Question #223143

Consider 3 distinct points M1, M2, and M3 with respective affixes z1, z2, and z3. Show that (Triangle M1, M2, and M3 is equilateral) ↔ ( z12 + z22 +z32 - z1z2 - z2z3 - z1z3).




1
Expert's answer
2021-09-27T15:55:05-0400

The problem is similar to this problem:

Given three complex numbers z1,z2,z3z_{1}, z_{2}, z_{3} prove that the points z1,z2,z3z_{1}, z_{2}, z_{3} are vertices of an equilateral triangle in C\mathbb{C} , if z12+z22+z32=z1z2+z1z3+z2z3z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1} z_{2}+z_{1} z_{3}+z_{2} z_{3}  .


Three points z1,z2,z3z_{1}, z_{2}, z_{3} are vertices of an equilateral triangle

 

z3z1z2z1=cos(±60)+isin(±60)=1±3i22z3z1z2=±3i(z2z1)(2z3z1z2)2=(±3i(z2z1))2z12+z22+z32=z1z2+z2z3+z3z1\begin{aligned} &\Longleftrightarrow \frac{z_{3}-z_{1}}{z_{2}-z_{1}}=\cos \left(\pm 60^{\circ}\right)+i \sin \left(\pm 60^{\circ}\right)=\frac{1 \pm \sqrt{3} i}{2} \\ &\Longleftrightarrow \quad 2 z_{3}-z_{1}-z_{2}=\pm \sqrt{3} i\left(z_{2}-z_{1}\right) \\ &\Longleftrightarrow\left(2 z_{3}-z_{1}-z_{2}\right)^{2}=\left(\pm \sqrt{3} i\left(z_{2}-z_{1}\right)\right)^{2} \\ &\Longleftrightarrow \quad z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1} z_{2}+z_{2} z_3+z_{3} z_{1} \end{aligned}

  


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