On schemes for learning to solve quadratic equations

So I’ve recently been teaching the Math2c course at a swedish upper secodnary school and had some reasons to be thinking about quadratics because that’s like one third of the content of the course. Or well teaching is actually a bit charitable this time as the students at this particular school are among the most consistently diligent at self-study in the country that peppering them with instruction they’ve already read actually slows them down, but the role of instruction and individual work in learning mathematics at different levels is a discussion for another time.

So the model of teaching quadratics in most of the books I’ve seen typically runs something like this.

  1. You start out some preliminaries about factoring techniques like the patterns a^2 - b^2 = (a + b)(a - b) and a^2 + ba  = a(a + b)
  2. You learn to solve the simplest quadratic x^2 = a by essentially explaining the pattern x = \pm \sqrt{a}
  3. You state the zero-product property (ZPP), that ab = 0 implies a = 0 or b = 0, and solve some quadratics that have already been factored like (x + 2)(x - 5) = 0.
  4. You learn the factoring method for solving quadratics without a constant term x^2 + px = 0 by factoring x(x + p) = 0 and then applying the ZPP.
  5. You (might) learn how to complete the square
  6. You quickly use the method of completing the square to run though the proof of the pq-formula x = -\cfrac{p}{2} \pm \sqrt{\left ( \cfrac{p}{2} \right )^2 - q}. (This does not use the ZPP nor the conjugate rule)
  7. You spend a couple of hours drilling the aforementioned methods possibly.

Now functionally this route of progression definitely works for conveying how to solve these kinds of equations, but there are still some things I don’t quite like about it. First of all the methods reduce to three separate cases which superficially seem to have nothing to do with eachother

  1. x^2 = a \Rightarrow x = \pm a
  2. x^2 - bx = 0 \Rightarrow x(x - b) = 0 \Rightarrow x = 0\, \text{or} x = b
  3. x^2 + px + q = 0 \Rightarrow -\cfrac{p}{2} \pm \sqrt{\left ( \cfrac{p}{2} \right )^2 - q}

So whenever one solves a problem like this you first have to use identification of type, which is actually a lot harder to do than people who can already do it think it is. Also pedagogically the arrow of progression isn’t clear. Step 1 as presented has nothing to do with 2. The knowledge of step 3 is not used beyond step 4, and if the only use of completing the square initially is to prove the pq-formula the student will forget about this super important pattern and will have to relearn it at uni if ever.

I’ve thus been thinking if the scheme for learning how to work with quadratics might instead be replaced with:

  1. Preliminaries on factoring, difference of squares, square of sum, distributive property.
  2. Zero product rule and applying it to factored equations (x - 1)(x - 2) = 0 (This should be complemented by graphing)
  3. Solving x^2 - a = 0 by factoring with the difference of squares and applying the zero product rule. x^2 - 16 = (x + 4)(x - 4) and x^2 - 3 = x^2 - (\sqrt{3})^2 = (x + \sqrt{3})(x - \sqrt{3}) and sneaking in problems the type (x + 1)^2 - 4 = 0
  4. Solving x^2 - bx = 0 by factoring with the distributive property
  5. Learning to complete the square x^2 + px= (x + p/2)^2 - (p/2)^2
  6. Solving general quadratics by completing the square and then factoring with the conjugare rule
    x^2 - 4x + 3 = (x - 2)^2 - 4 + 3 = (x-2)^2 - 1 = (x - 2 +1)(x -2 - 1) = (x -1)(x - 3)
  7. Derive the pq-formula from the factoring. (Now this is admitedly a lot messier than the normal way with \pm)

Now there are many downsides to this scheme, for example step 3 being harder than the normal way on account of the normal way of applying operations to both sides is more analogous to solving linear equatons ax + b = 0 ,which is why it’s probably not used but it would be interesting to try it (or hear about if these is a book or school system) which approaches it this way. The final step 6, the method of factoring a general quadratic directly with the difference of squares is a technique which I find would be a useful addition to the scheme as I think it neatly exposes the connection between the three different forms in which you can write a quadratic expression in an algorithmic way instead of in a more formally logical way (invoking the general factor theorem).

Finally I’d just also like to summarize the method outlined in step 2 as a full algorithm because it more or less superceedes all of those messy “6-different ways, factoring by grouping methods” which I get when I google factoring.

