Welcome to another Mathologer video.
Today's video is one I've been dreaming about making for a long long time. Today
I'd like to dazzle you with the solutions of some of the most famous
problems in the history of mathematics. These problems had remained unsolved for
more than 2,000 years after they were first puzzled over in ancient Greece. The
problems arose very naturally as part of the Greeks mathematicians quest to
determine the possible geometric constructions when all were permitted is
to draw lines and circles. That is what is possible using just the most basic
mathematical tools, the ruler and the compass. For example, given a square, we
could construct a new square of twice the area just using our basic tools.
However, unlike doubling a square like this, doubling a cube turned out to be
anything but easy. Nobody was able to figure out whether,
given a cube, we could construct a cube of double the volume by just using ruler
and compass. This is the first of those ancient problems I'll be tackling today.
Similarly, if you are given two lines making some angle, then halving that angle with ruler and compass is easy. Let us draw a couple of circles there and there
is the bisector, but nobody could come up with a ruler and compass construction
that would trisect an arbitrary angle. That's problem number two. Next, using
ruler and compass, to construct equilateral triangles, squares, regular
pentagon's and regular hexagons is not a problem at all. But, what about regular
heptagons? Nobody had a clue for thousands of years. That's problem number
three. Finally, most famously, and what turned out to be by far the hardest; Is
it possible to square a circle? That is given a circle, is it possible to use
ruler and compass to construct a square of exactly the same area? Well, as I said
it took over two thousand years to finally answer these questions. And
what's the answer? The answer is: Stop trying, you're wasting your time!
In the 19th century it was proved that it's impossible to use ruler and compass
to double a cube, or to trisect an arbitrary angle, or to construct a
regular heptagon, or to square circle. But who gets to see this impossibility
proofs? Not many people, except a few pure maths majors who may encounter them
immersed in a course on Galois Theory. Really heavy-duty stuff :) However, on close
inspection it turns out that those proofs only really require some of the
semi heavy-duty parts of Galois theory. Then if you are incredibly stubborn and
you try really, really hard, it's possibly possible to distill the essence of these
proofs into one short and not too hard YouTube video. I don't know about you, but
I find the idea of making a video like this super exciting. It's an opportunity
to make a small but tangible contribution to something that people
have been struggling with for thousands of years. And I hope that for you it's an
opportunity to get to the core of some classic and beautiful and hard
mathematics, to get some insight into a theory normally considered out of reach
of mere mortals. Okay, here's the plan for today. I'll first tell you exactly what
it means to construct things with ruler and compass, the precise rules of the
game. Then, I'll go very carefully through the proof that doubling the cube is
impossible. Our argument that doubling cube is impossible will be a proof by
contradiction and will run like this: If doubling the cube was possible, then we
can use that to show that the cube root of 2, a number closely related to our
problem could be written as an expression that only involves rational
numbers and square roots. However, as we'll also show, this would
imply that cube root of 2 is in fact equal to a rational number.
Since this is definitely not the case, doubling the cube cannot be possible
either. After the hard work of doubling the cube,
I'll then tidy up, by sketching how this wonderful proof by contradiction can be
modified to also take care of trisecting angles and constructing regular
heptagons. And, finally, squaring the circle can also be ticked off easily
because we already did most of the heavy lifting for the proof in a previous
video on the transcendence of pi. As you may already have guessed, this video is a
challenging one. So, as for my previous masterclass videos,
this one is broken into levels of enlightenment, six in all and each with
its own mathematical superhero guide. Of course, feel free to start skipping if it
gets too tricky, and let me know how far you make it in the comments. Okay, without
further ado, mathematical seatbelt on and on to level 1. This video is all about the things that
a ruler and compass cannot do. But first let's ask what can they do?
What can we construct this ruler and compass? To start, let me begin drawing to
just give you a feel for what's going on. Okay? let's start with two points in the
plane. Now draw the line through them. Okay,
draw a circle with this red point as it's center through the other point. So, now
we have a third point, the other spot where the circle and line intersect.
draw a circle with this red point as center through this new green point. That
gives us three new points of intersection, and our construction is
already getting pretty and also pretty interesting. Now, draw a line connecting
these two red points. Another intersection. Clearly this point is the
midpoint between our two starting points. And, our second line is clearly the
perpendicular bisector of our starting points. Plus, these three red points there
are the corners of an equilateral triangle.
Pretty impressive, by just drawing a couple of lines and circles you get all
this good stuff. What else is there? Let's draw another, smaller, circle, like so, okay.
