Four Points Two Distances Proof

   
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2017/10/03 - Four points - the proof

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The Four Points Puzzle challenge that Peter Winkler set me was this:

  • Find all configurations of four (distinct) points in the plane that determine exactly two distances.

As I remarked earlier, given four points there are six possible pairs, so in general, four points in the plane will determine six distances. But we can choose to place the points to create duplication, and the challenge here is to find all possible ways to do that and end up with only two distances.

That's reasonable, and lots of people have sent me their solutions. Peter had also said:

  • "Nearly everyone misses at least one, and for each possible solution, it's been missed by at least one person."

So the challenge isn't really to find all the configurations so much as to be able to prove that you have them all. That's what I'm going to do here, so if you haven't had a go at the puzzle, now is the time to bookmark the page, close the browser, get yourself a tea, coffee, or other beverage of your choice, and have a go.

Remember, it's the proof that matters.

Here's mine.

Preliminaries ...

We have four points, and therefore six pairs. But we only have two distances, so some are short, and some are long. The short ones are all the same as each other, and the long ones are all the same as each other. Obvious, I know, but we're just warming up.

So let's prove a few simple results.

Lemma: No three points are collinear.

These sorts of proofs, descriptions, and discussions can really stretch your visualisation skills. If you're an experienced mathematician you'll know that you can't just skim this sort of thing and try to "Read like a novel" - you need to take pencil/pen and paper and draw what's going on. And if not literally with pen and paper, then at least to work on explicitly visualising it.

You need to "Read like maths" - because it is.

If points $A$, $B$, and $C$ are collinear, with $B$ between $A$ and $C$, Then $|AC|$ is the long distance, and both $|AB|$ and $|BC|$ are the short distance. Consider point $D$. We have two possibilities.

If $D$ is the same distance from $A$ and $C$ then that must be the long distance, but then $|BD|$ is neither $|AB|$ nor $|AC|$. So that doesn't work.

So $|AD|$ and $|CD|$ are different lengths. We can compute exactly (up to symmetry) where $D$ must lie, then observe that again, $|BD|$ is neither of our possible lengths, so that doesn't work either.

So there is nowhere to put point $D$, and hence no three points can be collinear.

Observation: Each edge is in exactly two triangles.

Lemma: There are at least two short edges that share a vertex.

We cannot have no short edges, because then we'd have all long edges, and the points would have to be in a tetrahedron, which isn't in a plane.

We cannot have exactly one short edge, because then the other five would all be long, and have to form two equilateral triangles with a shared edge. The remaining edge would have to be longer.

We cannot have $|AB|$ and $|CD|$ as the only short edges, because both $C$ and $D$ would have to be on the perpendicular bisector of the line $AB$, and they would have to be on opposite sides of the line, making $|CD|$ too long. So if there are exactly two short edges they must share a vertex, which is what we need.

Any three or more edges must share at least one vertex.

So there are (at least) two short edges that share a vertex.


Now for the single insight that, I think, make the enumeration of all solutions quite straight-forward. The fact is simple, making certain it is a fact is a little tedious - feel free to skip this section if you're happy with the statement of the lemma.

Lemma: There is a triangle with two short and one long side.

We proceed by the cases of how many long edges there are.

We cannot have no long edges.

If there is exactly one long edge then either triangle it forms will suffice.

If there are exactly two long edges, consider one of them. It is part of exactly two triangles, at most one of which can contain the other long edge, so at least one of which has two short sides.

If there are exactly three long edges then we have three short edges. Take two that share a vertex - $AB$ and $BC$. If $AC$ is short then all distances to $D$ would have to be long, which is not possible. So $AC$ is long, and $ABC$ is our required triangle.

If there are exactly four long edges then the two short edges share a vertex and the remaining edge of that implied triangle is long.

We cannot have exactly five or six long edges.

So we are done.


Finding all solutions ...

http://www.solipsys.co.uk/images/TriangleWithTwoShortSides.png
Triangle with
two short sides
So now we know that there is at least one triangle with two short sides and a long side. So let's call that $ABC$ with $AC$ being the long edge. We can see that $B$ has to be on that blue line, above the line $AC$, and because the red lines are the shorter ones, $B$ cannot be as high as the grey point, which forms the equilateral triangle with $A$ and $C$.

Now let's consider the fourth point, $D$. Without loss of generality we can assume that $D$ is on or to the right of the blue line. If not, reflect around that line.

The distances from $D$ to each of $A$, $B$, and $C$ must be short or long, so let's write all the possible combinations of each of those distances.

Case
Distance
to $A$
Distance
to $B$
Distance
to $C$
0
Short
Short
Short
1
Short
Short
Long
2
Short
Long
Short
3
Short
Long
Long
4
Long
Short
Short
5
Long
Short
Long
6
Long
Long
Short
7
Long
Long
Long

Cases 1 and 3 are not possible, for then $D$ would be to the left of the blue line, which we have excluded.

Case 0 (sss): Point $D$ has to be the reflection of point $B$ across the line $AC$ giving a pair of equilateral triangles that share an edge.
http://www.solipsys.co.uk/images/FourPointsProblem_Case_0a.png
http://www.solipsys.co.uk/images/FourPointsProblem_Case_2a.png
Case 2 (sls): As case 0, point $D$ has to be the reflection of point $B$ across the line $AC$. In this case we get a square with $AC$ and $BD$ as diagonals.
Case 4 (lss): The triangle $BCD$ is equilateral, and $ACD$ is isosceles.
http://www.solipsys.co.uk/images/FourPointsProblem_Case_4a.png
http://www.solipsys.co.uk/images/FourPointsProblem_Case_5a.png
Case 5 (lsl): Here $ACD$ is equilateral, and $B$ is in the middle. Point $D$ must be where the grey point is on the diagram above.
Case 6 (lls): Here $D$ is a little below $C$, and the four points are the points of a regular pentagon.
http://www.solipsys.co.uk/images/FourPointsProblem_Case_6a.png
http://www.solipsys.co.uk/images/FourPointsProblem_Case_7a.png
Case 7 (lll): Now point $D$ forms an equilateral triangle with $A$ and $C$, and must be the opposite side of the line $AC$ from the point $B$. The points form a kite.

So there are 8 combinations, two of which are impossible, and the other six give exactly one solution.

And we are done.

  • Did you get them all?
  • Which one (or ones) did you miss?
  • Which one did you find last?
  • How did you prove that you had them all?

Send me email from the comment box below.


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