/sci/ - Science & Math

[Anonymous No.10561175]

The problem that stumped /sci/...

[Anonymous No.10561179]

>>10561175
arbitrarily close to B

[Anonymous No.10561188]

Hint: it is B

[Anonymous No.10561200]

>cube
>1x1x2

[Anonymous No.10561234]

>Graph theory

[Anonymous No.10561236]

>>10561175
(A+B)-B

[Anonymous No.10561240]

before we have another awful thread, everyone get a string and a box shaped object and see if, holding the string at the point that reaches the B corner, you can reach every point on the top and back of the object

[Anonymous No.10561273]

>>10561240
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA.
I was here last thread and have to admit I was one of the faggots that kept replying B.

[Anonymous No.10561279]

>>10561240
i did it is the longest distance

[Anonymous No.10561285]

>>10561273
some men just want to watch the world burn

[Anonymous No.10561345]

[Anonymous No.10561353]

For the Not-B-crowd: It's not specified that the bug must traverse walls and cannot fly. Thus, any answer that assumes this is automatically disqualified. Fuck you. It's B.

[Anonymous No.10561366]

Fuck you its B, but I pick an arbitrary point inside of the box where the bug cant reach but is still a point in a cube.

[Anonymous No.10561376]

How the fuck is it not B? Can someone give me a point with distance >4?

[1/2 No.10561377]

I'm actually not sure (yet) what the answer is,
but I'm going to show you that it's not B.

As you can see in my picture, you can reach point [math] B [/math] in a distance of [math] 2\sqrt2 [/math].

[2/2 No.10561380]

>>10561377
But can you reach point [math] P [/math] in this picture in a distance equal or less than [math] 2\sqrt2 [/math] ?

You could try
[math] 2 + \sqrt{(\frac{3}{4})^2 + (\frac{3}{4})^2} [/math]
or
[math] \sqrt{(2\frac{3}{4})^2+(\frac{3}{4})^2} [/math]
or
[math] \sqrt{(1\frac{3}{4})^2+(2\frac{1}{4})^2} [/math]

But none of those are smaller than [math] 2\sqrt2 [/math].

[Anonymous No.10561392]

>>10561377
>>10561380
Yeah, but it can travel straight to B, which is \sqrt5.

[Anonymous No.10561399]

>>10561392
I don't think flying is alowed
If flying is allowed, yeah then obviously it's B

[Anonymous No.10561412]

B
[math]1^2 + 1^2 = 2[/math]
[math]\sqrt{2}^2 + 1^2 = 3[/math]
[math]AB = 3[/math]
[math]1^2 + 2^2 = 5[/math]
[math]AC = AD = \sqrt{5}[/math]
[math]\sqrt{5} < 3[/math]

[Anonymous No.10561424]

>>10561412
Forgot pic.

[Anonymous No.10561434]

>>10561175
Any point in the cube, as long as it's on the inside if the bug is outside, or vice versa.

[Anonymous No.10561454]

Color me surprised, it's B.

[Anonymous No.10561474]

>>10561175
Assume that B is not the point with the farthest distance from A.
Let point C on the prism represent the point with the greatest distance from A.
Consider the prism formed from the points A and C, with C being on the opposite corner to A (this is always possible, some widths may be 0 though.
This is the same as the arbitrary prism with points A and B. But we already assumed that this point couldn't be the shortest distance.
This is a contradiction.
QED

[Anonymous No.10561478]

>>10561474
OP BTFO

[3/2 No.10561505]

>>10561377
>>10561380
Okay, I did some math and I think I've found the point which would take the longest time to reach.

Funnily enough it's the point that I drew here >>10561380, even though I just drew it as an example.

The shortest distance along the cube to this point is:

[eqn] \frac{\sqrt{130}}{4} \approx 2.8504 [/eqn]

What do I win?