Two more points of intersection which, together with the original two points,
from the corners of a square. Pretty easy right? A regular hexagon is also hiding
just around the corner. Let's go and find that one too. All child's play. But what about regular
pentagon's? That's not so obvious. Let me show you how you can do this, starting
with this part of the picture. There just that one. First, we locate the point
halfways between these two points. Now you all remember how to do that, right? So,
I'll skip the details. Nothing very five-ish yet. But just watch. Pretty cool right? But maybe I cheated? Is
that really a perfect regular pentagon? It's definitely is not obvious, but yep, it
is. And, your first real challenge today is
to work through the details and nail down the proof. Okay, so having played
with ruler and compass a bit we are ready to declare the rules of our game.
Here we go. At any stage of the game, we have before us a collection of points,
lines and circles. Right? Then there are two ways we can proceed. First way: choose
any two points and draw the line through them. As well as adding the new line to
our collection, it also adds all the intersection points of this new line
with the previous lines and circles. So there you go. All of them go in and there's
a second way and the second way is to again start by picking any two points.
And then you draw a circle centered at one of the points and passing through
the second point. The new circle is then added to the collection, as are all the
new intersection points. So, those are the rules of our game. Pretty easy right? Now,
what about those special problems, doubling the cube, and so forth? Okay, you have a cube and your aim is to
double its volume using just ruler and compass. What this means in our game is
that you start with two points the side length of our cube apart and your aim is
to construct, in a finite number of ruler and compass steps, two points that are
the side lengths of the doubled cube apart. For the purposes of our game we
can assume that our original cube has side length one, and so also volume of one, and
the volume of the double cube would have to be 2 of course, right? And so it's side
lengths would have to be the cube root of 2. Got it?
Starting with two points a distance 1 apart, our aim is to construct two points
that are a distant cube root of 2 apart. Similarly, a circle of radius 1 has area
pi. So constructing a square of the same area means the square should have side
lengths of root pi. As I already showed you, constructing squares is not a
problem. So, in our game "squaring the circle"'
amounts to a beginning with two point a distance 1 apart, and using them to
construct two new points a distance root pi apart. In exactly the same way, the
other two problems I showed you at the beginning boil down two starting again
with two points a distance 1 apart and, using these points, to construct two
specific new lengths. In fact, let's go all Cartesian and pin down things a little
bit further. Let's make our two beginning points the origin and (1,0) the point one unit to the
right along the x-axis. Now there's a little ruler and compass magic that I
won't go into, but which helps simplify our problems. It turns out that if we can
construct a segment of a certain length somewhere, then we can always translate
and rotate using ruler and compass to obtain a segment of the same length
starting at the origin and lying along the positive x-axis. That means our cube
doubling problem boils down to using a ruler and compass to locate the point
cube root of two along the x-axis. Similarly, squaring the circle boils down
to locating root pi on the x-axis. And there are two more mysterious numbers
which arise from our other two problems. Those two guys. Of course, faced with the
task of constructing numbers such as the cube root of 2 and root pi,
it's natural to just stare helplessly into space. Which is pretty much all that
happened for 2000 years. So let's look at it another way.
Let's ask what numbers can we construct? Well, obviously, all the integers can be
constructed like this. Then, as we've seen, we can construct midpoints. These are the
integer multiples of 1/2. There are more midpoints here. This gives all the
integer multiples of 1/4, and so on. What else?
Lots! It turns out that once you've constructed two numbers A and B, you can
also construct their sum, their difference, their product and a quotient by
executing four easy constructions. For example, to construct A+B, you just
need to transfer a distance which, remember, is no problem with ruler and compass. So there we go, just push it over and that's A+B there. And B-A is just as easy. Okay let's have a look. Let's push this over. That's B-A. Next, to construct A times B, we also need to be able to construct
perpendiculars, which we've already seen and to draw parallel lines, which is also
straightforward with ruler and compass. Ignoring the fiddly details, here is the
key idea of this product construction. Pretty nice, huh? And, constructing A/B
is similarly easy. Faantaaaastic!! Okay so we can add,
subtract, multiply and divide... Karl was here >:) ... constructible numbers and, with the
integers already constructed, this means that all quotients of integers are
constructible. In other words, we can construct all rational numbers with
ruler and compass. Anything else? Well, yes, we can definitely construct some
irrational numbers. For example, root 2 is the diagonal of a unit square which is
easy to construct. In fact, with a little Pythagorassing, it's easy to construct
the square root of any positive integer. But there's more. I'll now show you how,
taking any positive number we've already constructed, we can always
construct that number's square root. Suppose we've already constructed the
number A down there. First add 1 to A to get the green number here, using ruler and
compass of course. Now half the green number, draw this circle, draw a
perpendicular through the blue point. Then the pink segment has length root A.
