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 the opposite vertex lies on face \(ABC\). Let \(a = OA,b = OB,\) and \(c = OC.\) Show that the side length of the cube is \(\frac{abc}{ab + ac + bc}.\)
Hints to help YOU to guide to a full solution:
(Notes about this diagram: I constructed \(Q\) and \(R\) to be midpoints of their sides.)
-\(\triangle QTR\) might be a helpful triangle.
- Similar triangles are the key.
-\(\triangle COA \sim \triangle QTR\) might be a helpful pair of similar triangles.
- In your help/solution please designate that $s$ is the side length of the cube.
P.S. I don't NECESSARILY want a full solution/proof. But if you have the time and passion to do so, feel free! But hints/help/guidance is preferred.
+397 Let O be the origin of a 3D co-ordinate system with OA, OB and OC lying along the Ox, Oy and Oz axes respectively.
The equation of the ABC plane will be
\(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)
If h is the side length of the cube, then the co-ordinates of the point on the cube furthest from the origin will be
(h, h, h), and this point is to lie on the plane, in which case
\(\displaystyle \frac{h}{a}+\frac{h}{b}+\frac{h}{c}=1,\)
so
\(\displaystyle h\left\{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right\}=1 \\ \displaystyle h = \frac{abc}{ab+bc+ca}.\)
+397 I'll take you through an alternative method, but leave you to fill in the detail.
Referring to the diagram given in the question, start by drawing a line through C and the point of contact of the cube with the plane, (call that point D), onto its point of intersection with the line AB. Call that point E.
Draw a diagonal across the top face of the cube from OC to the point of contact with the plane. Call the point on OC, F.
You now have two (vertical) similar triangles, COE and within that the smaller CFD.
Let the side length of the cube be h and let k be the length of OE
You should then be able to use the similar triangles to show that
\(\displaystyle h = \frac{ck}{(c\sqrt{2}+k)}.\)
Moving on to the base triangle AOB, the angles AOE and BOE are both equal to 45 deg making it easy to write down the areas of the triangles AOE and BOE. (Area AOE = aksin(45)/2.)
Now say area AOB = area AOE + area BOE and you should be able to show that \(\displaystyle k= \frac{ab\sqrt2}{(a+b)}.\)
Finally, substitute that into the expression for h and tidy up.
Thank you so much for this answer and all the hard work you put into it!! It was life changing for me. I have been stuck on this problem for days and days. I have completed my work now and the weight has been lifted off my shoulders...THANKS AGAIN!!!
-Original poster :)
