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"Ridiculous!" you think "What can be the odds? Either I'm hallucinating or the amateur writing this story plunged to new depths of incompetence." Both being equally likely you don't bother finding out which but just accept that after drifting aimlessly in the ocean on a makeshift float for days you ran into someone else complete with their own float and not much else. "Let's pool our resources!" you cry with as much enthusiasm as your addled brain can muster. "What resources?" comes the reply. "Well," you answer detemined not to let your new "friend" ruin the moment "let's start with real estate! Let's join our floats together." -- "Join them together using what?" -- "We can use material from the floats themselves!" you counter even finding the energy to draw a little scheme in the mud that covers your float.

enter image description here

"With our primitive tools? The shapes are way too complex. The best we can hope for is cutting out simple convex shapes." The wretch started to seriously annoy you. He seemed to enjoy finding fault! But he had a point.

Will our two heros find a way to join forces or will they drift apart and die alone? Find out yourself, because this "story" will certainly not be continued...

ts; dr; (too silly, didn't read)

from two equal side-by-side rectangles cut out as few and as simple as possible convex shapes that can be rearranged such that the rectangles be interlocked. Or prove that this is not possible.

Hint:

It can't be done with two.

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  • $\begingroup$ Nice puzzle! I was so convinced it could not be done that I thought I'd almost completed a proof of impossibility, then realized a loophole in it and saw the solution suddenly. Fortunately, the proof could be tweaked to the proof of impossibility for any fewer pieces as found in my updated answer. This was certainly a good puzzle, and the answer is very mathematically pleasing! $\endgroup$
    – Anon
    Oct 14, 2021 at 4:23
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    $\begingroup$ @Anon, well done and glad you enjoyed the puzzle. Just one little thing: There is no need to make theta close to 90°. The construction works for any theta between 60° and 90° excluding the limits themselves. The best for physical stability would be something well in the middle. $\endgroup$
    – loopy walt
    Oct 14, 2021 at 4:37
  • $\begingroup$ @loopywalt I agree; I just couldn't be bothered calculating the precise limits on x and y, so I said that to guarantee the pieces wouldn't overlap... I've now made the answer more exact $\endgroup$
    – Anon
    Oct 14, 2021 at 4:52
  • $\begingroup$ Is it required that the two rectangles are initially touching as above and may not move during assembly? $\endgroup$
    – Magma
    Oct 16, 2021 at 14:46
  • $\begingroup$ @Magma yes, that's right. Only the cut out bits may move. But if you have an interesting solution for other rules I won't mind your posting it. I just cannot guarantee that others won't downvote. $\endgroup$
    – loopy walt
    Oct 16, 2021 at 22:39

2 Answers 2

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TL;DR

Our floating Robinsons can do this with $3$ pieces, but not with $2$

In this answer, I first show constructive examples with $4$ and then $3$ pieces, and then prove why $2$ pieces is impossible.


First, and easiest, we can do it in:

$4$ pieces

Using the principle of the:

Dovetail Joint

The method is as follows:

Cut along the outlines of 2 non-parallel congruent rectangles that both cross the dividing line between the two main rectangles, neither being perpendicular to the dividing line, thus forming 4 separate convex shapes to be manipulated:

Rob1

Apart from ensuring that the cut rectangles are the same length and width, you do not need to place them at precise locations or angles (as long as they are not perfectly perpendicular to the line dividing the two rectangles, and are not perfectly parallel, and as long as they extend sufficiently far to the right (wlog) the construction will work).

You then just need to rearrange your pieces.

Simply swap the pieces within each rectangle. We can do this with planar rotations and translations alone as follows:

Rob2

This works as long as we choose $|BC|<|DJ$ and $|GH|<|FK|$, in other words, our rectangles extend sufficiently far to one side of the join. There are other ways we could place these pieces to get the same result (including swapping between the upper and lower rectangles, and rotating the pieces so the front is now the back).

Why does this work?

Setting up our cuts so that the moved $b$ and $d$ pieces fully cross the join guarantee that the rectangle can only be pulled apart in a direction parallel to the long axis of our cut rectangles. By having two such cut rectangles that are not parallel to each other, we guarantee that there is no direction (within the plane) that the larger rectangles can be pulled to unlink them.


Now that we understand that method, how can we do it better?

We can do it in three pieces, by in some sense combining two of the pieces from the four-piece dissection.

Cut the wood as follows: Pick appropriate lengths $x$ and $y$ and any $\theta\in(60,90)$ and cut out the following three symmetrical trapezia:

Robinson16

Then rearrange as follows:

Robinson17

To see why this will work for any $\theta\in(60,90)$, consider the following diagram which shows the internal triangle formed by the dissection:

Rob5

We can then see that our trapezoids will overlap if and only if: $$(x-2y\cos\theta)\cos\theta\ge\frac{x-2y\cos\theta}{2}$$ In other words, $\cos\theta\ge\frac{1}{2}$, which happens for $\theta\le60$. Any other $\theta\in(60,90)$ will work.