[Anonymous No.10561510]

>>10561454
your proof is incomplete and doesn't properly take into consideration every path possible

[Anonymous No.10561564]

If the volume of the shaded area of the top 2 squares is more than the volume of 1 square. Them B is the farthest point.

[Anonymous No.10561584]

>>10561353
Even if the bug only walks it is still B. I can prove it, but OP is not worth my time.

[3/2 No.10561606]

>>10561564
>>10561584
IT'S NOT FUCKING B
STOP SAYING IT'S B

THIS >>10561380 IS THE FARTHEST POINT FROM A
IT HAS A MINIMUM WALKING DISTANCE OF [math] \frac{\sqrt{130}}{4} [/math] (WHICH IS LARGER THAN [math] 2\sqrt2 [/math])
IF YOU THINK I'M WRONG THEN FIND A PATH TO THIS POINT THAT HAS A DISTANCE SHORTER THAN [math] 2\sqrt2 [/math]

YOU CAN'T

[Anonymous No.10561639]

>>10561380
>>10561606
P is a shorter distance brainlets. Imagine P is on the same face as the second 1[sqrt]2 line segment and even someone of your IQ can understand why it's shorter.

[It's unambiguously B No.10561651]

>/sci/ - debating and getting confused by elementary geometry

[Anonymous No.10561656]

>assuming bug takes shortest path
How is this hard for the not-Bs to understand. It is moving in a straight line no matter where you're going.

[Anonymous No.10561660]

>>10561606
the minimum distance from A to B is not 2 sqrt 2
its sqrt 10, larger than sqrt 130 / 4

[Anonymous No.10561676]

>>10561606

[Anonymous No.10561684]

>>10561639
I have no idea what you're saying

We all agree that the shortest walking path from A to B is [math] 2 \sqrt 2 [/math] right? As illustrated here >>10561377

So what I'm asking you to do is to find a walking path from A to P (>>10561380) that has a distance equal to or shorter than [math] 2 \sqrt 2 [/math] .

If I'm wrong then you should be able to do that right?
Then fucking do it and stop making me try to decode your gibberish.

>>10561660
no it's not, it's [math] 2 \sqrt 2 [/math] as illustrated here >>10561377

JESUS CHRIST YOU PEOPLE ARE THICK

[Anonymous No.10561689]

>>10561676
Op eternally Btfo.
Checkmate mathlets.

[Anonymous No.10561698]

>>10561676
oh my god dude you have no spatial intelligence
pic related is where B is
and P is drawn wrong too

[Anonymous No.10561701]

this fucker is so retarded that he keeps deleting his posts...

[Anonymous No.10561757]

>>10561651
>>10561656

[Anonymous No.10561783]

>>10561505
It seems you are correct

Here's an desmos graph to help you guys visualize the different paths and their distances
Try fiddling around with the points
desmos.com/calculator/rqo0zrmpsa

[Anonymous No.10561806]

>>10561698
>>10561701
That is probably why they were deleted... You know, when you think about it.

[Anonymous No.10561821]

>>10561783
>It seems you are correct
thank you

not quite sure what I'm looking at though

[Anonymous No.10561833]

Yeah did my own calculations. It actually is that point, the math checks out. How the fuck...

[Anonymous No.10561836]

>>10561821
it's basically a net of the solid, the dots represent the center of the top face. All four squares to the right really represent the same top face, just that the points are rotated around the center if you are going a different path.

[Anonymous No.10561838]

Here's my desmos btw, https://www.desmos.com/calculator/dycppcz6f8.. It's a bit of a mess, but the x and y values are the distance away from B in that top square and the c value is the distance from A. As you increase c, the last point to be covered is 0.25,0.25 which is the point P.

[Anonymous No.10561846]

This version looks nicer: https://www.desmos.com/calculator/dycppcz6f8

[Anonymous No.10561849]

>>10561838
Wow, impressive.
I guess I was wrong then

[Anonymous No.10561852]

>>10561846
Am a moron, meant to post this: https://www.desmos.com/calculator/zs6rwbjj6e

[Anonymous No.10561859]

>>10561584
>>10561852
I guess I was wrong too. Well done all.

