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Question Number 164879 by ajfour last updated on 22/Jan/22

Commented by ajfour last updated on 22/Jan/22

$F i n d c o o r d i n a t e s o f A, B, C, T$ $i f O i s c e n t e r o f t h e r e g u l a r$ $t e t r a h e d r o n, a n d s i d e o f t e t r a h e d r o n$ $b e a (o r u n i t y) .$
	Answered by ajfour last updated on 22/Jan/22

	$l e t O A = r$ $C M = (\frac{A G}{2}) \sqrt{3} \Rightarrow$ $\frac{\sqrt{3}}{2} (r \sin θ) = \frac{a}{2} \Rightarrow r \sin θ = \frac{a}{\sqrt{3}}$ $A G = \frac{a}{\sqrt{3}}$ $N o w r + r \cos θ = h = \sqrt{\frac{3 a^{2}}{4} - \frac{a^{2}}{12}}$ $\Rightarrow h = r + \sqrt{r^{2} - \frac{a^{2}}{3}} = \frac{\sqrt{2} a}{\sqrt{3}}$ $\Rightarrow \frac{2 a^{2}}{3} + r^{2} - \frac{2 \sqrt{2} a r}{\sqrt{3}} = r^{2} - \frac{a^{2}}{3}$ $\Rightarrow r = \frac{a \sqrt{3}}{2 \sqrt{2}}$ $h - r = r \cos θ = (\frac{\sqrt{2}}{\sqrt{3}} - \frac{\sqrt{3}}{2 \sqrt{2}}) a$ $\sin θ = \frac{2 \sqrt{2}}{3}$ $D (0, 0, \frac{\sqrt{3} a}{2 \sqrt{2}}); A (0, \frac{a}{\sqrt{3}}, - \frac{a}{2 \sqrt{6}})$ $B, C \equiv (\mp \frac{a}{2}, - \frac{a}{2 \sqrt{3}}, - \frac{a}{2 \sqrt{6}})$ $s o m e o n e a t l e a s t h e l p c h e c k t h i s . . .$
	Commented by mr W last updated on 23/Jan/22

	${i t}^{'} s p e r f e c t s i r!$
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