HDU 5476 Explore Track of Point(几何)——2015 ACM/ICPC Asia Regional Shanghai Online

Explore Track of Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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 ΔABCand 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

10 1 -1 0 1 0

 

Sample Output

Case #1: 3.2214

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

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给出等腰三角形ABC，AB=AC，M为BC中点。P点为三角形内使min{∠MPB+∠APC,∠MPC+∠APB} 最大的点。求P点轨迹。 
则容易找到中线AM上的P点都满足使得∠MPB=∠MPC，∠APC=∠APB，则∠MPB+∠APC=∠MPC+∠APB=180° 
故轨迹包含中线AM。 
并且所有满足的P点都应满足：∠MPB+∠APC=∠MPC+∠APB=180°

任取中线上的一点P，可找到与之对应的一点Q，使得∠MQB=∠MPB，且∠AQC=∠APC，则∠MQB+∠AQC=180°成立，故Q点形成另一条子轨迹。 
Q点的作法： 
 
ΔBPM与ΔAPC的外接圆的交点。另一个交点即为P点。（由此可证明没有其他多余的轨迹）圆弧BM、AC分别对应圆上的∠MQB=∠MPB，且∠AQC=∠APC。

P与Q重合时，两者在ΔABC的内心上。 
 
证明：P与Q重合，即ΔBPM与ΔAPC的外接圆相切。作切线PD，则： 
∠DPM=∠APE=∠ACP（对顶角相等，弦切角相等）……[1] 
又∠AMB=∠BPD=90° 
则∠CPM=∠BPM=∠PDB（左右对称，余角相等） 
则∠PCM=∠DPM（余角相等，对顶角相等）……[2] 
由[1][2]得到∠ACP=∠PCM即PC是角平分线，又AP也为角平分线，故此时的P点为ΔABC的内心I。

猜测Q的轨迹是个圆弧 
已证明 o(^▽^)o看code后面的段落 
这一点是猜的，还没证明。但可以尝试证明∠BQC是个常量入手，得到Q的轨迹是个圆弧。 
猜的依据是：轨迹关于AM左右对称，且经过ΔABC的内心时为最高点。

圆弧为ΔBCI的外接圆的一部分 
前面已经证明了Q的轨迹过ΔABC的内心I。由对称得圆心在AM的延长线上，现在直接求该圆心的位置即可。

计算过程： 
由三点坐标得到AB的距离为b，BC的距离为2a，AM即高为h,则内接圆的半径r=2ah/C=2ah/(2a+2b)=ah/(a+b) 
ΔBCI的外接圆半径为R，且设M到圆心距离为x,a2+x2=(r+x)2，则x=a2−r22r，R=r+x 
圆弧BIC张角为2β=2arcsin(a/R) 
圆弧BIC长为2Rβ

附code:

#include <cstdio>#include <iostream>#include <cstring>#include <queue>#include <algorithm>#include <cmath>using namespace std;struct node{    double x, y;}A, B, C, M;double dis(node a, node b){    return sqrt(pow(a.x-b.x, 2)+pow(a.y-b.y, 2));}int main(){    int T, o = 0;    scanf("%d", &T);    while(T--)    {        scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);        M.x = (B.x+C.x)/2, M.y = (B.y+C.y)/2;        double a = dis(B, C)/2, b = dis(A, B), h = dis(A,M);        double r = a*h/(a+b);        double R = (a*a-r*r)/r/2+r;        double ans = 2*R*asin(a/R);        //printf("%f %f %f %f %f %f\n", a, b, h, r, R, ans);        ans += h;        printf("Case #%d: %.4f\n", ++o, ans);    }    return 0;}

补充证明

重新描述一遍题 
 
在圆O外一点作切线AB、AC,M为BC中点，连AM交圆O于P、F。在劣弧BC上取一点Q，连AG延长交圆O于D。 
求证：P为ΔABC的内心，∠BQM+∠AQC=180° 
证明：你可以先了解一点关于调和点列和调和线束的知识。 
因为APMF为调和点列，所以CA、CP、CB、CF为调和线束。 
CP垂直于CF，则CP平分∠ACB。 
又AP平分∠BAC，则P为ΔABC的内心。

因为APMF为调和点列，所以DA、DP、DM、DF为调和线束。 
DP垂直于DF，则DP平分∠ADM。∠ADP=∠MDP. 
又∠CDP=∠BDP,则∠CDQ=∠BDM 
又∠CQD=∠CBQ，则∠BDM=∠CBQ……[3]

因为CA、CP、CB、CF为调和线束，所以AQED为调和点列，所以MA、MQ、ME、MD为调和线束。 
MA垂直于MC，则MC平分∠QMD。 
则∠QMC=∠DMC，又由[3]，则∠BQM=∠MBD=∠CQD

则∠BQM+∠AQC=∠CQD+∠AQC=180°

证毕#

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