Full algorithm for factoring a quadratic expression over the reals

  1. Rearrange all the terms so that you have expression = zero
  2. Complete the square of the expression (ignore if there is no linear term)
  3. Rewrite the constant term as a square (possibly by) a = (\sqrt{a})^2
  4. Apply the conjugate rule to the difference of two squares

x^2 -2x-15 = (x - 1)^2 - 1 - 15 = (x - 1)^2 - 16 = (x - 1)^2 - 4^2 = (x - 1 + 4)(x - 1 - 4) = (x + 3)(x - 5)

An application of continued fractions to composite electrical resistor design

I was teaching Ohms laws law and introductory circuit calculations bout two days ago at a Swedish gymnasium, a subject  which I’m about as adept as now as I was when I myself took the class 7 or so years ago. There are basically only two formulas one has to learn in that course and those are the two effective resistance formulas for resistors in series and resistors in paralell

R = R_1 + R_2 + ... + R_n, \qquad \cfrac{1}{R} = \cfrac{1}{R_1} + \cfrac{1}{R_2} + ... + \cfrac{1}{R_m}

Not sufficient to describe circuits with more complicated  typologies such as bridges like the Wheatstone bridge but sufficient for at least engaging in some basic design.

The problem which came to mind as I was reviewing the material was how one would set out to build a circuit element with a specified resistance from a given set of elements. Usually it’s the other way around. You get the component and  you compute the resistance but the inverse problem of getting a specified resistance and then designing the circuit from some pieces if of course a lot more interesting and as we shall see in this special case surprisingly straightforward.

Let us study this particular problem: “Given a set of resistors with resistance R how does one combine these to form an element with a specified fractional resistance \cfrac{p}{q}R“. If we want an integer multiple or integer fraction or a specific resistance then things are straightforward, just chain p together to get a resistance pR or put q in parallell to R/q and this is the smallest number of equal number of components you need. seriesnadparalellresistors.png

For fractional resistances of a more general form however there are clearly very different ways you can go about designing the circuit and there are different benefits to them

Solution one: Construct $latex $q$ chains with p resistors each and put those chains in parallel

mWastefulResistor1.pngThe chains have resistance p R and putting $latex $q$ of them in paralell reduces the total resistance by a qth to p R / q.

Solution two: Another is to first create p blocks of $latex $ resistors in paralell and then put those in series.paralellfirstthenseries.pngHowever both of these solution has the disadvantage that it requires  a total number of pq resistors to complete it which for most fractions becomes very impractical.

My question was therefore if there was a simple way to get a component with this resistance but which wouldn’t require quite so many resistors. Let us just start off by making clear that it is possible. These two components you see below both have the same resistance 2R/3


but the right one was built using only 3 resistors instead of 6 which is (if you count by means of manufacturing cost) is a more effective solution. And the process or computing the resistance of the latter composition betrays the coming idea

\cfrac{2}{3} = \cfrac{1}{1 + \frac{1}{2}}

(I will henceforth just set R = 1) Usually one would have used ^(-1)-notation for the parallell coupling but not doing that reveals the continued fraction expression associated with this construction.

The central idea can be laid out as follows inductively. If you want to design a component with resistance $p/q$ (where p/q is a reduced fraction such that p < q) perform euclidean division and rearrange it according to


\cfrac{p}{q} = \cfrac{p}{dp + r} = \cfrac{1}{d + \cfrac{r}{p}} = \cfrac{1}{\cfrac{1}{1/d} + \cfrac{1}{p/r}}

We now recognize the right hand side as the expression for the resistance of two resistor with resistance 1/d and p/r in parallel. The idea can diagrammaticaly be represented by


The 1/d resistor is constructed by putting d resistors in parallel and the p/r resistor is constructed by repeating the induction step.**

This can be streamlined by first computing the continued fraction of a fraction and then move backwards to form these construction steps . Take for example the continued fraction for 7/24

\cfrac{7}{24} = \cfrac{1}{3 + \cfrac{1}{2 + \cfrac{1}{3}}}= [0; 3,2,3]

which now contains a recipe for designing the composite resistor


Where the top 3-parallell component corresponds to [0;3,2,3], the 2-series at the bottom to [0;3,2,3], and the 3 in series at the bottom to [0;3,2,3].

Not all continued fractions will however have so small numbers in the expansion and especially when the numerator and denominator are fibonnacci numbers we’ll still end up with having to use a very large number of resistors.

