Hard problem HDU - 5858 （数学几何题或几何中的两个圆相交的面积求法）

obsorb_knowledge

于 2018-05-04 19:07:01 发布

阅读量305

点赞数

CC 4.0 BY-SA版权

分类专栏：几何题

本文链接：https://blog.youkuaiyun.com/obsorb_knowledge/article/details/80198461

几何题专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一道针对小学生的奥林匹克数学题目，该题目要求计算特定几何图形中阴影部分的面积。通过数学方法，利用正方形边长和内切圆特性进行辅助线构造，最终求得准确答案。

cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult for cjj. Can you solve it?

Give you the side length of the square L, you need to calculate the shaded area in the picture.

The full circle is the inscribed circle of the square, and the center of two quarter circle is the vertex of square, and its radius is the length of the square.

InputThe first line contains a integer T(1<=T<=10000), means the number of the test case. Each case contains one line with integer l(1<=l<=10000).OutputFor each test case, print one line, the shade area in the picture. The answer is round to two digit.Sample Input

1
1

题意：求阴影部分的面积，给你正方形的边长；

方法一：

纯数学题：做辅助线，因求的是类似扇形的面积，那么一定会借助扇形解的，扇形的两条线是相等的那么一定会借助正方形的中心和圆心，很好他们两个是重合了，做辅助线，因给出的是边，所以要找边的关系；

下面做辅助线：

其实求一个角的弧度时，可以借助于反三角函数；

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#include<math.h>
const double PI = 4*atan(1.0);   // pai的弧度；

double  hu(double l1,double l2,double l3)    // 由边长求 cosa，再由反函数求弧度 
{
	double cosa = (l1*l1+l2*l2-l3*l3)/(2*l1*l2);
	return acos(cosa);
}
int main()
{
	int i,j,t;
	double l;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf",&l);
		double db = sin(PI/4)*l;    //对角线的一半剩去了*2 /2;
		double hd1 = hu(l,db,l/2);
		double shan1 = l*l*hd1/2;
		double san = (l*db*sin(hd1))/2;
		double bshan  = shan1 - san;
		double hd2 = hu(l/2,l,db);
		double hd3 = hd1+hd2;
		double shan2 = (l/2)*(l/2)*hd3/2;
		printf("%.2f\n",4*(shan2-bshan));
	}
	return 0;
}