MathB.in
New
Demo
Tutorial
About
\begin{align*}z&=\bigg|a+b(-\frac12+i\frac{\sqrt3}2)+c(-\frac12-i\frac{\sqrt3}2)\bigg|\\&=\bigg|\frac{2a-b-c}{2}+i\frac{\sqrt3}{2}(b-c)\bigg|\\&=\frac12\sqrt{(2a-b-c)^2+3(b-c)^2}\\&=\frac12\sqrt{4a^2+b^2+c^2-4ab-4ac+2bc+3b^2-6bc+3c^2}\\&=\frac12\sqrt{4a^2+4b^2+4c^2-4ab-4bc-4ca}\\&=\frac1{\sqrt2}\sqrt{(a-b)^2+(b-c)^2+(c-a)^2}\end{align*} It is neccesary that $(a-b)^2+(b-c)^2+(c-a)^2=2$, hence without the loss of generality, we have $a=b=c\pm1$.<br> In the case $a=b=c+1$, we have $(2,2,1),(3,3,2), \cdots, (10,10,9)$ totally $9$ solutions, and in the case $a=b=c-1$, we have $(1,1,2),(2,2,3),\cdots, (9,9,10)$ also $9$ solutions. The pattern times $3$ times more, the final answer is $18\times3=54$.
ERROR: JavaScript must be enabled to render input!
Sat, 06 Oct 2018 05:22 GMT