A pretty easy recipe to follow. I'll leave it as homework for you and your friend
Pythagoras to check out that this actually works. But, just for fun, let's
calculate root 2 this way. That should be root 2 but root 2 is also the length of
the diagonal of a unit square. There's the square, well let's check. Yep that
looks about right which means it must be true. Well of course not but as you will
show in your homework it is right. RIGHT? RIGHT, you will show this! So, it turns out
we can construct a lot of numbers. Starting with the numbers 0 and 1, we can
add, subtract, multiply, divide and square root lengths any finite number of times.
And here then are just a few examples of numbers we can construct. That's a ton of
constructible numbers, but are there even more? What do you think? The answer to
this question is... NO! These nested square rooty creature's
based upon the integers are the only kind of numbers we can construct with ruler and compass and it's actually very easy to see that this is the case. Suppose you
have two points with square rooty coordinates. Then the equation of the line
through those points will have square rooty coefficients. Pretty
obvious that they are square rooty right? And, a circle with square rooty center
and passing through a square rooty point will have a square rooty radius, and so
will also have an equation with square rooty coefficients. There that's the
equation. (as well as Descartes). :) And then how do we get new points? By
intersecting lines with lines, lines with circles and circles with circles. That
corresponds to solving pairs of linear and quadratic equations. And the
solutions of those systems of equations are of course all square rooty again.
And so, to summarise. Starting with our two numbers 0 and 1 on the x-axis, we can use
ruler and compass to produce any square rooty coordinates we want, but only square
rooty coordinates. Nothing else. So you're back for more. (laugh) Very
good!! So, now we know exactly what type of of numbers we can construct, those square
rooty monsters. But what about the cube root of 2? Can that number be written as
a mess of square roots? It seems unlikely but how do we go about proving it?
Well, some of those square rooty monsters are more monstrous than others, right? So,
let's start easy consider some of the less scary monsters first and work our
way up. Hopefully, we'll detect a pattern which
will suggest a plan of attack. So, what are the very simplest square rooty
numbers? Well, there would be the fractions, with no roots at all. So, let's
first ask whether the cube root of 2 can be written as a fraction. That is, our
base level question is `Is cube root of 2 a rational number? And, as all regular
Mathologerers would know the answer is a big NO. The cube root of 2 is not a
rational number. I recently did a whole video about how you can quickly prove
the irrationality of this and a bunch of other rooty numbers using the amazing
rational root theorem. So, let's take the irrationality of the cube root of 2 as
given. We'll also take it as known that squared of 2, the square root of 3, the square
root of 5, and so on, are all irrational. Okay? Not a big deal. Okay, now, as a warm-up
exercise, let's think about all square rooty numbers of this form there, where
both a and B are rational numbers. There's absolutely nothing special about
7 just that it's a rational number whose
square root is irrational. We could have used lots of other numbers there. Now,
what we'd like to prove is that the cube root of two cannot be written as one of
those root seveny numbers over there. That will be progress, right?
Little, tiny, special progress but progress. Now, if you're worried that this
tiny progress will be too tiny don't. Later there will be a humungous
plot revelation that will more than compensate for the incy wincyness here.
Anyway, to prove our special case, we set up the usual proof by contradiction. So,
we start by assuming the cube root of 2 is equal to one of these special
numbers. In other words, we are assuming that our number is the solution of this
equation here. Cubing the cube root of 2 gives 2, right? In a moment we'll show
that the assumption that a plus b root 7 solves the equation implies that
the conjugate of this number a minus b root 7 does so as well.
However, ignoring complex number possibilities, this equation has one and
just one solution, the cube root of 2 which means that that our plus solution
and our minus solution must be the same and b must therefore be 0. But that would
mean that the root 7 stuff is irrelevant and that the rational number a itself would
solve our equation. In other words, the conclusion at the end of this chain of
consequences is that cube root of 2 would have to be rational which, as you agreed,
YOU AGREED (pointing finger) is nonsense. And therefore we've arrived at the
contradiction you're chasing. So, again, the assumption that cube root of 2 is of
the form a plus b root 7 implies a contradiction which proves that the cube
root of 2 is not of this form. But, of course, to make the proof complete, I
still have to fill in that crucial step of the argument. I have to show that if
a plus b root 7 solves this equation, then a minus b root 7 does so too. Okay,
you're ready? Then here we go. So when in doubt
calculate, right? So moving the 2 to the left, plugging our route 7 thing in and carefully expanding we get, let's see, ... all
of this mess. Okay, but since we assumed a and b are rational numbers, the blue and
the yellow numbers must be rational, right? But then much more than being
rational, the yellow number must be exactly 0.
