XL OM - II - Task 3

Given is a trihedral angle $ OABC $ with vertex $ O $ and a point $ P $ inside it. Let $ V $ be the volume of the parallelepiped with two vertices at points $ O $ and $ P $, whose three edges are contained in the rays $ OA^{\rightarrow} $, $ OB^{\rightarrow} $, $ OC^{\rightarrow} $. Calculate the minimum volume of the tetrahedron, whose three faces are contained in the faces of the trihedral angle $ OABC $, and the fourth face contains the point $ P $.
Let $ \mathbf{Q} $ denote the considered parallelepiped with volume $ V $. Let $ OA $, $ OB $, $ OC $ be its three edges. We draw a plane through point $ P $ intersecting the rays $ OA^\rightarrow $, $ OB^\rightarrow $, $ OC^\rightarrow $ at points $ K $, $ L $, $ M $ respectively (thus forming the tetrahedron $ OKLM $; its volume is the quantity we need to minimize).
Consider the following three numbers (ratios of triangle areas):



Of course, $ x, y, z > 0 $, $ x + y + z = 1 $.
Let the distances from points $ P $ and $ M $ to the plane $ ABC $ be denoted by $ d $ and $ h $, and the distances from these points to the line $ KL $ - by $ d $ and $ h $. Due to the parallelism of the line $ CP $ to the plane $ OAB $, the proportion holds:




Similarly, we can interpret the numbers $ x $ and $ y $. Thus,






From the formula for the volume of a pyramid and the relationships (1) and (2), we obtain the equality:




Twice the area of triangle $ OAB $ is the area of the face of the parallelepiped $ \mathbf{Q} $ contained in the plane $ \mathbf{OAB} $; the number $ d $ is the length of the corresponding height of this parallelepiped. Therefore, $ d \cdot 2\mathrm{area}(OAB) = \mathrm{volume}(\mathbf{Q}) = V $. From this and the arithmetic mean-geometric mean inequality for the numbers $ x $, $ y $, $ z $, it follows that:




The obtained inequality becomes an equality when $ x = y = z = 1 $, i.e., when the segments $ OK $, $ OL $, $ OM $ are three times longer than $ OA $, $ OB $, $ OC $, respectively. This happens when the plane $ KLM $ is drawn through point $ P $ parallel to the plane $ ABC $.
The number $ \frac{9}{2} V $ is thus the sought minimum.
Note. When the segments $ OA $, $ OB $, $ OC $ are pairwise perpendicular and have equal length, they can be taken as the unit vectors of a Cartesian coordinate system. The entire reasoning can then be elegantly expressed in the language of analytic geometry. For: $ A = (1,0,0) $, $ B = (0,1,0) $, $ C = (0,0,1) $, $ P = (1,1,1) $; any plane passing through point $ P $ is given by an equation of the form $ ax + by + cz = 1 $, with positive coefficients $ a $, $ b $, $ c $ chosen so that $ a + b + c = 1 $. (It is not hard to notice that the numbers $ a $, $ b $, $ c $ are nothing else but the numbers $ x $, $ y $, $ z $ from the previous solution method; this time we need the letters $ x $, $ y $, $ z $ to denote the variables of the coordinate system; hence the change in notation.) The considered plane intersects the coordinate axes at points $ K = (1/a, 0,0) $, $ L = (0,1/b,0) $, $ M = (0,0, 1/c) $.
The parallelepiped $ \mathbf{Q} $ is now a unit cube; its volume $ V $ equals $ 1 $. The tetrahedron $ OKLM $ has a volume of $ (1/6)(1/a)(1/b)(1/c) $. As before, applying the inequality between means, we observe that (with the constant sum $ a + b + c = 1 $) the product $ abc $ reaches its maximum when $ a = b = c = \frac{1}{3} $; the volume of the tetrahedron $ OKLM $ then reaches its minimum, equal to $ \frac{9}{2} $.
Thus, the thesis of the problem is proven in a very special case, where $ |OA| = |OB| = |OC| = 1 $, $ OA \perp OB \perp OC \perp OA $. This "very special" case is, however, in a certain sense, quite general. It should be obvious to those familiar with the concept of affine transformations and their simplest properties. Any parallelepiped can be affinely transformed into a unit cube; and the ratio of volumes is an invariant of affine transformations of three-dimensional space. And it is precisely these two observations that suffice to deduce the thesis of the problem in full generality from the special case considered above.