/sci/ - Science & Math

>>9251755
15 brainlet

>>9251774
how?>

>>9251774
explain

24

>>9251777

You just realign the deflector to the matrix of the hyphothenuse that you get when you use pythagoras on the triangles that you get when you connect the anode and the kathode. Basic neurology

Consider the bottom left corner as the origin of the complex plane and let the the segment of length 12 make an angle a with the real axis. It is easily seen that the other end of the diagonal passing through the origin is [math] 12e^{i a} + 3ie^{i a} + 9e^{ia} = (21 + 3i)e^{ia}[/math], so the area of the square will be (21^2 + 3^2) / 2 = 225 = 15^2, so the side length is 15.

>>9251789
in english doc

>>9251755
15

It's easy to see that
x = 12 cos(a) + 3 cos(pi/2 + a) + 9 cos(a)
x = 12 sin(a) + 3 sin(pi/2 + a) + 9 sin(a)

solve this system to get x=15.

>>9251775
>>9251777
Draw the diagonal and then continue either the line of length 12 or of length 9 until you can create a right triangle with sides 21 and 3, with the diagonal as the hypotenuse. So the diagonal is 15sqrt2.

B R A I N L E T S

>>9251755
Do your own homework brainlet

May or may not be correct

>>9251794
>>9251810
>>9251815
Alternatively an elementary schooler can imagine that there are two 3x3 squares in the middle, calculate that 12-3 = 9 and 9-3 = 6, and sum that it comes out to 15.

20 hours in MS paint.

>>9251755
It's there on the left side and on the bottom.

>>9251755
what makes me even more curious...why would one want to know? Is there a practical example of making use of it?

>>9251827

the lines aren't parallel dummy

It's a square you retards
You cannot draw two parallel diagonals
The diagram is wrong

>>9251755

OPs pic is wrong too. those angles can't be 90*

>>9251838
>2 right angles
>not perpendicular

>>9251789
>Basic neurology
kek

>>9251844
who said about diagonals?

>>9251826
Fixed

>>9251838
How would you go about drawing two parralel lines using a straightedge ruler?

>>9251852
>using approximations

>>9251854

Use the corner of the ruler to draw a right angle to the line. Then draw another line at a right angle to that line; the resulting line is parallel to the first line.

>>9251838
Holy shit, dumbest post on /sci/

>>9251856
I truncated for the sake of drawing, but not on the Google calculator. It gave me 15 exactly.

>>9251860
Calculators use approximations too

>>9251857
Good job. Now you know the lines are parallel.

>>9251865
what if a right angle is 91 degrees in this space?

>>9251913
really trigons my metrics.

12 *[v1,v2] + 9*[v1,v2] + 3*[-v2,v1] = [x,x]
21*[v1,v2] + 3*[-v2,v1] = [x,x]
(21)v1 - (3)v2 = (21)v2 + (3)v1
(0.75)v1 = v2

sqrt(v1*v1 + v2*v2 ) = 1
=> v1 = 0.8
=> v2 = 0.6
=> x = (21)*0.8 - (3)*0.6 = 15

x = 15

>>9251827
How exactly do you know that the sum is x?

>brainlet here

All of these solutions just seem to be nuking the issue to me. Can't you just do,
(sqrt(12^2+1.5^2)+sqrt(9^2+1.5^2))cos45
which comes out to 15.003

>>9251980
That's just an approximation. The diagonal doesn't bisect the 3cm line. The actual ratio was calculated by >>9251852

>>9251852
what is Y?

>>9251755
>Find X
Here it is.

ITT: Brainlets

>>9251992
The angles are both 90 degrees, the diagonal must perfectly bisect the 3cm line.

>>9252044
no body promised that it is a square.

>>9252044
Where did the 21 come from?

>>9252053
Look at the problem again, but closer, nimrod

>>9252055
>doesnt know that 12 + 9 = 21
>calls me nimrod
what are you stupid

>>9252055
12+9=21

>>9252044
Was just about to post this

Hilarious how /sci/ tries to solve this by fucking trigonometry, it is just simple geometry.

>>9252056
maybe he didn't instantly realize it

btw if it wasn't a square the sides would have been called x and 'another letter'

Line '9' and line '12' are not parallel. No line '3' exists that is perpendicular to both. The diagram is impossible as presented.

>>9252057
Thank you

>>9252063
Nah, I drew it IRL with a ruler, it checks out

>>9252009
Y is the length of the short side of the top triangle created by connecting the square's corners.
>>9252050
If both diagonals were the same length, yes... but they're not. There's a lot of confusion in this thread because as drawn, OP's diagram is misleading. It's actually drawn as a rectangle and taller than it is wide, which is why both diagonals can be depicted as parallel.

>>9251940
It's basically the same as >>9251810 except dumbed down for children who don't understand how to solve systems of equations/what they're doing.

>>9251940
>>9252169
Or wait, maybe it's even clearer like this.

Here's the video where the image came from
https://www.youtube.com/watch?v=v80jDho0emQ

Okay, so in the question the numbers are all clearly multiples of 3, so you can divide through to get the numbers on the zigzag diagonal as (4,1,3) giving x=5.

My question to you is, what is the next smallest integer x such that you can have diagonal (a,1,b), with a and b both integers?