Why? Because if it weren't, we could solve for root 7 like this. That would mean
root 7 is a quotient of rational numbers and so would also be rational. We
already know that this is not the case, so the yellow number is definitely equal
to 0. But with the whole left side being equal to zero that means that the
blue number has to be equal to zero, too. And now everything's clear, right? Well
maybe not but we're basically done. The key observation is that the blue
expression only contains an even power of the number b and both terms in the
yellow part have b to an odd power. So what is the trick? Let's replace b with
minus b. What changes? Well, the blue number is left unchanged since the one
minus sign gets squared away. And because of the odd powers the yellow becomes
negative, like this. So there, the minuses goes out, that's it. So, the
overall effect of replacing b by minus b in this left side of our equation was
simply to turn the plus sign in the middle into a minus. Overall that's the only
change, right? But since both blue and yellow are zero the overall expression
with the minus is still equal to 0. How does that help?
Remember the original blue and yellow expression was obtained by plugging our
original root seveny number into x^3-2. But that means we get the new
blue and yellow expression by replacing b in our number by minus b and then
again plugging this new number into x^3-2. But since the new blue and
yellow expression is also equal to 0 this means that our new number is also a
solution of our cubic equation. Tada!!!! Well maybe you'll have to go through this
stuff again but that's basically it. Once again, the assumption that our original
number a plus b root 7 solves x cubed minus 2 is equal to 0 implies that it's
conjugate, a minus b root 7 solves this equation, as well. And this then
implies the contradiction and we conclude that cube root of 2 cannot possibly
be of the form a plus b root 7. Great!!! So you're still here?! Fantastic!!! Now I
promised you a big plot revelation. We're getting there.
It turns out that we can iterate our a plus b root 7 proof by contradiction
to conclude that the cube root of 2 is not equal to any square rooty number.
To make the iteration work we need to isolate the essential ingredients of our
proof. It turns out that these ingredients all have to do with the rational numbers
which served as a foundation for what we're doing. There are four ingredients.
First, we know that cube root of 2 is an irrational number.
Second, we needed the property that adding, subtracting, multiplying, and
dividing of rational numbers always gives rational numbers. Right? we need
this to be able to conclude that the blue and yellow numbers being made up of
rational numbers are rational themselves. We also needed this ingredient when we argued that the
ratio of two rational numbers is rational. The second ingredient the
property of being "closed" under the four arithmetic operations, is described by
saying that the set of rational numbers is a subfield of the field of all real
numbers. It tells us that the rationals from a self-contained universe of
numbers. Okay, the third important ingredient was the fact that 7,
inside the square root is a rational number. Right? We needed 7 to be
rational because at various spots we use that squaring the square root brought
us back to a rational number. For example, when we expanded right at the beginning.
Let's do it again! So there expand, expand, expand. That 7 there is the result
of squaring the square root and we need that to be rational.
The last ingredient of our proof was the fact that although 7 is a rational
number, it's square root is not. We use this in two spots. Okay, to summarise, our
magic ingredients are: the cube root of two is not a rational number, rational
numbers form a subfield, 7 is rational, square of 7 is not.
And from all that follows that cube root of 2 is not one
of the red numbers. Okay, now it turns out that the red numbers themselves are also
a subfield, that is, if you add, subtract, multiply or divide two root seveny numbers,
then you again get a root seveny number, there like this. All straightforward except for the division which involves a conjugate trick that some of
you may remember from school. And you can worry about division by zero, if you're
the worrying type. :D Challenge for you is to fill in the details in the comments.
Ok, ingredients again: cube root of 2 is not a rational number, rational numbers
are a subfield, 7 is rational, square root of 7 is not. From this it follows that cube root
of 2 is not one of the red numbers. Now the red numbers themselves are subfield,
and now things are supposed to repeat. So what comes next?
