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/sci/ - Science & Math


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File: 27 KB, 873x604, fitsquare.png [View same] [iqdb] [saucenao] [google]
4948013 No.4948013 [Reply] [Original]

I saw this problem on tv the other day and my IQ was too low to solve it. I wonder if /sci/ can solve it.

>> No.4948032

Fuck man, I have an IQ for 157 and I can't do it. At least I tried.

>> No.4948038

Not possible, something about sqrt(2) being an irrational number.

>> No.4948048

>>4948038
lol all it is asking is to make the rectangle into a square.

>> No.4948051

>>4948038
wrong

I would first cut the 8 by 1 into 8 1 by 1 squares. Then I would cut each square in half diagonally giving me a bunch of triangles with hypotenuse being sqrt(2). After that it's just moving them, it will fit perfectly...

>> No.4948056
File: 46 KB, 1604x883, 1344407386645.png [View same] [iqdb] [saucenao] [google]
4948056

>> No.4948058

It's possible if you cut the rectangle into a lot of small squares and rectangles but this question asks you to separate it into 4 pieces.

>> No.4948059

>>4948058
I meant triangles instead of rectangles

>> No.4948062

>>4948051
you are only allowed 4 peices.

>> No.4948063

>>4948051
>>4948058
Yup, that was my impression. The main issue is that it's simply not possible to do this if you're only allowed to cut it into 4 pieces...

>> No.4948066

>>4948051
that's more than 4 pieces. This isn't a hard question if you aren't limited.

>> No.4948069

>>4948051
you have 4 cuts it says pic related solved in 3

>> No.4948073
File: 75 KB, 2397x1544, solved.png [View same] [iqdb] [saucenao] [google]
4948073

>>4948069
oops

>> No.4948076

It is impossible if you stick to what it says and only try to use four pieces.

Easiest way would be to chop it in half (into 2 1x4pieces) and lay them on either side of the diagonal. The ends that stick out form two triangles that will then fit into each corner
This only takes 6 pieces

>> No.4948079

The cool thing about this question is it only allows 4 pieces but doesn't limit the cuts. It also doesn't say anything about reforming the rectangle. As 'tricky' language is often the root of these problems I've no problem using literal language to ignore it. Therefore, I cut it into infinitely small squares of delta_x that I reform into for squares of root 2 and place them the form of the larger square.

>> No.4948081
File: 5 KB, 382x399, hint.png [View same] [iqdb] [saucenao] [google]
4948081

OP here, my picture isn't drawn to scale so don't try to cut and assemble it. Here is a hint on one of the pieces if you're having trouble.

>> No.4948085

>>4948073

As per the thing you're limited to four pieces, not four cuts.

>> No.4948087

It might be easier to figure out how to turn the square into the rectangle instead...

>> No.4948089

hmm allow me to try

so the sides of the square of 4/sqrt(2)...

so I ideally I need 4 pieces of [4/sqrt(2)]/4 = 1/sqrt(2)

So lets fold the rectangle to the point where the width 1 becomes 1/sqrt(2).

Now we fold until the length becomes 4/sqrt(2)

and now we cut the folds, which should give us 4 pieces.

Doesn't that work?

>> No.4948095

Same guy as >>4948089

So I made 2 folds, that should give me 4 pieces... I think I've won?

>> No.4948096

I am probably very dumb,

but how would you turn a figure of area 8 into a figure of area 2 using only pieces of said figure?

>> No.4948101

>>4948096
oh wow

>> No.4948104

>>4948096

The sides of the square are 2 root 2. twice 2 root 2 is 8, they're the same area.

>> No.4948113
File: 31 KB, 873x604, 1344407386645.png [View same] [iqdb] [saucenao] [google]
4948113

>> No.4948115

>>4948104

Wouldn't the sides be sqrt(2),

\[ a^2 + b^2 = 4 \]

>> No.4948119

>>4948051
actually that is the correct solution. What you didn't realize is that you CAN cut into 8 1x1 squares and diagonally cut them all in just 4 cuts.

>> No.4948121
File: 177 KB, 1022x392, FatalityRedContourClearBackground.png [View same] [iqdb] [saucenao] [google]
4948121

Guysssss did I win?

>>4948089

If not explain to me my fatal error

>> No.4948122

>>4948115

fuck, I'm dumb, back to /lit/ for me.