What is kind of nice though is that irrational numbers which have a quickly convergent continued fraction can be approximated pretty accurately. \pi for example has a continued fraction [3;7,15,1,292,1,1,…] which means you can use the approximation \pi \approx [3;7,15,1] = [3;7,16] which using only 26 resistors gives 6 digits of pi which is pretty neat. A


I’ve got half a mind to build this circuit to see if it checks out but unfortunately cheap hobby resistors only have about 1% accuracy to them so you’d only really end up with 3 digits tops.

EDIT: Many of the diagrams in the original publications had errors in them which made them not agree with the text and which were corrected around one hour later.

Magnet assisted one way crank

This evening I took a trip by the Museum of Technology in Stockholm. Its somewhat recently undergone a redesign since last I was there maybe 7 years ago and I was interested in seeing how its turned out. General judgement is that it is now absolute heaven for families with small children with lots of novel contraptions and controller less video games while there is a little less on offer with respect to showcasing historical mechanism but still a fair trade.

I had gone primarily to get some stimulus and inspiration for mechanical configurations and mechanism which are simple enough to turn into elementary physics problems but which aren’t completely disjointed from applications. Like the hand crank wheel turning mechanism on a quaint car-like thing where the crank moves a rack back and forth which is linked to an elongated arm connected to the wheel joint.


The principal novelty however which I would like to point to a little was one I found in a machine that wasn’t even working. It was another crank mechanism where by turning a crank you set in motion a sequence of chains and gears which lifted some balls up to some slides of different curvature illustrating the principles of Brachistochrone problem. Everything about this machine was delightfully improvised which is why the electronics had broken down rendering the thing inoperable but what I want to talk about instead is a curious element in the crank mechanism.


It is admittedly hard to make out in this image but the odd parts were the two rubber elements which are in contact with eachither in the middle.

When the crank is pushed forward nothing of note happens. The whole axis, rubber parts and all, move together, turn the chains and everything works.

It is when you try to push the crank in the reverse direction that something novel happened. Since the machine isn’t supposed to be run in reverse reasonably there are two possible things that might happen, either there is a ratchet somewhere which prevent reverse motion entirely or you put in something similar to a freewheel that the crank becomes disengaged and you move it backwards with zero resistance.

Neither of those things happened here however. Now at first if you pulled it in the reverse there is a full stop, seemingly there is a rachet somewhere down the line, however if you try to turn the crank back harder the right rubber element somewhat dramatically begins to slip and you can turn back the crank almost freely.


This feels really really weird when this happens but also oddly familiar. A snapping you shouldn’t encounter in a purely mechanical system. The reason for how these rubber elements stuck together when the torque of the crank was low such as in the forward motion and the gentle reversed motion is that they weren’t sticking together by friction at all. There was pair of magnets embedded in the rubber keeping them together.


When the torque on these rubber elements becomes too high the magnets are pulled apart and the crank can move backward freely.


This was just so bafflingly weird this must have been an improvised fix to some problem they realized too late was present in the overall design. My guess is that the problem is that the ratchet preventing backwards motion of the system (not explicitly the crank) is located too far way from the crank somewhere down the line in a way that the reverse torque puts strain on something in the fixture, wood or plastic, which risks breaking if pulled back too hard. This is for children after all and if they can break it they will. The magnets guarantee that the torque on the axis has a manageable maximum at which the magnetic link snaps instead of something else in the machine..


Greatest common divisor of two consequetive triangle numbers

Currently thinking of simple problems which showcase how the Euclidean algorithm can be applied to deducing more theoretical results. It’s a largely abandoned algorithm in the Swedish primary and secondary educational systems and I suspect it’s mostly because it is a)hard to teach other than out of habit and b) the computational skill of students is too unreliable for them to work enough problems in a short enough time for it to be taugh within a few classes. (and c) it’s not as useful as other things like elementary geometry)

The problem which I’m going to review here was just inspired by the pretty well known result that two consecutive Fibonacci numbers are relatively prime where a proof by the Euclidean algorithm is delightfully elegant but once you’ve shown that one in an educational setting I only think it fair to have available some problems showcasing the same method and which are of equivalent difficulty.

Finding the greatest common divisor of two triangle numbers only superficially applies the Euclidean algorithm and its division into two cases makes it necessary to identify the two cases first.