Well something like this! So what we need is a number, that is red, but whose square
root is not. Okay and I'll just give you one okay? So 1 plus root 7 is such a
number and root 1 plus root 7 is not a red number. And now with exactly the same
arguments as before we can conclude that, all these blue numbers are not
candidates for anything that can be equal to cube root 2. And now things
repeat again because the blue numbers can also be seen to be a subfield, with
exactly the same argument as before. And now? Well, repeat so we need a line here
so we need a number that is blue but whose square is not. And there are lots of
possibilities so let me just give you one again. 86 divided by 5 is such a number and now
with exactly the same arguments as before it follows that cube root of 2 is not
one of the pink numbers and we can go on like this. At this stage, perhaps you're feeling a
little "rooted", as we say in Australia. :D Time to catch our breath and take stock.
The field property of the subworlds of numbers that we've been looking at shows
that the final pink numbers are exactly the numbers that can be constructed by
combining the rational numbers and these three rooty expressions with the usual
arithmetic: adding, subtracting, multiplying and dividing.
For example, this complicated rooty number is one of the pink numbers. Same
for this one. And this one. And, because they are all pink numbers none of them
can equal cube root of 2. Of course, any individual pink number can be ruled out
using a calculator. But the point is that we've ruled out ALL the pink numbers of
that subfield. And, we can go to bigger and bigger subfields. Given any square
rooty number, we can start from the rational numbers and by iterating the
extension process, we can construct, in a finite number of steps, a subfield that
contains that particular square rooty number. And, so that square rooty number,
ANY square rooty number, cannot be cube root of 2. And that's it. That completes
the proof that cube root of 2 is not a square rooty number, and therefore that
the cube cannot be doubled with ruler and compass. Very hard work, but also very
very nice. Don't you think? Now before going to the next level,
there's just a little tidying up to do. Remember that we knew or we assumed or you
simply trusted me that root 1 plus root 7 is not part of the a plus b root 7 subfield and that route 86 divided by 5 is not part of the last subfield, and so on.
but even if root 1 plus root 7 was part But even if 1 plus root 7 was part of the a plus b root 7 subfield that
would have been a problem. Why? Because that supposedly new subfield would be
just the same as the previous subfield. So there would
just be no new numbers to worry about, that would just mean that to get to this
point we would need one less step of the extension process. It's the same for all
the square roots that go into building our subfield extensions. If the subfield
is enlarged, we know how to handle it and if not then nothing is changed. All that
matters is that whenever our subfield is enlarged we are guaranteed that the new
subfield still cannot contain cube root of 2. And this really finishes the proof
that no square rooty number is equal to the cube root of 2. In turn this shows
that it is impossible to double a cube just using ruler and compass. What an ingenious argument, don't you
think? But there were four problems that I promised to solve in this video. One
down after quite a fight and still three to go. How long a video is this going to be?
Well, not that much longer. It turns out that proving the impossibility of
trisecting angles and constructing a regular heptagon with ruler and compass
can be taken care of in a very similar fashion. Just like doubling the cube
boiled down to showing that the solution of the cubic equation x^3-2=0 cannot be a square rooty expression, showing the impossibility of
trisecting and heptagoning reduces to showing that the roots of two other
cubic equations cannot be square rooty expressions. First trisecting angles. Of
course, some angles can be trisected. The angle 90 degrees, for example, is easy to
trisect. But the point is that not all angles can be trisected and to show this
we just have to prove that one particular angle cannot be trisected
and we've chosen 60 degrees as our victim. Here's a quick run-through of how you show that ruler and compass cannot be
used for constructing 20 degree angles. That then proves the impossibility of
trisecting the constructible angle 60 degrees. And, you guessed it, it will be
a proof by a contradiction. So, let's assume that starting with our two points
it was possible to construct a 20 degree angle. Then we can transfer the angle to
the origin like this and then constructing this perpendicular also
gives us the cosine of 20 degrees. Time to dust off your high school trig.
Remember your double angle formulas? Well there are also triple angle formulas and
the formula for the cosine is this. Substituting 20 degrees for theta we get
this. The cosine of 60 degrees is one-half and this means that the cosine
of 20 degrees is a solution of this cubic equation here. Now we can use the rational root theorem
from two videos ago to prove in the blink of an eye that this equation has
no rational roots. And because we are again dealing with a cubic equation things
proceed very much along the lines of the cube root of 2 proof. In particular,
assuming that a solution of this equation is contained in a square rooty
extension field forces the conclusion that there is also a solution in the
smaller subfield and this then gives the usual contradiction. Challenge for you:
fill in the details. One slight difference you have to deal with: the
cubic for the trisection has three real solutions rather than just one. For those
who get lost or are just plain exhausted I can understand. I'll also provide some
links to a write-up. And on to heptagons. If it was possible to
construct a regular heptagon with ruler and compass, then starting with the usual
two points it would be possible to transfer the heptagon here. This would mean that
it was possibly to construct the number cosine 360/7 degrees.