We now need to prove that we cannot do better.

The next step would be to get a solution in two pieces.

However, this is impossible. The following is a sketch of a proof:

Suppose I have performed a dissection, cutting out two pieces and rearranging them such that the two rectangles are locked together.

Then there must be at least a single piece (call it $A$) in the rearrangement which crosses the dividing line $MN$ (otherwise our two rectangles could be slid past each other vertically). Now consider this piece by itself and imagine we have the following construction:

Rob3

Now, take arbitrarily small left and right tangents to $P$ about the points $R$ and $S$ as follows, and consider the angles formed $w, x, y, z$:

Rob4

(Note that the left and right tangents about $R$ and $S$ need not be parallel, since $P$ may not be differentiable). Now suppose $w+x<180$, then we could construct a line as follows (arbitrarily close to the point $R$):

Rob5

This line would be a straight line connecting two points in the shape $P$, but it would pass through the exterior of $P$. This would contradict convexity. Therefore $w+x\ge180$. By an identical argument, we can prove that $y+z\ge180$.

Now suppose $w+y<180$ and $x+z<180$. Then $w+x+y+z<360$ which is impossible, since $w+x\ge180$ and $y+z\ge180$. Therefore either $w+y\ge180$ or $x+z\ge180$. Without loss of generality, let $w+y\ge180$. Now extend these tangents as follows:

Rob6

These tangents thus define a region $Q$ to the left of the line $MN$, which is either an infinite rectangle (if $w+y=180$), or a triangle (if $w+y>180$). The portion of $P$ that lies to the left of $MN$ must be wholly contained within $Q$. To see this, suppose without loss of generality that $P$ extends above $Q$. Then a line could be drawn through $R$ as follows:

Rob7

Which contradicts convexity of $P$.

Now my claim is that if we pull our left hand large rectangle in the direction shown by the following arrow, where the angle $u=\frac{w+y}{2}$, then it can be pulled away from $P$ and the right hand larger rectangle.

Rob7

To prove this, assume the converse, i.e. that the left hand rectangle cannot be pulled away parallel to the arrow. Now consider all lines intersecting $P$ that can be drawn parallel to the arrow as follows:

Rob8

For there to be something preventing the left rectangle being pulled away, at least one of these lines (going left to right) would have to pass through part of $P$, and then pass through part of the rectangle again before reaching $MN$. But at $MN$, all these lines pass through $P$ again, so if such a line existed, it would contradict convexity of $P$. Thus there is no line on which there is anything to stop the left rectangle being pulled away at an angle $u$.

Now, let us return to our final dissection (with pieces rearranged to lock the squares together). If only one piece in this dissection crosses $MN$, then we have shown there is some angle $u$ at which the left rectangle can be pulled to detach it. Thus there must be at least a second piece (call it $O$) that crosses $MN$ in the final dissection. Let us look at these two pieces alone (ignoring all others):

Rob10

Now consider the cuts required to form these pieces. In making the initial cuts to form the slot that $P$ fits into, an area larger than $P$ may have been cut (to see why have a look at my example four-piece dissection, where $P$ corresponds to the piece $b$, which was formed as part of a larger rectangular cut). However, any cut that would generate the slot that $P$ fits into, must have crossed $MN$, and therefore have produced at least two different pieces (we label the parts that intersect $P$ as $P'$ and $P''$ - the actual pieces produced may have been larger). The same is true of $O$.

Rob11

Thus for two pieces to be used, $P'$ and $O'$ must be part of one larger piece $\#$, and $P''$ and $O''$ must be part of one larger piece $\#'$. In the initial dissection (before rearrangment) all pieces must be entirely on one side of $MN$. Let us look at $\#$ by itself:

Rob13

We can look at the angles produced by the tangents to $\#$:

Rob14

If $r+s\ge180$ then our argument from earlier shows that the left-hand rectangle can be pulled away in a certain direction. Therefore we must have $r+s<180$.

Now, let us rearrange our pieces. Since $\#$ was the only piece cut out of the left-hand rectangle, it is obvious that $P$ and $O$ (one of which must be identical with $\#$) must fit perfectly into the space formed when $\#$ is removed. Let us ignore what happens underneath the grey square:

Rob16

Now by convexity, we can show (as we showed earlier) that $r+r'\ge180$ and $s+s'\ge180$, which implies (since $r+s<180$) that $r'+s'\ge180$.

However, we could have made exactly the same argument with the piece $\#'$ to show that $r'+s'<180$:

Robinson18

This is a contradiction. Thus it is impossible with two pieces.

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Congratz to @Anon for finding the intended solution. I'm posting this just to show off a to scale picture I happened to have lying around:

enter image description here

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