The two distinctive cases are when the index of the largest of two consecutive triangle numbers is even and the other when it is odd. A triangle number being alternately defined by

T_{n} = 1 + 2 + 3 + ... + n = \cfrac{n(n + 1)}{2}\qquad \text{or} \qquad T_{n} = T_{n-1} + n.

Case 1. The larger one has an odd index. T_{2n}, T_{2n +1}. Applying the Euclidean algorithm we have

T_{2n + 1} = T_{2n} + 2n + 1

T_{2n} = 2n(2n + 1)/2 = n(2n + 1) + 0

After two steps we get that the greatest common divisor was 2n + 1 and we summarize

\boxed{\gcd(T_{2n}, T_{2n + 1}) = 2n + 1}

Case 2: The larger one has an even index. T_{2n-1}, T_{2n} Applying the Euclidean algorithm again we have

T_{2n} = T_{2n-1} + 2n

T_{2n-1} = (2n-1)2n/2 = (2n-1)n = (n - 1)2n + n

(n -1)2n = 2n + 0

and we have thus found that if the larger triangle number has an even index then the greatest common divisor is n.

\boxed{\gcd(T_{2n-1}, T_{2n}) = n}

Interesting way droplets hang on the tip of a needle

A nonmathematical post but nevertheless something I novel I noticed this evening and which made me think that I should probably investigate this a bit further.

I recently bought a cheap digital microscope and  have been randomly looking up close at things to verify some of my previous explanations for various phenomena. One being how small droplets of liquid produce pretty vivid dots when they lie on the surface of the screen of your phone. The basic physics is that the droplets forms a semispherical lense which magnifies the individual pixels of the screen making them partially visible even from afar as illustrated by this image of a ~1mm droplet on a Samsung Galaxy S3 screen.  0120-26.jpgand here slightly more magnified.0120-25.jpg

As I was applying the droplets however dipping various small thin things like needles I realized dropping the droplets from needles was no good because the droplet hangs on the tip of a needle in a rather curious way. 0120-8.jpg

Or rather it doesn’t hang from the tip at all. This is to be compared with how droplets would hang from things which arent sharp at the tip like a pencil lead0120-14.jpg

and sharpening the pencil lead at the tip with a knife the same effect is produced 0120-23.jpg

In a way this isn’t that strange given that one of the effects which retains the droplet is that the tip isn’t inside the droplet for if it was there would be no (discontinous) barrier towards decreasing the surface tension by simply sliding while with a blunt tip releasing to form a spherical droplet requires a discontinuous increase in surface area and I suppose the probability of it sticking at all would go down.

(Details on the formation of these droplets by the way is by means of dipping it pretty deep and have an originally even coating slide down and accumulate at the tip.)

Nevertheless I can’t quite make out the details of what the origin of the force that is fulling the droplet upwards is coming from, whether it is driven principally by surface tension or adhesion. Because it the needle is shaken a bit and the droplet pushed down a fraction of a millimeter it will climb back up.

The 7 Tetris block circumference problem

Fancified name in the title aside I was checking out some worksheets for geogebra yesterday and came across one which had predefined tetris block elements. I played around building different shapes, originally wondering if you could make a periodic pattern out of an even ditribution of tetris blocks but I eventually landed on a different question which will be the topic of this prost.

So you start with the 7 basic tetris blocks as featured in the image below.sevenbasictetrisSo these pieces you can stick together into chunks in an amazing number of ways and each new figure naturally has some circumference. Assuming all the pieces fit rogether in one chunk the maximal circumference such a figure can have is 56 (think about it) but the question quickly becomes: What is the least circumference such a composite piece can have?

Here are some of the initial trial and error solutions which I played around with

Tetriscompositesnominimalcircumferencebut there nevertheless is a packing with even smaller circumference and it is proving the existence and minimality of this packing that will be the theme of this post.

Continue reading

The connection between diameter and pi

So I was having the worst headache and lying in bed thinking about simple geometry problems waiting for the pain killer to kick in. One of the things I did was compute the perimeter/circumference of the Reuleaux triangle. Okay so that one’s parameter is composed of three sixths of a circle with radius twice the widthof the triangle so if the width of the triangle is D that’s O = 3 \cdot 2\pi D / 6 = \pi D. Similarly for all the other Reuleaux polygons the (n = 2k+1)-sided reuleaux polygon O = n \cdot 2\pi/(2n) D = \pi D. Huh… I thought. That’s kind of odd. All of these figures with constant widths have the same circumference formulas as the circle which of course has constant width called the diameter.

