Hey /sci/ I got an easy question for you.
No, it's not my homework, I've already solved it. I'll post the answer 41 minutes after this is posted.
Again, I'm not saying this question is hard, it's just simple but fun.
16
when the side of the small corner square is equal to 1/4 the side of the big square.
>>8039461
How did you figure that out?
>>8039466
u friggin sirius m8
>>8039479
Yes. Did you eyeball it? I stated it's not to scale, you need to prove it algebraically.
When the ones on the outside are 25% of the inside. Are you even trying?
>>8039485
jesus...
say the side of the small sq = a
and one in the middle is 2b
then for the equality:
4a^2 +4b^2 = 8ab
a^2 -2ab +b^2 =0
(a-b)^2 =0
since length cannot be negative a=b
one side of the big sq = 2a+2b =4a
etc...
>>8039489
>dat image
every. single. time.
>>8039485
Well let the side of the big square be s. Then the side of the small square is s/4.
The gray area is:
4 (s/4)^2 + (s - 2*s/4)^2 = (s^2)/ 4 + (s^2)/4 = 1/2 s^2
The rectangles in white have dimensions s/4 by (s - 2*s/4) = s/2. They have area 1/8 s^2 . Now there are 4 of them. So the white area is s^2/2.
You can also start out with the the assumption that the area of the big square is is s^2, and try to find the size of the small square that makes the grey or white area (s^2)/2, since in order for the two to be equal they must be half the area of the big square.
>>8039485
It's honestly so trivial that "by inspection" would probably be accepted.
>cut into 4
>ask yourself when grey = white, remembering they're squares
>obviously when the two greys are equal
>>8039423
Area of gray: (a-2b)^2+4b^2
Area of white: 4(a-2b)*b
(a=side of large square, b=side of small gray squares)
Equating,
[math] b=-\frac{1}{16} (a-8)a [/math]
>>8039505
The problem only ask you to show that it's true, not to derive it.
>>8039506
Crap, sorry inputted into wolfram wrong, it should be
[math] b=\frac{a}{4} [/math]
>>8039499
Did you miss the big bold red text saying "not to scale"? Sure it's trivial but it's just a simple math exercise, not asking you to solve an unsolvable problem.
>>8039513
What the fuck do you think "show that it's true" means?
>>8039423
small grey = a^2
big grey = b^2
white rectangle = ab
4a^2 + b^2 = 4ab
4a^2 -4ab + b^2 = 0
(b-2a)^2 = 0
b-2a = 0
b = 2a
>>8039423
a : Small grey square (area of all of them is 4a^2)
b: Center grey square (area of it is b^2)
white square area = 4ab
4a^2+b^2=4ab
b=2a
>>8039499
>algebraically
>by inspection
>>8039524
I did show that it's true though. I showed that if the side of the small rectangle is 1/4th the size of big rectangle, then the areas are equal.
>>8039548
>*sighs*
>only person in the thread to fuck it up
>>8039543
that methodology (induction) will almost never work elsewhere, where it isn't so easy to make educated guesses
>>8039555
Actually it's how most of science is conducted since the times of Francis Bacon and David Hume and before.
>>8039562
that's pretty much why science took baby steps until descartes/cartesian/deductive reasoning and then boomed after that
>>8039559
Did you miss the part where I used the symmetry of the problem to cut it into 4?
>brainlets
>>8039423
let [math]x[/math] be the length of the sides of the larger grey square, and [math]y[/math] be the length of the sides of the smaller grey squares.
As the white rectangle shares two sides with smaller grey squares and one with the larger square, the area must be equal to [math]xy[/math].
Given there are 4 white rectangles, 4 smaller grey squares and one larger grey square; under the circumstances that the white and grey areas are equal
[eqn]x^2+4y^2=4xy[/eqn]
[eqn]x^2-4xy+y^2=0[/eqn]
[eqn](x-2y)^2=0[/eqn]
[eqn]x-2y=0[/eqn]
[eqn]x=2y[/eqn]
Hence the grey area is equal to the white area when the sides of the larger square are double that of the smaller square.
>>8039572
I get the feeling you don't know very much about science and the history of science.
>>8039574
>substituting the problem and forgetting to undo the substitution before giving an answer
>everyone is stupid but me for incomplete thoughts
>>8039576
ah, shite, missed the 4 off the [math]y^2[/math] on the second equation, but the rest is correct.
>>8039577
nice try spidey
>>8039423
Length of corner grey squares = x
Length of large centre grey square = y
[math]
A_w = 4xy\\
A_g = 4x^2 + y^2\\
A_g = A_w\\
y^2-4xy+4x^2=0\\
(y-2x)^2=0\\
y=2x
[/math]
big square edge twice the small edge
more easily done by just cutting the whole square into 4x4 grid
>>8039423
The gray ad white aren't necessary equal, though. Is it asking to find proportions (line lengths) so that the two areas are equal?
>>8039538
This is the answer. Originally I set it up with L being the side of the large square, and s the side of each of the small, corner squares. Therefore, L - 2 is the length of a side of the center square. Then it turned into a cluster fuck. This is the easiest way to do it for sure.