There that creature. And then, it's fiddly but it can be shown, that this cosine is
a solution to this cubic equation there. And it's straight sailing from there.
Again I'd say first try to fill in the details yourself in the comments and if
you get desperate follow the links in the description. Finally, finally, finally what about
squaring a circle and what about using ruler and compass the the number
root Pi. Well if root Pi was constructible, then
it's square the number Pi would also be constructible. However, it turns out that
all square rooty numbers are algebraic, that is, all square rooty numbers and
in fact all rooty numbers are solutions of polynomial equations with
integer coefficients. But Pi is not as we showed you in a previous masterclass
Mathologer video. Pi is a transcendental number, it is not a solution of such a
polynomial equation. This proves that Pi cannot be
constructed and that's all there is to it. But to nail down the last little bit
of the proof, how does one prove that at all square rooty numbers are algebraic. It
turns out to be pretty easy. The idea is to start with a square rooty expression,
set it equal to x and then unravel the resulting equation, successively
isolating and squaring away the square roots. Eventually all that is left are
integers and powers of x. To give you some intuition I'll finish this video
with an animation of constructing such an equation for one of our previous
scary square rooty expressions. But before the pretty ending, first let me finish
with a few thoughts on impossibilities. Today's video was a tricky tour through
some difficult mathematics. It easily took me 200 hours to put it together and
still, even after me and Marty there behind the camera have agonized and
reagonzed over every detail of the video, it's tough going and I don't expect
that everybody who watches this will get everything in a first viewing. Don't feel
bad if you don't, just give it another go and remember it took mathematicians a
couple thousand years to sort out these ideas and it took that long for a reason.
And, so, if it takes you a few viewings and maybe a question or two in the
comments that's perfectly
fine. But there's one more thing. Every year I get at least a few people writing
to me with the news that after devoting anything from a few seconds to half a
lifetime of study they have managed to square the circle or have achieved some
other impossibility. Of course, they are all wrong. It's important to realize that
if you change the rules of our game just a tiny, tiny little bit, it's then no
problem at all to square the circle, and so on. And this has been known since the
days of Euclid. The plan is actually to make another video about this
fascinating topic of rule changing in the near future.
Let's see. Anyway, it's very easy to misinterpret the rules of the Greek game
and end up with false solutions of those ancient problems, especially if you base
your understanding on a bare-bones exposition of the rules such as the one
I have given in this video. Just like those people who write to me, there have
been many thousands of people who've made this mistake over the years and
whole books have been written about their pointless endeavours. To make
absolutely sure that you really, truly understand the rules and most definitely
before you embark on a pointless quest of squaring the circle, I recommend you
very carefully study those rules and the common mistakes people make interpreting
them. The wiki page on ruler and compass constructions is a great starting point.
Ok enough of that. To finish off in a much prettier style here's the
construction of a polynomial equation with integer coefficients that has this
square rooty expression as a solution. Ok, let's call this guy, oh I don't know, how
about x. Now you can watch the polynomial materialize. And that's all for today.
IIRC, it is because of pi being transcendental and doubling cubes requiring you to solve a cubic. Fun fact: Origami can solve cubics, so you could in theory double a cube if you are allowed to fold the paper, in addition to using the classical construction tools.
Tangentially related but it reminded me of this game where you have to construct random figures, split angles etc with a circle and a compass - euclidea is the name, it was pretty fun
Loved this video! Learned this in Abstract Algebra 2 in college and thought it was really challenging but you explained it very well!
Covering the doubling cube and, within the same theme, trisecting angle problems in my Galois Theory class remains one of my favourite moments in undergraduate mathematics.
Someone tell /r/SquaredCircle
No one is talking about that shirt?
(pi x r2 ) /l = w
If the square have 1 side length The circle encompassing the square have the diameter of the hypotenuse of the square.
(pi x 0.5 ) / 1 = pi x 0.5 = w
Area of the circle = area of the rectangle
pi x 0.5 = 1 x (pi x 0.5)
area of circle = area of rectangle
2pi
2pi and pi width of rect
2pi = square root of ?
2pi