And looking up the wiki-article on shapes of constant widths and this is apparently a thing proven in 1860. All curves with constant width have a perimeters which is \pi times it’s width. and it’s called Barbier’s theorem.

I also get the feeling that this should be possible to be reworked into a pro-pi argument in the tau-vs-pi debate. Of course by the law of the equivalence between pi and tau implies it could be made into a pro-tau-argument as well but since I’ve never seen it come up in one of those debates I would like to see what happens when a tauist brings up that defining the circle constant via the diameter of a circle was a mistake since the radius is a more fundamental property and then have the piist counter that at least there is a whole class of objects with diameters and they all have perimeters which are \pi times the diameter : P

Folding polyhedra and a novel connection between folding a tetrahedron and a 2D result.

So I’ve been folding a lot of polyhedrons out of paper this past week. Like a lot… But since the types of interesting polyhedrons I have left to built take a bit too long to construct adn I already wasted enough time folding the (small) stellated dodecahedron.


Instead I’m rather reviewing the ‘legitimacy’ of the folding schemes and figure out what relations actually govern weather something fits together or not. The prototypical example is the tetrahedron, the polyhedron with the fewest number of faces. The explicit construction of the regular tetrahedron (all sides equilateral triangles) is easy to show directly and is found in the elements and elsewhere but you can fold generic tetrahedrons by just making 4 triangles such that every triangles has common sides with the other triangles in a natural way. Made from one piece they are then flded together to form a possibly irregular tetrahedron. foldingapyramid

Empirically and intuitively this basically defines natural, a result in no need of further explanation. Nevertheless mathematically it almost seems coincidental. Just because two face can be glued pairwise doesn’t immediately mean all the faces should fit together collectively even though it’s about as natural a thing to postulate as the equality of the two base angles in an isosceles triangle.

The novel thing is that you can turn this 3 dimensional problem into a 2 dimensional one and use the vast machinery available to you in that space. You can note that folding an individual side is equivalent to rotating it about it’s common side with the bottom triangle where the it’s outer corner traces out a circle lying in a plane parallel to the altitude of the triangle and perpendicular to the common side.

rotatingtrianglesinpyramidIf the three faces are to come together their corresponding circles must intersect at a single point which means 1. the planes of rotation intersect along a common line (which is perpendicular to the plane and forms the height of the tetrahedron and 2. with the altitudes of the faces and satisfying relations to guarantee a consistent height.

Especially 1. is interesting because it forms a more ‘physical explanation’ for why dropping and extending altitudes in the original net intersect at a common point which can be taken as a purely classical 2 dimensional geometric result.

pyramidtriangletheoremThe thing I find neat is that this result seems kind of arbitrary in 2D but as part as a 3-dimensional picture it makes perfect sense.

I’ll have to see what’s the best way of extending this to folding schemes with a general polygonal base.

The cuboid volume formula and the dissection of a cuboid into a cube

So at the end of the last post I gave the outline of the idea of how you might extend the dissection of a rectangle into a square and the definition of area to an analogous procedure for the cuboid and it’s volume. The idea is really simple but I thought I’d practice my diagram writing skill by providing an illustration and expand the definition of volume.cuboiddissection0

Okay so we say that we have a cuboid with dimensions a \times b \times c where we conventionally say that it has volume abc meaning it has the same volume as a cube with side \sqrt[3]{abc}. Let us use dissection to motivate this; that is describe a procedure of cutting up the cuboid and reassembling it into the prescribed cube.

We will essentially perform regular rectangle to square dissections face-wise splitting the cuboid into prism.

Let us start with the a \times b face and dissect it into a rectangle having one side \sqrt[3]{abc} and the other (by the area formula) being ab / \sqrt[3]{abc}.Applying this to the whole cube the cuts through the face are extended through the solid along planes perpendicular to the face forming prism which are rearranged to form a new cuboid with \sqrt[3]{abc} as one of it’s side.

Note the depth c of the cuboid is unaffected by this procedure.


(You might need more pieces than the image indicates)

Next we turn out attention to the face with sides ab/\sqrt[3]{abc} and c and perform the same procedure where this face is transformed into a rectangle with one side being \sqrt[3]{abc} by construction and the other side

\cfrac{\text{Area of face}}{\sqrt[3]{abc}}= \cfrac{abc / \sqrt[3]{abc}}{\sqrt[3]{abc}}=\sqrt[3]{abc}

And so this face is square! All the sides of this cuboid are therefore equal and we’ve ended up with a cube.