>> No.4948125

>>4948115

a^2 +b^2 = c^2, not c

>> No.4948128

Guys it is not 4 cuts

it is 4 pieces

>> No.4948133
File: 4 KB, 590x306, accurate.png [View same] [iqdb] [saucenao] [google]
4948133

accurate drawing for you guys to manipulate

>> No.4948141

>>4948115
No. Diagonal is 4, sides are 4/(sqrt2)=2*sqrt(2)

>> No.4948150

I swear if no one disproves my proof...

I'm done with this post>>4948089

>> No.4948152

>>4948121
Put it in picture form, its too hard to accurately understand what you are trying to say.

>> No.4948155

The more I think about it, the more I think this is a troll made by a 10th grader. I mean, why would an IQ test be testing your knowledge of trig? Why wouldn't they say a square of length 2sqrt(2) instead of saying it has a diagonal of 4? Or why doesn't it just say "form a rectangle of length 8 and width 1 into a perfect square by cutting it into 4 pieces and re-arranging"? I think OP is 15.

>> No.4948157 [DELETED] 
File: 41 KB, 873x604, 1344407386645.png [View same] [iqdb] [saucenao] [google]
4948157

>> No.4948160

>>4948150
it's wrong because you cannot evenly fold 1 into pieces of 1/sqrt(2) so you would have more than 4

>> No.4948162

>>4948157
Nice try, but that's 5 pieces rather than 4. We already knew we can do it in 16 (cut it into 1 x 1s, and then cut each 1x1square into two triangles). This approach is exactly the same but merges a number of the redundant triangles.

>> No.4948167

>>4948150

To many pieces or pieces too long or too short

>> No.4948176
File: 5 KB, 466x354, Untitled.png [View same] [iqdb] [saucenao] [google]
4948176

like this? no?

>> No.4948177 [DELETED] 

I think there's no way to do it. I'd start trying to prove that there is no solution, with >>4948038
as a big hint probably.

>> No.4948183

lol, I just skimmed this thread and it's obvious everyone fails at geometrical constructions like this because they spent all their time learning axiomatizations and none of it learning constructions.

>> No.4948188

>>4948176

No, the diagonal of the square is 4 units, not the sides. The rectangle you made there has slightly different dimensions because of that.

>> No.4948193

>>4948176

That gives a 2x4 rectangle.

>> No.4948199

Aside from my "fuck you puzzle" in >>4948079 The only way I've figured out how to do it with 4 legitimate pieces is to connect the 8x1 so it forms a circle and to cut it into 4 diagonals such that one has 2 triangles of h=1 and 2 trapazoids of h=1

>> No.4948202
File: 31 KB, 666x452, Untitled.jpg [View same] [iqdb] [saucenao] [google]
4948202

This is the solution(again, not drawn to scale). You can cut up your own rectangles and try it if you don't believe me.

>> No.4948203

>>4948076
In picture form, its 6 pieces not 4, but it works

>> No.4948205
File: 31 KB, 873x604, solved.png [View same] [iqdb] [saucenao] [google]
4948205

>>4948203
dammit

>> No.4948206

>>4948150
> if no one disproves my proof
You need to learn what a "proof" is. If you had a proof, the thread would be over. If you think your answer is right, then work it out and show it.

>> No.4948212
File: 205 KB, 300x300, thales_theorem.gif [View same] [iqdb] [saucenao] [google]
4948212

>>4948038

There are tons of ways to construct sqrt(2). It's one of the first things you learn in constructivist geometry. Here is a neat method that can be used to construct the sqrt of any number.

Draw a circle with a diameter of 3. Now draw a line cutting the circle in half. From the new line, at 1 unit in from the edge (remember, this line is 3 units long because it is the diameter), draw a line perpendicular extending vertically till it reaches the edge of the circle. Draw the two new lines from the intersections of the diameter line with the circle to the intersection of the vertical line with the circle. These two new lines with the first diameter line now make a right angle in accordance to thales theorem. The vertical line splits said right triangle into two congruent right angles. You can now do the algebra fairly trivially with the congruent triangles to show that the length of the vertical line is sqrt(2). By repeating the process with a circle of diameter n it is possible to construct a line of length sqrt(n). I'll leave this up to you to show so that you may hopefully learn something.

>> No.4948218
File: 72 KB, 1221x1896, solved.png [View same] [iqdb] [saucenao] [google]
4948218

k someone tell me what i did wrong

>> No.4948222

>>4948202
Looks right. Nice one.

