The following puzzle bears resemblance to some Moscow Mathematical Olympiad Problems. I believe Hoeflin himself drew as much inspiration from Martin Gardner's puzzles as from USSR Olympiad problems for the gifted. Although some of the Mega tests' items are indeed known math results, there is some ingenuity in their presentation and in asking new questions based on prior formulated problems.
In other words, they may be seen as good follow-up questions to classical problems with known elementary solutions. At first glance, this problem mirrors the one discussed in the previous article, which seeks to determine the optimality of a spatial arrangement. Still, instead of asking for 3D space regions, it asks for the lines that unite the planes dividing space. However, it contains additional surprises I would not like to give away yet.
It is interesting to speculate on the extent to which math literature and Western classics influenced Ron in creating or reformulating pre-existing problems for his tests. For it may seem to privilege a certain cultural mastery. But what else can be done if we are expected to follow Wechsler's approach to intelligence testing?
How can we effectively test general intelligence if we cannot use specific content? Regardless of how universally recognized the content may be, it will inevitably provide some individuals with an advantage over others. This tangential issue deserves further discussion in future articles. For now, I hope to spark the reader's curiosity on this topic. Without further ado, here's the puzzle:
The Dots in Space Puzzle
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Let \(A, B, C, D, E\) be the points given. We begin by noting a very basic but nonetheless important fact: a plane is uniquely determined by \(3\) points in three-dimensional space. Thus, the total number of planes determined by every group of dots is*,
$$\binom{5}{3}=10$$
Since any pair of planes would determine a distinct line. Nevertheless, in this case, it is given that for each set of three points, there is a plane that contains them. Note what this means: Whenever two of the given points, say \(A\) and \(B\), are contained in a line of intersection of two planes, a third point must form a plane with the other two points in the line by hypothesis.
Consequently, there are only two possibilities: either the third point is coplanar with the other two, or it is not. The former case implies that this point is in one of the two intersecting planes. The latter case entails that there must be a third plane intersecting the other two at this precise line (since any three given points have a plane passing through), and thereby creating an intersection of three planes.
To account for the cases in which three planes intersect, rather than two, observe that for any possible pair of points in a given line of intersection, we are counting three times as many lines as there are. This is because the three intersecting planes form a single line rather than three lines. Therefore, we must subtract the "two redundant lines" that result from considering the \(3\) possible pairs of planes at a given intersection as forming three distinct lines.
$$\binom{10}{2}-2\cdot\binom{5}{2}=45-20=25$$*For all practical purposes, hereafter, the terms dots and points are used interchangeably.
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