If there are four non-coplanar points, find the number of planes such that all four of them are equidistant from the plane. Sorry one of those problems where I don't know what to do. How should I do this sum?


2 Answers 2


There are two distinct scenarios. (This answer is not entirely rigorous, but it provides help with visualisation. Also, images are generated from GeoGebra, an amazing free and open source application. Note the specificity of the coordinates are irrelevant, picked for visual coherence.)

First scenario: choose any three points from the four, and form a plane through them. Then construct a perpendicular line from the fourth point to the plane, and find the midpoint of this line. The red plane through this midpoint, parallel to the blue plane, is equidistant to all 4 points. Note: There are $\binom {4}{3}$ occurrences of this scenario, since you are choosing three points from four. First Scenario

Second scenario: Split the points into pairs. Then form parallel planes (blue) through each of the pairs. A third plane parallel to these two and in the center of them is equidistant to all 4 points. There are $1/2 \times \binom {4}{2} $ occurrences of this scenario, since you are choosing 2 pairs from four. Here's a link on choosing pairs: Combination of splitting elements into pairs Second Scenario

$$ \binom{4}{3} + \frac{1}{2} \binom{4}{2} = 4+3 = 7 $$

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    $\begingroup$ Nice figure but in the first one the required plane is not shown (rather some unnecessary plane is shown) and that is confusing. Also it'd have been better had you explained your combinations. Specially the $1/2$ in second case. $\endgroup$
    – Hritik
    Commented Feb 10, 2018 at 9:45
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    $\begingroup$ Thanks for your suggestions! I have amended those faults mentioned. $\endgroup$ Commented Mar 21, 2018 at 15:11

This is a problem in enumerative geometry. We don't have to compute these planes.

Basically all admissible arrangements of four points in ${\mathbb R}^3$ look the same: Imagine three different points in the $(x,y)$-plane $z=0$ and a fourth point anywhere in a parallel plane $z=c$, $\>c\ne0$.

Considering the planes $\pi$ from which all four points are equidistant you have to distinguish three cases: (a) all four points on the same side of $\pi$, (b) three points on one side and one point on the other side of $\pi$, (c) two points on each side of $\pi$.

Case (a) can obviously not occur.

Case (b): You can choose three of the four points in four ways. They determine a plane $z=0$, say. Let $c$ be the $z$-coordinate of the fourth point. Then the plane $\pi:\ z={c\over2}$ is equidistant from all four points. This gives four such planes $\pi$.

Case (c): You can pair up the four given points in three ways. Let $(a,b)$, $(c,d)$ be one such pairing, and consider the lines $g:=a\vee b$ and $h:=c\vee d$. By assumption these two lines are skew, i.e., they do not intersect in ${\mathbb R}^3$, nor are they parallel. Visualize now $g$ and $h$ as axes of cylinders with increasing radius $\rho$, where $\rho$ grows from $0$ until the two cylinders touch at a point $P$. Parallels $g'$ and $h'$ to $g$ and $h$ through $P$ together span a plane $\pi$, and one easily convinces oneself that all four points are equidistant to $\pi$, two of them on each side. This gives three such planes $\pi$.

It follows that the answer tou your problem is $7$.

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    $\begingroup$ I tried visualising all cases but am getting confused , can u explain how should we go about it $\endgroup$
    – avz2611
    Commented May 11, 2014 at 19:04

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