>> No.4948230

>>4948218


Took me a bit to figure out what you were doing. The thing you did wrong was that the diagonal is 4 units, not the sides.

In other words, it is a right triangle with a hypotenuse of 4 and two equal sides.

A^2 + A^2 = 4^2
2A^2 = 16
A^2 = 8
A = sqrt(8) (ignoring the negative result because it doesn't apply in this scenario)
A = (2)(sqrt(2) = 2.82842712474619009760....

In other words, you have a square with sides 2sqrt(2) (about 2.8) and a diagonal of 4.

This thread is a perfect example with what's wrong with geometry education across the world.

>> No.4948235
File: 10 KB, 827x416, toscale.png [View same] [iqdb] [saucenao] [google]
4948235

>>4948218
What you did wrong was that you drew like shit and no one can tell what you're trying to show.

Here is the solution that this guy showed:
>>4948202
... drawn roughly to scale using the awful copy/paste in MSPaint.

>> No.4948241

>>4948218
That forms a 16/3 by 3/2 rectangle, not a square.

Also, hold shift in MS Paint to draw straight lines.

>> No.4948246

>>4948230
the question doesn't to do that, all it says is to make a rectangle into the shape of a square which I did

>> No.4948250

>>4948241
The actual grid was done pretty fast and not evenly If someone has the time to make each box even itll come out as a rectangle

>> No.4948260

>>4948246
Read the first line on OP's picture and then refer to where the 4 is located on the square.

>Depicted is a 1 by 8 rectangle and a square of diagonal 4..

>>4948235
>>4948202
This looks legit.

>> No.4948263

This puzzle is clever but I don't think it belongs on /sci/. There's no way anyone could have come up with that solution by just drawing and writing geometric shapes and equations.

>> No.4948267 [DELETED] 

post more of these OP

>> No.4948300

>>4948263

Yes it does, and there is. What's great about mathematics is that one can still learn a lot just by trying, without necessarily having to get to their destination. Not to mention that one mathematician's journey may be completely different from another's especially considering that there are often times many different ways to prove the same thing.

Personally, I much prefer these math threads to all of the science bullshit that constantly plagues the front page.

>> No.4948304

>>4948128

I loled

>> No.4948308

>>4948263
Who said you were supposed to try to solve it by drawing and writing geometric shapes and equations?

Welcome to math faggot.

>> No.4948459

>>4948212
oh god i am stuck trying to figure out how to do this am i retard

>> No.4948470

>>4948459

No, you're not. It's actually really clever. I'll draw it but I have to unpack my touchpad (moved a month ago or so and haven't needed it up until now).

>> No.4948472

>>4948235
So how did you manage to cut the yellow and red pieces? Both of those have an irrational length.

>> No.4948473

>>4948470
i've been drawing triangles and circles for over an hour :(
i was about to go to sleep damn you

>> No.4948484

>>4948470
actually does it have to do with how the congruent triangles make rectangles?

>> No.4948490

>>4948473

I'll skip some of the more technical constructivist parts then (a formal thorough constructivist proof would require every process to be shown by only straightedge and compass).

>> No.4948502

>>4948490
if i am to fall asleep i will surely check this thread again when i wake up

also, the
>damn you
was meant that your problem intrigued me and made me stay up instead of sleeping

>> No.4948504 [DELETED] 

coincidentally, if you do a diameter 3 circle and make a triangle with one of the sides being 1 unit, the other side ends up as 2root2 which is used in this problem

>> No.4948524
File: 153 KB, 1920x1158, constructing sqrt2.png [View same] [iqdb] [saucenao] [google]
4948524

>>4948502
>>4948490
>>4948484
>>4948473
>>4948470

Okay here, sorry it took me a while to set up my touchpad because I couldn't find the USB cable for it. There are actually lots of very interesting results that come from constructivist geometry, but I won't mention them any further or you'll be up for weeks. lol

>> No.4948532
File: 5 KB, 295x300, thale.png [View same] [iqdb] [saucenao] [google]
4948532

bonus root 3 and root 6

>> No.4948533
File: 16 KB, 480x360, hqdefault.jpg [View same] [iqdb] [saucenao] [google]
4948533

I have the better solution

>> No.4948538

>>4948524
so cool wow
tell me everything :(

>> No.4948541

>>4948532
i had this drawn previously and i thought it was really interesting, could probably do a lot of fun math with that

>> No.4948563
File: 10 KB, 525x431, root5.png [View same] [iqdb] [saucenao] [google]
4948563

i guess this is how you would make root5 with straightedge and compass

>> No.4948572
File: 91 KB, 1920x1158, constructing sqrt2 2.png [View same] [iqdb] [saucenao] [google]
4948572

>>4948538
>>4948541

Well, it can be expanded for any sqrt of n. You can also construct a really neat spiral out of the simple right triangle with two sides of length 1.

