+101 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 \(\frac{abc}{ab + ac + bc}.\)
+193 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).
