题解 | #圆与三角形#

诈骗题。

所谓的 ${\sum }_{cyc}\tan {\displaystyle \frac{A}{2}}\cdot \tan {\displaystyle \frac{B}{2}}\backslash sum\_\{cyc\}\backslash tan\backslash dfrac\{A\}\{2\}\backslash cdot\backslash tan\backslash dfrac\{B\}\{2\}$ 实际上是一个定值，等于 $11$。

考虑证明，首先 $A+B+C=\pi A+B+C=\backslash pi$，据此 $\tan {\displaystyle \frac{C}{2}}=\tan ({\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{A+B}{2}})\backslash tan\backslash dfrac\{C\}\{2\}=\backslash tan\backslash left(\backslash dfrac\{\backslash pi\}\{2\}-\backslash dfrac\{A+B\}\{2\}\backslash right)$

接着，考虑合并：

所以 ${\sum }_{cyc}\tan {\displaystyle \frac{A}{2}}\cdot \tan {\displaystyle \frac{B}{2}}=1\backslash sum\_\{cyc\}\backslash tan\backslash dfrac\{A\}\{2\}\backslash cdot\backslash tan\backslash dfrac\{B\}\{2\}=1$，另外众所周知 $\sin \theta \in [-1,1]\backslash sin\backslash theta\backslash in[-1,1]$，所以原式最大值为 $r+1r+1$。

#include<cstdio>
int main(){
    double r;
    scanf("%lf", &r);
    printf("%.2lf\n", r + 1);
}