In constructivist geometry people became concerned with what could and what could not be constructed. This was all formalized using field extensions. So you have the set of integers which can be constructed pretty easily, and then as we showed we can construct any square root, so that's a field extension to the integers. Similarly we can take it further and get some instances where we can take the fourth root (square root twice) and then add it as a field extension. If a figure required a value that did not exist in the field of what can be constructed, then that figure could not be constructed. One of the conclusions of this is that some regular polygons cannot actually be constructed with straightedge and compass, they can only be approximated. The heptagon is one such example. Gauss discovered the construction of the 17-gon which led him to become a mathematician and later on proved that any polygon that could be constructed had to have it's number of sides broken down into its prime factors and then each one of those primes had to be of the special case 2^(2^n) + 1, so a triangle for example would have 3 sides (which is already factored) and can be written as 2^(2^0) + 1. So the first five prime polygons that can be constructed have side lengths 3, 5, 17, 257, and 65,537. A mathematician named Johann Hermes worked on the construction of the 65,537-gon for a span between 10-20 years eventually publishing a huge dissertation. The proofs that something can or can't be constructed can actually become really difficult and involved. Proving that the heptagon can't be constructed requires 5 or so corollaries even using complex numbers.

>> No.4948586

>>4948572

I should add, when I said prime polygons, I didn't mean to imply that the number of sides was prime. As it turns out this was a conjecture by Fermat (they are also known as fermat numbers), but even 2^(2^5) is not prime, though it is huge. What I meant though was that one can overlay two polygons on top of each other in some really fancy ways and connect their vertices to make other regular polygons that have multiples of the number of sides.

>> No.4948607

>>4948472
^this, once you get to this point it's sufficient to say it cannot be done.

>> No.4948619
File: 90 KB, 1920x1158, constructing sqrt2 3.png [View same] [iqdb] [saucenao] [google]
4948619

>>4948563
Yeap, also if you continue the spiral in >>4948572 for one more step then you'll see that the next hypotenuse is also sqrt(5).

Here is a more detailed diagram showing the sqrt(5) using the other method I posted. Note how counting is done with the compass and how forming the perpendicular line is accomplished by the intersection of two circles. It's a shame that you actually see lot more of this type of stuff in autocad courses than you do in geometry courses.

>> No.4948656
File: 110 KB, 1920x1158, constructing sqrt2 4.png [View same] [iqdb] [saucenao] [google]
4948656

>>4948607
>>4948472

I thought it was pretty straightforward. I only skipped the circles necessary for showing the perpendicular blue lines, but that's because it was getting cluttered.

captcha: sketch effedc

>> No.4948660
File: 19 KB, 1152x648, Fără titlu.png [View same] [iqdb] [saucenao] [google]
4948660

too lazy,but i think you get the idea

>> No.4948668

touchpad guy here

By the way, if anyone has more cool stuff I'd be totally interested in tackling it. It's been a while since I've used my touchpad and I forgot how much it kicked ass doing math with it. If you have one and don't know what good apps to use, I recommend you pirate AutoDesk Sketchbook Pro (don't buy it, it's waaaay overpriced and geared at artists and designers, unfortunately no one makes good apps for geometers besides geogebra which isn't really for sketching). There is also a version that does vector graphics called Sketchbook Pro Designer. They both have their flaws though, like, the normal one doesn't snap to edges/intersections/etc and the vector one does sometimes but clumsily and I don't think it does intersections. Also, simple things like drawing circles can be totally unintuitive at times. AutoCad is actually not bad either, but much more bloated and with tons of functions geared specifically at drafting people.

>> No.4948714
File: 180 KB, 400x500, lol wut - sale.png [View same] [iqdb] [saucenao] [google]
4948714

>>4948660
yea..what?

>> No.4948761 [DELETED] 

took me 40 seconds: Cut the 8x1 strip into 2 4x1 strips and glue them together to get a 4x2 rectangle. With a 4x2 rectangle, choose one of the sides of length four and connect the vertices of the side to the midpoint of the opposite side with lines. Cut along those lines and glue the resulting pieces together. voila, you guys are retards.