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 \(\frac{abc}{ab + ac + bc}.\)

Vxritate Dec 18, 2023

#1**0 **

Here's how we can show that the side length of the cube is π = πππ/(ππ + ππ + ππ):

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 π . 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 β3 times the side length:

AB = β3 * 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 = β((3a^2 + c^2)/2)

Now, multiply both numerator and denominator by 2b:

s = β((6ab + 2bc)/4b)

Simplifying further:

s = β((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 π = πππ/(ππ + ππ + ππ).

Boseo Dec 18, 2023