>>9252204
If I'm any good at pattern recognition, you can get x=4 with (3,1,2).

>>9252231
Wait no, that's x=3, I'm retarded

>>9252201
Is this a troll? The shape cannot be constructed
Draw a diagonal and break it like in the OP image, there is no instance of those angles are equal to 90

>>9252231
>>9252234
You can get x=4 with (4,1,2) then.

>>9252231
No.
[math](3,1,2) \implies x=\sqrt{13}[/math]
[math](4,1,2) \implies x=\frac{1}{2}\sqrt{74}[/math]

I tried to solve it like pic related and got x=15. I'm not sure if it is calculated in the correct way.

>>9252273
That's great. So is >>9251852
And>>9252044
Intuitive physically-correct explanations.

>>9252236
There is, do it yourself, draw the center lines first and then the square around it

>>9252312
>>9252236

I tried to draw it after calculating it like >>9252273. Sorry, it is quite messy, but I hope you can see that it's 90 degrees.

Assuming this is a flat square and not some bullshit:

The red diagonal can be found using Pythagoras and is equal to root(3^2 + 21 ^2)

this is also the hypotenuse of the right triangle with sides x and x therefore

(3^2 + 21 ^2) = 2x^2
225 = x^2
15 = x

>>9252325
amirite?

not a square?

>>9252328
Yes, that is correct.

>>9252339
it can be a square or it can not be a square, to find x you have to assume it is a square because there isn't enough angle information otherwise

it CAN be a square,and 3 lines with measurements on them can be simplified to a diagonal line across that square, you can make a square with any measurements on those lines in the middle

if you assume this is a square, you can find that x= EXACTLY 15

>>9252350
Yeah, I realized after posting that point E could be positioned differently.

>>9252339
Imagine the point A as a pivot for the three segments AC, CB, BD. As you rotate them down, AE(=i) will decrease and ED(=h) will increase.

At some point i will equal h

>>9252328
>>9252325
ye, that works.

>>9252236

>mfw this asshole has tons of videos and views

>>9251755
x/y + y/x = (x^2 + y^2)/xy = m
(n/m * x/y)^2 , (n/m * y/x)^2
^ s , t
s + r^2 , t + p^2
^ a , b
sqrt(a) + sqrt(b) / sqrt(2)
^ all possible desc. x

>>9251755

>>9252320
those lines CANNOT be parallel if they both start for opposite vertices, therefore the angles CANNOT be 90 degrees

>>9253239
Yes they can.

Proof: Think.

Nigga. 18. It is 18.

>>9253253
>18

Fucking finally. OP here, study guide agrees. Still asking how.

>>9251755
It is 12

Anyone who still thinks it's not 15 look at>>9253178

>>9253244
obviously he doesn't accept the parallel axiom you brainlet

>>9253377
Sorry, I forgot it was drawn on a non-Euclidean piece of paper.

14.859

15*sqrt(2)=x*sqrt(2) x=15

>>9252263
wtf (3,1,2) means?

>>9253239

that's what i thought at first but yeah, they can. if you break the diagonal at some point in the middle and pivot each line around their respective corner, they can be made to have right angles by increasing their lengths. the problem is figuring out how much extra length.

t. brainlet with poor spatial reasoning.

>>9253259
Your study guide is wrong.

sum the squares together to get a rectangle with length=21 and width=3 and then find the diagonal of the rectangle (which is also the diagonal of the giant square) using Pythagoras. And then again use Pythagoras to find the sides of the giant square.

pic related took me ages on paint, you brainlets better appreciate

>>9253575
The lengths of the three segments in the diagonal.
In the original question they would be (12,3,9) with x=15, but obviously you can divide everything by 3 to get (4,1,3) with x=5.

The question is, assuming the middle segment still has length 1, what is the next possible integer value of x that can have (a,1,b) with a and b integers?

>>9251755
63

>>9251755
Sqrt(909)

15 is correct

>>9252236
>The shape cannot be constructed
let [math]x = 15[/math]
then it's easy to see
[eqn]
x + i x = 12e^{i\phi} + i 3 e^{i \phi} + 9 e^{i\phi}
[/eqn]
where
[eqn]
\phi = \arctan\left(\frac{3}{4}\right)
[/eqn]
which completes the construction in the complex plane

>>9253931
(a,1,b) would have
[eqn] x = \sqrt{\frac{(a+b)^2 + 1}{2}} [/eqn]

>>9254299
True, but that's not the question.

>>9253931
>>9254299
Now if you define y=a+b you get
y^2 - 2 x^2 = -1
which is the negative Pell equation with n=2 which only has the solutions
(x,y) = (1,1)
(x,y) = (5,7)
(x,y) = (29,41)

Visually you get this, pure mathematically you get exactly 15.

It is 15 niggas, obviously

>>9254319
Correct. x=29 is the next number.

Although there are actually an infinite number of solutions. The next one after that is x=169.

>>9252273
Lmao i did this after i demonstrated in my head that the middle point of such segment made by two paralel lines and a sq angle, the diagonal of a square always coincides with the middle point of such segment, while walking my pibull on the park. I'm superior to all of you eurangutan whitey boys.

>>9251827
This is the only solution I didnt understand.

Easy. 12 + 9 divided by root 2 of course.

>>9252178
That's just a coincidence, no solution here.