2020_III_3_Pure

2020_III_3_Pure

(i) From the beginning we have \frac{k-a}{b-a}=e^{-i\frac{\pi}{3}}=\frac{1}{2}-\frac{\sqrt{3}i}{2}, then as \omega=e^\frac{i\pi}{6}=\frac{\sqrt{3}}{2}+ \frac{i}{2} we have:
k=a(\frac{1}{2}+ \frac{\sqrt{3}i}{2})+b(\frac{1}{2}-\frac{\sqrt{3}i}{2})\to g_{ab}=\frac{k+a+b}{3}=\frac{1}{\sqrt{3}}(a(\frac{\sqrt{3}}{2}+ \frac{i}{2})+b(\frac{\sqrt{3}}{2}-\frac{i}{2}))=\frac{1}{\sqrt{3}}(\omega a + \omega^* b)

(ii) As XYZW is a parallelogram if and only if x+z=y+w then:
Q_2 is a parallelogram \iff g_{ab}+g_{cd}=g_{bc}+g_{da}
\iff \frac{1}{\sqrt{3}}(\omega (a+c) + \omega^* (b+d))=\frac{1}{\sqrt{3}}(\omega (b+d) + \omega^* (a+c))\iff \frac{1}{\sqrt{3}}i(a+c)=\frac{1}{\sqrt{3}}i(b+d)
\iff a+c=b+d\iff Q_1 is a parallelogram.
Then Q_1 is a parallelogram if and only if Q_2 is a parallelogram.

(iii) We have that w^4-w^2+1=0 and w^{-1}=w^* then, \omega-\omega^*=-\omega^3 and w^6=-1\to w^5=-w^*.

Now let’s see that g_{ab}-g_{bc}=\frac{1}{\sqrt{3}}(a\omega+b(\omega-\omega^*)-c(\omega^*))=\frac{1}{\sqrt{3}}(a\omega-b(\omega^3)-c(\omega^*)).
Analogously we have g_{bc}-g_{ca}=\frac{1}{\sqrt{3}}(b\omega-c(\omega^3)-a(\omega^*))
\to w^2(g_{bc}-g_{ca})=\frac{1}{\sqrt{3}}(b\omega^3-c(\omega^5)-a(\omega))=-(g_{ab}-g_{bc})
\to \frac{g_{bc}-g_{ab}}{g_{bc}-g_{ca}}=w^2=e^{\frac{i\pi}{3}}. Therefore T_2 is an equilateral triangle.

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