(m1==INF)

{

px=x1;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

else

{

px=(py-c1)/m1;

printf("POINT %.2lf %.2lf\n",px,py);

return;

}

if(m1!=m2)

{

px=-(c1-c2)/(m1-m2);

py=m1*px+c1;

printf("POINT %.2lf %.2lf\n",px,py);

}

int main()

{

cin>>n;

cout<<"INTERSECTING LINES OUTPUT\n";

while(n--)

{

cin>>x1>>y_1>>x2>>y2>>x3>>y3>>x4>>y4;

comparing the slopes of the st. line x1y1-x2y2, x3y3-x4y4,

st. line x1y1-x3y3, x2y2-x4y4,

st. line x1y1-x4y4, x2y2-x3y3,

d=(y_1-y2)*(x3-x4)-(y3-y4)*(x1-x2);

d1=(y_1-y3)*(x2-x4)-(y2-y4)*(x1-x3);

d2=(y_1-y4)*(x2-x3)-(y2-y3)*(x1-x4);

if(d==0 && d1==0 && d2==0)

{

cout<<"LINE\n";

continue;

}

else if(d==0)

{

cout<<"NONE\n";

continue;

}

else

{

intersect();

}

cout<<"END OF OUTPUT\n";

return 0;

}

Another approach: (but gives wa)

#include<stdio.h>


int main()
{
   printf("INTERSECTING LINES OUTPUT\n");
   int n,x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4;
   double d,d1,d2,t1,t2,x,y;

   scanf("%d",&n);


   while(n>0)
   {
      scanf("%d%d%d%d%d%d%d%d",&x_1,&y_1,&x_2,&y_2,&x_3,&y_3,&x_4,&y_4);
      d=(double)((x_1-x_2)*(y_3-y_4))-((y_1-y_2)*(x_3-x_4));
      d1=(double)((x_3-x_4)*(y_3-y_1))-((y_3-y_4)*(x_3-x_1));
      d2=(double)(-((x_3-x_1)*(y_1-y_2))+((y_3-y_1)*(x_1-x_2)));


      if((int)d==0 && (int)d1==0 && (int)d2==0)
      {
         printf("LINE\n");
      }
      else if((int)d==0)
      {
         printf("NONE\n");
      }
      else
      {
         t1=d1/d;
         t2=d2/d;
         x=x_1+t1*(x_2-x_1);
         y=y_1+t1*(y_2-y_1);
         printf("POINT %.2lf %.2lf\n",x,y);
      }
      n--;

   }

   printf("END OF OUTPUT\n");
   return 0;
}

#include<stdio.h>

int main()

{

printf("INTERSECTING LINES OUTPUT\n");

int n,x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4;

double d,d1,d2,t1,t2,x,y;

scanf("%d",&n);

while(n>0)

{

scanf("%d%d%d%d%d%d%d%d",&x_1,&y_1,&x_2,&y_2,&x_3,&y_3,&x_4,&y_4);

d=(double)((x_1-x_2)*(y_3-y_4))-((y_1-y_2)*(x_3-x_4));

d1=(double)((x_3-x_4)*(y_3-y_1))-((y_3-y_4)*(x_3-x_1));

d2=(double)(-((x_3-x_1)*(y_1-y_2))+((y_3-y_1)*(x_1-x_2)));

if((int)d==0 && (int)d1==0 && (int)d2==0)

{

printf("LINE\n");

}

else if((int)d==0)

{

printf("NONE\n");

}

else

{

t1=d1/d;

t2=d2/d;

x=x_1+t1*(x_2-x_1);

y=y_1+t1*(y_2-y_1);

printf("POINT %.2lf %.2lf\n",x,y);

}

n--;

}

printf("END OF OUTPUT\n");

return 0;

}

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