radius of circumsphere of tetrahedron (2020 Euler Teachers MO II p5)

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parmenides51

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parmenides51

#1 h Mar 31, 2021, 6:17 PM • 1 Y

Y by FaThEr-SqUiRrEl

All faces of a tetrahedron are triangles, the lengths of the sides of each of which are are equal to $a, b$ and $c$ . Find the radius of the ball circumscribed around the given tetrahedron.

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natmath

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natmath

#2 h Mar 31, 2021, 6:56 PM • 3 Y

Y by FaThEr-SqUiRrEl, Mango247, Mango247

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

This post has been edited 1 time. Last edited by natmath, Apr 1, 2021, 12:17 AM

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HamstPan38825

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HamstPan38825

#3 h Mar 31, 2021, 6:57 PM • 1 Y

Y by FaThEr-SqUiRrEl

natmath wrote:

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

Do all tetrahedrons have the said circumscribed rectangular prism?

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thanosaops

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thanosaops

#4 h Mar 31, 2021, 7:00 PM • 1 Y

Y by FaThEr-SqUiRrEl

natmath wrote:

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

Very close, the real formula is
$\sqrt{\frac{a^2+b^2+c^2}{8}}$

And yes, 2017 AIME II #15 was absolutely wrecked with this formula.

Also, only isosceles/equihedral tetrahedra (the type of tetrahedron with all four faces congruent and opposite edges equal) can be inscribed into a rectangular prism with all four tetrahedron vertices as prism vertices.

This post has been edited 1 time. Last edited by thanosaops, Mar 31, 2021, 7:03 PM
Reason: more info

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HamstPan38825

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HamstPan38825

#5 h Mar 31, 2021, 7:03 PM • 1 Y

Y by FaThEr-SqUiRrEl

thanosaops wrote:

natmath wrote:

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

Very close, the real formula is
$\sqrt{\frac{a^2+b^2+c^2}{8}}$

I think they meant $\frac{\sqrt{a^2+b^2+c^2}}2$ which makes sense.

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thanosaops

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thanosaops

#6 h Mar 31, 2021, 7:05 PM • 2 Y

Y by FaThEr-SqUiRrEl, Mango247

HamstPan38825 wrote:

thanosaops wrote:

natmath wrote:

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

Very close, the real formula is
$\sqrt{\frac{a^2+b^2+c^2}{8}}$

I think they meant $\frac{\sqrt{a^2+b^2+c^2}}2$ which makes sense.

The side lengths of the rectangular prism that has four vertices as tetrahedron vertices are not a, b, and c. Try the regular tetrahedron case.

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natmath

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natmath

#7 h Apr 1, 2021, 12:17 AM • 1 Y

Y by FaThEr-SqUiRrEl

thanosaops wrote:

natmath wrote:

less than a sketch

On a side note: this gives a really easy method to find the circumradius of a regular tetrahedron.

Very close, the real formula is
$\sqrt{\frac{a^2+b^2+c^2}{8}}$

And yes, 2017 AIME II #15 was absolutely wrecked with this formula.

Also, only isosceles/equihedral tetrahedra (the type of tetrahedron with all four faces congruent and opposite edges equal) can be inscribed into a rectangular prism with all four tetrahedron vertices as prism vertices.

Oh yeah, I computed the diameter.

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