已知地面上矩形abcd区域，ab中点经纬度坐标中点o1 ,cd中点经纬度坐标o2,ab长度是6米，求abcd四个顶点坐标，c++编程

这道题需要知道一些地理计算的知识，包括经纬度和平面坐标的转换、地球半径等等。下面是一个可能的解法，仅供参考。

#include <iostream>
#include <cmath>

using namespace std;

const double R = 6371000; // 地球半径，单位是米

// 经纬度转平面坐标
void ll2xy(double lat, double lon, double& x, double& y) {
    x = R * cos(lat) * lon;
    y = R * lat;
}

// 平面坐标转经纬度
void xy2ll(double x, double y, double& lat, double& lon) {
    lat = y / R;
    lon = x / (R * cos(lat));
}

// 计算两点距离
double distance(double lat1, double lon1, double lat2, double lon2) {
    double x1, y1, x2, y2;
    ll2xy(lat1, lon1, x1, y1);
    ll2xy(lat2, lon2, x2, y2);
    double dx = x2 - x1;
    double dy = y2 - y1;
    return sqrt(dx * dx + dy * dy);
}

int main() {
    double lat1 = 31.2314; // ab中点纬度
    double lon1 = 121.4701; // ab中点经度
    double lat2 = 31.2316; // cd中点纬度
    double lon2 = 121.4705; // cd中点经度
    double len = 6; // ab长度，单位是米

    double dist = distance(lat1, lon1, lat2, lon2); // 计算cd和ab中点的距离
    double angle = atan2(lon2 - lon1, lat2 - lat1); // 计算cd和ab中点之间的方位角
    double lat3 = lat1 + len / (2 * R); // 计算ac线段中点的纬度
    double lon3 = lon1 + len / (2 * R * cos(lat3)); // 计算ac线段中点的经度
    double lat4 = lat3 + dist / (2 * R) * cos(angle + M_PI / 2); // 计算ad线段中点的纬度
    double lon4 = lon3 + dist / (2 * R) * sin(angle + M_PI / 2) / cos(lat4); // 计算ad线段中点的经度

    double x1, y1, x2, y2, x3, y3, x4, y4;
    ll2xy(lat1, lon1, x1, y1);
    ll2xy(lat2, lon2, x2, y2);
    ll2xy(lat3, lon3, x3, y3);
    ll2xy(lat4, lon4, x4, y4);

    double dx = x3 - x1;
    double dy = y3 - y1;
    double ux = dx / len;
    double uy = dy / len;
    double vx = -uy;
    double vy = ux;

    double ax = x3 - ux * len / 2;
    double ay = y3 - uy * len / 2;
    double bx = x3 + ux * len / 2;
    double by = y3 + uy * len / 2;
    double cx = ax + vx * dist / 2;
    double cy = ay + vy * dist / 2;
    double dx = bx + vx * dist / 2;
    double dy = by + vy * dist / 2;

    double lat_a, lon_a, lat_b, lon_b, lat_c, lon_c, lat_d, lon_d;
    xy2ll(ax, ay, lat_a, lon_a);
    xy2ll(bx, by, lat_b, lon_b);
    xy2ll(cx, cy, lat_c, lon_c);
    xy2ll(dx, dy, lat_d, lon_d);

    cout << "(" << lat_a << ", " << lon_a << ")" << endl;
    cout << "(" << lat_b << ", " << lon_b << ")" << endl;
    cout << "(" << lat_c << ", " << lon_c << ")" << endl;
    cout << "(" << lat_d << ", " << lon_d << ")" << endl;

    return 0;
}

这个程序中用到了经纬度和平面坐标的转换、计算两点距离、计算方位角等等地理计算的知识。最终输出的是abcd四个顶点的纬度和经度。注意这里的坐标系是平面坐标系，与实际地球表面的形状不完全一致，但可以近似地表示一个小区域的形状。