UVa 12165 - Triangle Hazard

链接

传送门

题意

如图，给出\(P\)、\(Q\)、\(R\)三个点的坐标和\(m1:m2\)、\(m3:m4\)、\(m5:m6\)的值。求\(A\)、\(B\)、\(C\)的坐标。

思路

由梅涅劳斯定理可得：

\[\frac{AR}{RP} \cdot \frac{PQ}{QB} \cdot \frac{BD}{DC} = 1\]

即：

\[\frac{AR}{RP} \cdot \frac{PQ}{QB} = \frac{m_5}{m_6}\]

设

\[k_1 = \frac{m_5}{m_6}, \space k_2 = \frac{m_1}{m_2}, \space k_3 = \frac{m_3}{m_4}\]

化简得：

\[\frac{AR}{RP} = k_1 \left( \frac{BP}{PQ} + 1 \right)\]

同理可得，

\[\frac{BP}{PQ} = k_2 \left( \frac{CQ}{QR} + 1 \right), \space \frac{CQ}{QR} = k_3 \left( \frac{AR}{RP} + 1 \right)\]

解得：

\[\frac{AR}{RP} = \frac{k_1k_2k_3 + k_1k_2 + k_1}{1 - k_1k_2k_3}\]

\[\frac{BP}{PQ} = \frac{k_1k_2k_3 + k_2k_3 + k_2}{1 - k_1k_2k_3}\]

\[\frac{CQ}{QR} = \frac{k_1k_2k_3 + k_3k_1 + k_3}{1 - k_1k_2k_3}\]

然后利用向量乘法很容易解决了。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

struct Point {
  double x, y;
  Point() {}
  Point(double x, double y): x(x), y(y) {}
} P, Q, R, A, B, C;

typedef Point Vector;

Vector operator + (const Point& a, const Point& b) {
  return Point(a.x + b.x, a.y + b.y);
}

Vector operator - (const Point& a, const Point& b) {
  return Point(a.x - b.x, a.y - b.y);
}

Vector operator * (const Point& a, const double& k) {
  return Point(a.x * k, a.y * k);
}

int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    double m1, m2, m3, m4, m5, m6;
    scanf("%lf%lf%lf%lf%lf%lf", &P.x, &P.y, &Q.x, &Q.y, &R.x, &R.y);
    scanf("%lf%lf%lf%lf%lf%lf", &m1, &m2, &m3, &m4, &m5, &m6);
    double k1 = m5 / m6, k2 = m1 / m2, k3 = m3 / m4, k = k1 * k2 * k3;
    A = (R - P) * ((k + k1 * k2 + k1) / (1 - k)) + R;
    B = (P - Q) * ((k + k2 * k3 + k2) / (1 - k)) + P;
    C = (Q - R) * ((k + k3 * k1 + k3) / (1 - k)) + Q;
    printf("%.8f %.8f %.8f %.8f %.8f %.8f\n", A.x, A.y, B.x, B.y, C.x, C.y);
  }
  return 0;
}