题解 | #Task Computing#

L题(直接求五种正多面体的新边长通项公式，而无分类讨论)

下文证明凸正多面体有且仅有五种：

设凸正多面体一个顶点向外有 $n(n\ge 3)n(n\backslash ge\; 3)$ 条棱，设一个面是正 $m(m\ge 3)m(m\backslash ge\; 3)$ 边形。每个顶点附近的空间角加起来必须满足小于 $360\mathrm{°}360°$，不然这个点不复存在(平了)，正多边形外角和是 $360360$ 度，故每个外角为 ${\displaystyle \frac{360}{m}}\backslash dfrac\{360\}m$ 度，每个内角为 $180-{\displaystyle \frac{360}{m}}180-\backslash dfrac\{360\}m$，固有：

观察可知可以枚举法得到该方程的唯五整数解为 $(3,3),(3,4),(3,5),(4,3),(5,3)(3,3),(3,4),(3,5),(4,3),(5,3)$

根据欧拉公式 $V+F-E=2V+F-E=2$，每条棱属于两个点即 $2E=nV2E=nV$，每条边属于两个面，即 $2E=mF2E=mF$，解三元一次方程组得，分别是 $V=4,6,8,12,20V=4,6,8,12,20$；边数 $EE$ 为 $4,8,6,20,124,8,6,20,12$。证毕。

参考这个文献，可知经过题意变换后，正四面体还是正四面体，正方体与正八面体相互转换，正十二面体和正二十面体相互转换。

设每个面中心到边的距离为 $dd$，边长为 $aa$，边所对的圆心角为 $\alpha ={\displaystyle \frac{2\pi }{m}}\backslash alpha=\backslash dfrac\{2\backslash pi\}\{m\}$，根据三线合一 $\tan {\displaystyle \frac{\alpha }{2}}={\displaystyle \frac{d}{{\displaystyle \frac{a}{2}}}}\backslash tan\backslash dfrac\; \backslash alpha2=\backslash dfrac\{d\}\{\backslash dfrac\; a2\}$ 得 $d={\displaystyle \frac{a}{2}}\tan {\displaystyle \frac{\alpha }{2}}d=\backslash dfrac\; a2\backslash tan\backslash dfrac\backslash alpha2$

参考这个文章，可证正多面体二面角大小为 $\delta =2\arcsin (cos{\displaystyle \frac{\pi }{n}}\csc {\displaystyle \frac{\pi }{m}})\backslash delta=2\backslash arcsin(cos\backslash dfrac\backslash pi\; n\backslash csc\backslash dfrac\backslash pi\; m)$，根据二面角正弦定理，解得新边长 $xx$ 为 $\sin {\displaystyle \frac{\delta }{2}}={\displaystyle \frac{{\displaystyle \frac{x}{2}}}{d}}\backslash sin\backslash dfrac\backslash delta2=\backslash dfrac\{\backslash dfrac\; x2\}\{d\}$ 得 $x={\displaystyle \frac{\cos {\displaystyle \frac{\pi }{n}}a}{\sin {\displaystyle \frac{\pi }{m}}\tan {\displaystyle \frac{\pi }{m}}}}x=\backslash dfrac\{\backslash cos\backslash dfrac\backslash pi\; na\}\{\backslash sin\backslash dfrac\backslash pi\; m\backslash tan\backslash dfrac\backslash pi\; m\}$

#include <bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%lld", &x)
typedef long long ll;
typedef long double db;
map<ll, pair<ll, ll>> h = {{4, {3, 3}}, {6, {3, 4}}, {8, {4, 3}}, {12, {3, 5}}, {20, {5, 3}}};
map<ll, ll> to = {{4, 4}, {6, 8}, {8, 6}, {12, 20}, {20, 12}};
ll t, v, a, k;
const db pi = acos(-1);
signed main()
{
    for (sc(t); t; --t)
    {
        sc(v), sc(a), sc(k);
        if (to.find(v) == to.end())
        {
            printf("impossible\n");
            continue;
        }
        db x = a;
        while (k--)
        {
            ll n = h[v].first, m = h[v].second;
            x *= cos(pi / n) / (sin(pi / m) * tan(pi / m));
            v = to[v];
        }
        printf("possible %lld %.12Lf\n", v, x);
    }
    return 0;
}