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Given a regular tetrahedron of edge length $a$, how do I prove that the circumradius of the tetrahedron is equal to $\frac{\sqrt 6}{4}a$?

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  • $\begingroup$ What approach(es) have you tried? $\endgroup$ – Lubin Mar 9 '18 at 21:20
  • $\begingroup$ I've found that the height of the tetrahedron is $\frac{\sqrt 6}{3}a$ but I don't know how to go from here. $\endgroup$ – Matan Haller Mar 9 '18 at 21:25
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    $\begingroup$ So the altitude segment, drawn from the peak of the pyramid down to the base, is of length $\frac{\sqrt6a}3$, and the center is somewhere on this segment. Can you reckon where? Once you have the location of this point, you can get the radius of your sphere. $\endgroup$ – Lubin Mar 9 '18 at 21:28
  • $\begingroup$ What are you allowed to use? For instance, once you have the coordinates of the vertices, there’s a direct way to construct the equation of the sphere that passes through those point which involves a $5\times5$ determinant $\endgroup$ – amd Mar 9 '18 at 23:51
  • $\begingroup$ Fun Fact: The vertices of a regular tetrahedron are half of the vertices of a cube. The relationship between the edge of the tetrahedron and the edge of the cube is easy to determine, as is the circumradius of the cube. $\endgroup$ – Blue Mar 10 '18 at 4:36
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In Elements XIII, 13 Euclid proves that the square of the diameter of the circumsphere is one and a half times the square of the side of the tetrahedron.

So$$d^2=\frac{3}{2}a^2$$Then if $r$ is the radius$$4r^2=\frac{3}{2}a^2$$making$$r^2=\frac{3}{8}a^2=\frac{6}{16}a^2$$and$$r=\frac{\sqrt 6}{4}a$$

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