**(1)** Find the side of a regular tetrahedron which is inside a sphere of radius ' r ', Exactly fitting.

(Just as a triangle has circumcircle, say the Tetrahedron has a Circumsphere)

**(2)** A solid tetrahedron is sliced off a wooden cube of side 'a' by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices.

The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down.

What is the height of this object?

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## Comments

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TopNewestWhat is 2d and 3d geometry?

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I feel 1st one will be easy if someone tells what is the formula for Height of a regular tetrahedron.......

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2nd problem is really good I had to use the \(\text{Colours}\) \(\text{Cube}\) (played as a puzzle) for the imagination :P....Ans can be found by subtracting \(\text{height}\) \(\text{of}\) \(\text{tetrahedron}\) from \(\text{body}\) \(\text{diagonal}\) \(\text{of}\) \(\text{cube}\) so ans is \(\frac {2}{\sqrt{3}}a\)

Worked solution...Consider the tetrahedron with vertical angle \(\frac{\pi}{2}\) steradian (angle at vertex of cube) and base equilateral triangle of side \(\sqrt{2} a\) (which is face diagonal of cube )............This equilateral triangle has \(C\) as circumcentre and it's circumradius will be \(\frac{2}{\sqrt{3}} (\frac {1}{2} \sqrt{2}a)\)= \(\frac{2}{\sqrt{3}} \frac {a}{\sqrt{2}}\)=\(\sqrt{\frac{2}{3}}a\)

Then apical vertex \(A\) joined to \(C\) say circumcentre(=centroid=incentre) of base will be height of tetrahedron .

In the right angled \(\Delta ABC\) formed by joining a base vertex say \(B\) , apical vertex \(A\) and \(C\) circumcentre of base, The hypotenuse is \(AB\) of length 'a' which was side of cube. The segment \(BC\) is of length circumradius of base=\(\sqrt{\frac{2}{3}}a\) so \(AC\) (height of Tetrahedron) is \(\sqrt{a^2 - \frac {2}{3}a^2}\)=\(\frac{a}{\sqrt{3}}\) by Pythagoras' theorem

As the body diagonal of Cube is \(\sqrt{3} a\) and height of Tetrahedron is \(\frac{a}{\sqrt{3}}\),

The answer is \(a (\sqrt{3}-\frac{1}{\sqrt{3}})\)= \(\frac {2}{\sqrt{3}}a\)

Actually this is perhaps the same way which \(Felix\) wanted to say but i hadn't read it before so didn't post as reply to \(Felix\)

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For question 1 the centre of the tetrahedron will also the the centre of the sphere find the point then the distance from this point to the vertex will be the spheres radius (or work backwards from this length if you find reverse proofs easier)

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Answer for (2):

A better description of the problem is a plane passing through all the vertexes connected to a vertex by sides.

The diagonal of this cube is \(sqrt{2}\) a which we shall call b for simplicity.

Now we have a tetrahedron with base length of b and side length of a

Now we proceed to calculate the height of this shape:

1) the base of the triangle:

Imagine a top view of the tetrahedron, an equilateral triangle split into three equal isosceles triangles (sides b, c and c). (30, 30, 120 degrees) (as they bisect the 60 degree angle of the equilateral triangle)

First we find the length from the mid point of b to the centre of the triangle. (cosine rule works) but i shall use area for simplicity.

Area of equilateral triangle is (b^2 root(3))/4

area of one isosceles is one third this (b^2 root(3))/12

area is also half bc sin(30)

solve for c and you get (b root(3))/3

using pythag you get (b root(3))/6 for the distance from midpoint of b to the centre

2) the face of the triangle

the length obviously = b/2

by pythag we have the height of the tetrahedron as 2 root{2}b

pythag gives us root(2)b as the diagonal length

now the length of the diagonal - the hight of the tetrahedron will give the answer to the question, but i have made an error somewhere along the way.

i currently do not have the time to re-do but i shall do so later. sorry.

I shall post this anyway as it shows the process (minus algebraic manipulation) to reach the answer

also sorry i cannot figure out formatting and i'm pressed for time

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thanks Felix

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https://docs.google.com/drawings/d/1oHYGjPz9EkEbH41fZXW98-3_94A0R6x5jJP2ollIOwk/edit?usp=sharing

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Comment deleted Mar 03, 2015

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From which 2D view should this be Done?? because if we consider for a Triangle(Face of Tetrahedron) with it's circumcircle in 2D, then we will get radius of the circle in which ONE FACE OF TETRAHEDRON is inscribed not ' r ' which will be actually more than the radius obtained from 2D view mentioned above.....I am stuck up here....

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i can't see how it can be done

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my intuition tells me that the solution lies in calculating the percentage of the spheres volume the tetrahedron takes up have been unable to do this however

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