/sci/ - Science & Math

First one. I counted the squares.

>>10594260
looks like left one to me

>>10594260
I'm on mobile so don't want to solve it out but you can use inscribed angle theorem to solve it. First find one of the inscribed angles, double it and make a isoceles triangle and then solve for R of each circle that way.

>>10594260
Whichever one uses an extra decimal digit of approximation for pi of course :^)

>>10594260
They're the same size.

>>10594331
they are VERY close to the same size
I chose them that way

>>10594336
nope they are the same size i analyzed them

>>10594260
The left one, all you have to do is see that its diameter only extends 2.5 squares beyond the major grid lines while the once on the right extends a bit more than that.

>>10594260
Left traces (1,9) and (5,8). Right traces (-3,9) and (3,7), which we can flip to be (3,9), (6,2) by mirroring across the x axis and repicking the origin. This gives us midpoints and slopes. Eh.

>>10594372
but their left sides are also different distances away from the major grid lines...

you cannot tell with the naked eye which of these is bigger
trust me

>>10594384
>but their left sides are also different distances away from the major grid lines...
Doesn't matter, since the gridlines are parallel and evenly spaced all you have to do is find the farthest point of the circle from each gridline and count boxes.

>>10594384
Their diameters differ by about a quarter of a box, trust me.

>>10594395
>all you have to do is find the farthest point of the circle from each gridline and count boxes
Right, and if you do that, you will count the same amount of boxes.

As I said before, you cannot tell with the naked eye.
The difference between their radii is less than 1/100000

>>10594406
are you trolling?

>>10594425
>Right, and if you do that, you will count the same amount of boxes.
False. The farthest top point on the left is very close to a half box from the major gridline. The bottom point is close top 2 boxes away from the gridline. This gives us a total diameter of 9.25. On the right, the topmost point is very close to 2 boxes from the same gridline. The bottom then should be very close to a half box from the gridline but instead is clearly above a half box, about three quarters. So the diameter on the right is about 9.275. The resolution of the picture and is much higher than these differences so it's clear that the right is larger.

The solution is:
Draw triangles using the points he fixed on the image, for each circle, use Pythagorean’s theorem to find the length of each side of the triangle, then find any angle with cosine’ law, and then use this angle, plus the length of the side opposite to that angle, to find the radius of the circle.

>>10594441
>This gives us a total diameter of 9.25
no that clearly gives you a diameter of 9.5
I'm not even gonna bother reading the rest

Look I'm just going to spoil the thread because you're being retarded:
The left one has a radius of [math] \frac{\sqrt{\frac{62033}{2}}}{37} \approx 4.7598653 [/math]
The right one has a radius of [math] \frac{5\sqrt{\frac{29}{2}}}{4} \approx 4.7598581 [/math]

You can verify that here:
https://www.wolframalpha.com/input/?i=circle+(10,10)+(15,18)+(11,19)
https://www.wolframalpha.com/input/?i=circle+(23,10)+(26,17)+(20,19)

>>10594441
The small boxes are actually 0.2 so the diameters are approximately 9.5 and 9.55, clearly different.

>>10594470
oh yeah and here is the geogebra file:
https://www.geogebra.org/graphing/dxrqzt3t

The right one is bigger.

The radius of the first one is the square root of 90.51257648, and the radius of the second one is the square root of 90.625

>>10594486
Solution:
The distance between two points in any circle is equal to the radius of that circle multiplied by the sin of the angle formed by those two points and any point in the circle edge.

The OP gave use the coordinates of 3 points in each circle. We use that and the cosine’s law to find the radius, and the result is that.

The right one is bigger.

>>10594486
I think you mean diameter
You are right about the second one, but I think you made a calculating error on the first one

I already posted the answer >>10594470

>>10594470
>>10594512
You posted the answer before I could but I had fun with it.

>>10594260
The two wheels look the same to me.

This is actually pretty easy but most posters are justifiably too lazy to waste their time doing the boring calculations to get the answer.
You have A1(10, 10), B1(11, 19), C1(15, 18) and A2(23, 10), B2(20, 19), C2(26, 17). Those give you P1(10.5, 14.5), Q1(12.5, 14), P2(21.5, 14.5), and Q2(24.5, 13.5), because of ((x1+x2)/2, (y1+y2)/2).
To get the equation of p1, the perpendicular bisector of A1 and B1 going through P1, you just have to do f(x)=-1/(slope of A1, B1)*x + some bullshit based on the location of P1. That gives you
p1: y = -1/9 * x + 141/9
q1: y = -5/8 * x + 349/16
p2: y = 1/3 * x + 22/3
q2: y = -3/7 * x + 24
Now you have to solve p1 = q1 and p2 = q2 to get O1 and O2.
O1: (885/74, 1061/74)
O2: (175/8, 117/8)
Now you can just calculate the distance between A1 and O1 to get r1 and so on:
r1 = 4.75987 (rounded)
r2 = 4.75986 (rounded)
So now we know that the circle on the left is bigger.

>>10594524
sorry I didn't mean to post the answer, but that dude was really annoying me

>>10594529
nice, there we go

>>10594529
>r1 = 4.75987 (rounded)
>r2 = 4.75986 (rounded)
>So now we know that the circle on the left is bigger.

your monitor can resolve a difference of 0.00001 inches? I call BS.

they're the exact same size by any reasonable measure, and you just proved it.

>>10594548
>monitor
???
>inches
???

the state of /sci/ these days...

>>10594260
>post two circles the same size
>everyone is arguing about whether or not they are different sizes

>>10594281
can't see, too close even with a paint prog

>>10594260
Import opencv

Cumcum=Opencv.read.image(“c://users//faggotretaed//downloads//kikel(420).png”
Cumcum.get_circumference()
Circumference.get

>>10594319
Sorry I'm back now
R=c/(2sin(arccos((a^2+b^2-c^2)/(2ab)))
Looks like you guys already figured it out
I'll do first circle
http://m.wolframalpha.com/input/?i=sqrt%2817%29%2F%282sin%28arccos%28%2889%2B82-17%29%2F%282sqrt%287298%29%29%29%29

>>10594470
god what a shithead, you just made this thread to show off how much of a big brain you are eh?

>>10594260
thx opie had fun

>>10594902
oh yeah I need such a big brain to input the circles into wolfram alpha

>>10595529
you are

>>10594548
Brainlet