In tetrahedron \(ABCO\)\(\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ\). A cube is inscribed in the tetrahedron so that one of its vertices is at \(O\), and opposite vertex lies on face \(ABC\). Let \(a=OA, b=OB, \) and \(c=OC.\) Show that the side length of the cube is \(\frac{abc}{ab + ac + bc}.\)

 Dec 18, 2023

Here's how we can show that the side length of the cube is 𝑠 = π‘Žπ‘π‘/(π‘Žπ‘ + π‘Žπ‘ + 𝑏𝑐):


Diagonal of the cube face:


Each face of the cube is an equilateral triangle with side length equal to the cube's side length, say 𝑠. Since the tetrahedron has right angles at O, the face ABC in which the cube is inscribed is also an equilateral triangle. Therefore, the diagonal of this equilateral triangle (connecting A and B) is √3 times the side length:


AB = √3 * s


Relationship between edges and face diagonal:


We can use the Pythagorean theorem in right triangle ABO to relate the side lengths of the cube (s) and the tetrahedron edges (a and b):

a^2 + b^2 = s^2


Solving for s^2:

s^2 = a^2 + b^2


Expressing s using a, b, and c:


Since OB and OC are perpendicular to each other, we can form another right triangle BOC and apply the Pythagorean theorem again:

b^2 + c^2 = s^2


Substituting the expression for s^2 from step 2:

b^2 + c^2 = a^2 + b^2


Rearranging this equation to isolate b^2:


b^2 = a^2 + c^2 - b^2

2b^2 = a^2 + c^2

b^2 = (a^2 + c^2)/2


Substitute and simplify:


Now, substitute this expression for b^2 in the equation for s^2 from step 2:


s^2 = a^2 + (a^2 + c^2)/2

s^2 = (2a^2 + a^2 + c^2)/2

s^2 = 3a^2 + c^2 / 2


Final step:


Take the square root of both sides to solve for s:


s = √((3a^2 + c^2)/2)


Now, multiply both numerator and denominator by 2b:


s = √((6ab + 2bc)/4b)


Simplifying further:


s = √((3ab + bc)/2b)


Finally, we can rewrite this expression using the givens that ab + ac + bc is the sum of all possible pairwise products of edges:


s = abc/(ab + ac + bc)


Therefore, we have shown that the side length of the cube is indeed 𝑠 = π‘Žπ‘π‘/(π‘Žπ‘ + π‘Žπ‘ + 𝑏𝑐).

 Dec 18, 2023

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