拿破崙三角形

任意三角形各邊長分別向外側作正三角形，則此三個正三角形的重心連線構成一個正三角形

證明：

$$\triangle ABD,\triangle ACE與\triangle BCF均為正三角形，其邊長分別為c=\overline{AB}, b=\overline{AC}, c=\overline{AB}；\\及其重心分別為G_1,G_2,G_3；因此\cases{\overline{AG_1}=c/\sqrt 3\\ \overline{AG_2}=b/\sqrt 3 } 及\cases{\angle G_1AB=60^\circ \div 2=30^\circ \\ \angle G_2AC =60^\circ \div 2=30^\circ} \\\cases{\cos \angle G_1AG_2 ={\overline{AG_1}^2 +\overline{AG_2}^2-\overline{G_1G_2}^2 \over 2\times \overline{AG_1} \times \overline{AG_2}} ={( b^2+c^2)/3-\overline{G_1G_2}^2 \over 2 bc/3}\\\cos \angle G_1AG_2 =\cos(\angle A+60^\circ) = (\cos \angle A-\sqrt 3 \sin \angle A)/2} \\ \Rightarrow {( b^2+c^2)/3-\overline{G_1G_2}^2 \over 2 bc/3} ={\cos \angle A-\sqrt 3 \sin \angle A\over 2} \\\Rightarrow {1\over 3}(b^2+c^2)-\overline{G_1G_2}^2= {1\over 3}bc\left( \cos \angle A-\sqrt 3\sin \angle A \right) \\ \Rightarrow \overline{G_1G_2}^2={1\over 3}(b^2+c^2)-{1\over 3}bc\left( \cos \angle A-\sqrt 3\sin \angle A \right) \\={1\over 3}(b^2+c^2)-{1\over 3}bc\left( {b^2+c^2-a^2\over 2bc}-\sqrt 3\sin \angle A \right)\\ ={1\over 3}(b^2+c^2)-{1\over 6}(b^2+c^2-a^2)+{\sqrt 3\over 3}bc\sin \angle A ={1\over 6}(a^2+b^2+c^2)+ {2\sqrt 3\over 3}\triangle ABC\\ 同理，\overline{G_2G_3}^2 =\overline{G_3G_1}^2 = \overline{G_1G_2}^2 ={1\over 6}(a^2+b^2+c^2)+ {2\sqrt 3\over 3}\triangle ABC \\ \Rightarrow \triangle G_1G_2G_3為一正\triangle$$