易知直线$y=k_1x,y=k_2x$之间的夹角为$\arctan\left(\frac{k_1-k_2}{k_1k_2+1}\right)$
所以,不妨设 $k_1 > k_2 >k_3 >0$ 解方程
- sol=Values@Solve[(k1-k2)/(1+k1 k2)==(k2-k3)/(1+k2 k3)&&10^7>k1>k2>k3>0,{k1,k2,k3},Integers];{Length@sol,sol}
复制代码
返回:
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