ACM学习历程—HDU5476 Explore Track of Point（平面几何）（2015上海网赛09题）

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Problem Description

In Geometry, the problem of track is very interesting. Because in some cases, the track of point may be beautiful curve. For example, in polar Coordinate system,ρ=cos3θ is like rose, ρ=1−sinθ is a Cardioid, and so on. Today, there is a simple problem about it which you need to solve.

Give you a triangle ΔABC and AB = AC. M is the midpoint of BC. Point P is in ΔABC and makes min{∠MPB+∠APC,∠MPC+∠APB} maximum. The track of P is Γ. Would you mind calculating the length of Γ?

Given the coordinate of A, B, C, please output the length of Γ.

Input

There are T (1≤T≤104) test cases. For each case, one line includes six integers the coordinate of A, B, C in order. It is guaranteed that AB = AC and three points are not collinear. All coordinates do not exceed 104 by absolute value.

Output

For each case, first please output "Case #k: ", k is the number of test case. See sample output for more detail. Then, please output the length of Γ with exactly 4 digits after the decimal point.

Sample Input

1

0 1 -1 0 1 0

Sample Output

Case #1: 3.2214

题目稍微转换一下就变成求∠MPB+∠APC=∠MPC+∠APB=180的点p的轨迹了。

这最后结论是一道平面几何题，高中数竞虽然搞过平面几何，不过基本全部忘光了，定理也只记得一个梅涅劳斯定理了。。。虽然当时就很弱。。

高中数竞时小烈平几就很强@hqwhqwhq，果然赛后题解交代了轨迹寻找的过程。。

http://blog.csdn.net/u014610830/article/details/48753415

虽然找的过程没怎么看懂，不过证明过程基本看懂了。

如果能猜出轨迹的话题目也就解决了，剩下就是怎么证明这个轨迹满足条件了。

首先三角形的高AM是满足条件的，基本是没问题的。

其次B和C点在极限情况下发现也是满足条件的，由于对称性，基本上剩余轨迹就是过B和C的一种图形。。。

运气好的话猜到它是个圆就能解决。。。

盗一张图：

结论是剩余的图是AB过B的垂线与AC过C的垂线交于点M，以M为圆心，BM为半径的圆弧。

接下来证明：

对于圆弧上某一点P，AP延长交圆于点D，

目测的话，∠BPM = ∠CPD。结论就是这个，接下来就是证明这个。

由于B、P、C、D四点共圆，根据托勒密定理：

CP*BD+BP*CD = BC*DP

由根据割线定理：

AB*AB = AP*AD

于是可得，三角形APB相似于三角形ABD

于是BP/BD = AB/AD

同理得：CP/CD = AC/AD

又AB=AC

于是BP/BD = CP/CD

即BP*CD = CP*BD

联合上面的托勒密得2BP*CD = 2CP*BD = BC*DP = 2BM*DP

提取BP*CD = BM*DP

即BP/BM = DP/CD

又∠MBP = ∠CDP（同弧所对圆周角相等）

于是三角形MBP相似于三角形CDP

于是结论得证。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long

using namespace std;

const double PI = acos(-1);

inline double dis(double xA, double yA, double xB, double yB)
{
    double ans = (xA-xB)*(xA-xB) + (yA-yB)*(yA-yB);
    return sqrt(ans);
}

void work()
{
    double xA, yA, xB, yB, xC, yC;
    double a, h, d, ans, v, r;
    scanf("%lf%lf%lf%lf%lf%lf", &xA, &yA, &xB, &yB, &xC, &yC);
    d = dis(xB, yB, xC, yC)/2;
    h = dis(xA, yA, xB, yB);
    a = asin(d/h);
    v = PI-2*a;
    r = h*tan(a);
    ans = sqrt(h*h-d*d)+v*r;
    printf("%.4lf\n", ans);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        printf("Case #%d: ", times);
        work();
    }
    return 0;
}

ACM学习历程—HDU5476 Explore Track of Point（平面几何）（2015上海网赛09题）

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原文地址：http://www.cnblogs.com/andyqsmart/p/4857623.html