*This procedure is not original but my interpretation hasn’t been checked so some caution should be taken

Sidenote: Just as triangles and polygons can be dissected to rectangles a prism can therefore be disected into a cube as well and paying attention to the details of such a procedure you recover the conventional Bh formula for general prism. For the cylinder we don’t have any elementary means of dissecting into a cube but just as you may approximate a cirvle by triangles you can approximate the cylinder by triangular prisms and obtain the standard formula.

Dissection of a rectangle into a square and a definition of area

This post is simply a reworking of my previous post on this subject as I realized I had missed a special case which is accounted for in a single line but which nevertheless had gone unmentioned. The problem was to start with a rectangle of dimensions a \times b, naturally having area ab and cutting it apart into pieces which are then rearranged to form a square with the same area.

This procedure involves abusing the knowledge that the resulting square will have as the lengths of its sides \sqrt{ab} where you start by cutting a ‘pseudo-diagonal’ having this length across the rectangle, having sliced of a triangular corner. Displacing this corner to the other side of the rectangle you obtain a parallelogram with \sqrt{ab} as its base and implicitly also \sqrt{ab} as its height by virtue of the known area. This parallellogram is then cut along on of it’s altitudes normal to the side with length \sqrt{ab} and rearranging this pieve the resulting shape is a square. Diagrammatically the procedure is shown below.

BisectingFatRectangleNote 1: There were two things I wanted to expand on regarding this procedure. Firstly that it needs to be supplied an additional step if the rectangle is too wide. More precisely if the base is more than twice the height. 2a < b If that is the case an additional cut will be required. We may first note the limiting case 2a = b where the second cut now  intersects with one of the corners


and if the width is increased beyond this limit our desired seond piece is only partially contained in the actual paralellogram. You see pretty much immediately that all you need is to move this piece and then slice of a tip segment and move that but it’s nevertheless an additional step which should have been noted upon.


Note 2: The second thing I wished to expand upon was simply argument for the fact that this actually forms a square. Knowing all the formulas for area it is pretty easy to fill in all the blanks such as that parallelogram with area ab and and base b must have height $a$ or if a rectangle has area A and base $b$ then its height would be \sqrt{A} / b and so on.

However it would be much neater to take this rearrangement as the proof that the area of a recangle with sides $a\times b$ has area ab. To outline this proof I will make use of this particular definition of area:

If a geometrical figure can be cut into a finite number of smaller pieces and rearranged into the shape of a square with side \sqrt{A} we say that the figure has area A

This is not a complete definition of area as what constitutes a ‘piece’ and ‘rearranged’ is unclear and not all figures are necessarily rearrangeable and like all casual definitions it might invite paradoxes but it allows us to form a class of figures which we say have equal area and make more practical sense of that means instead of just memorizing formulas. For example to say an A4 paper has an area of 623.7 cm^2 now means it can be cut and rearranged into a square with a $\sqrt{623.6} = 27.97$ cm side.

The part of the dissection which I would like to prove without references to area that the resulting shape really is a square. With reference to the diagram below we say we have a parallelogram height a and base b so such that it’s slanted side has length sqrt{ab} and wish to show that the altitude normal to DC goign through B, BE, also is \sqrt{ab}


This is most easily proven using the similarity of the two triangles AEB and EDB which follows from the angles A and C in a paralellogram being equal. From this we get

BE / b = a/\sqrt{ab}

and thus

BE = \sqrt{ab}

Therefore it is truly a square that the rectangle is being rearranged into and just to be needlessly pedatic lets once again note that all this meant that a rectangle with dimensions a \times b was possible to rearrange into a square with side \sqrt{ab} meaning the area of the rectangle really is ab!

Sidenote: What jogged my memory on was that I’ve been thumbing through this Peter R: Cromwells rather charming book Polyhedra where the problem of how you might define volume partially (but ultimately not exclusively) though dissection sprang to mind. On page 45 I also recovered another outline of how to dissect a general polygon to a square though with a different final step than my own and for whatever reason omitting the step outline above. In principle you should be able to perform this dissection ‘face wise’ to prove the volume formula of a prism (or rather the a\times b \times c cuboid) but I’ve yet to complete the details of the proof.