[Anonymous No.10561882]

>>10561838
What do the three circles represent?
>t. brainlet

[Anonymous No.10561917]

I'm sorry guys.

[Anonymous No.10561941]

>>10561757
>>10561783
>>10561838
>>10561846
niggers

[Anonymous No.10561974]

>>10561941
pajeet 100

[Anonymous No.10561977]

>>10561941
you have to stay on the surface, big poppa

[Anonymous No.10561996]

>>10561836
IS THIS RIGHT

[Anonymous No.10562014]

>>10561996
Yes.

[Anonymous No.10562016]

>>10561977
>assuming bug takes shortest path
Staying on the surface is not the shortest path
gb2 high school

[Anonymous No.10562019]

>>10562016
hopefully I'll see you back their when you learn reading comprehension

[Anonymous No.10562022]

>>10561175
In the center of the square which has B as a vertex.

[Anonymous No.10562025]

>>10562019
>bug is in [sic] A
>bug travels shortest path
>lines are the shortest path between points
>find point that takes longest [why are we maximizing time instead of distance]
What is the issue?

[Anonymous No.10562031]

>>10562025
see >>10561917

[Anonymous No.10562034]

>>10562031
Ah, so I'm supposed to add parameters to the problem that were not stated. Fuck off.

[Anonymous No.10562050]

>>10562034
it was stated and you were reminded.
That's checkmate guy. I'm tired of watching you humiliate yourself and will no longer be replying to your low IQ posts, having thoroughly outclassed you on all levels.

[Anonymous No.10562054]

What a crazy thread.

[Anonymous No.10562055]

The image shows the shortest path from A to B and from A to P. The distance from A to P is larger than the distance from A to B.

[Anonymous No.10562139]

>>10562055
This

[Anonymous No.10562332]

>>10562054
It aint B. I'm not trolling

[Anonymous No.10562356]

>>10562332
It's looking that way, but I need another minute with my pythagorean theorem.

[Anonymous No.10562554]

>>10561917
>not a cuboid
>cuboid is not empty

[Anonymous No.10562602]

>>10561175
The distance from A to B is obviously sqrt(1^2+3^2), but I have no idea where you could get more than that, I suppose it will be somewhere on the farthest edge, since sqrt(2)+2>sqrt(10), but I'm not sure where exactly. It seems it coudl be ((sqrt(2)+2)-sqrt(10))/2 down from B, but I'm not sure.

[Anonymous No.10562642]

>>10561454
So why did op say it wasn't B. Did he just lie to us?

[Anonymous No.10562674]

>>10562055
In your diagram: but AP is not the shortest path to P from A. There is even a shorter path to P along a different path which is even shorter than AB. Making point B still the longest point to reach.

[Anonymous No.10562681]

>>10562674
I retract that statement. AP is the shortest path to P.

[Anonymous No.10562790]

>>10562554
>i cant read

[Anonymous No.10562802]

>>10561175
How should I know how the bug decides to move?

[Anonymous No.10562973]

>>10561175
That's not a cube.

[Anonymous No.10563036]

Hopefully we all learned something ITT

[Anonymous No.10563076]

Give us a hint OP. Do you have to use calculus of variations to solve this problem?

[Anonymous No.10563080]

>>10562642
it ain't B
it be P >>10561505

[Anonymous No.10564692]

Not so sure about p but it is better than b for sure

[Anonymous No.10564711]

>>10563080
It would be P if the problem actually stated that the surface can't be left. It doesn't state that, so your answer is unequivocally *wrong*.

[Anonymous No.10564911]

>>10564692
Yeah its p

[Anonymous No.10564981]

>>10561454
How will op ever recover

[Anonymous No.10565141]

alright fags, funs over. so whats the solution?

[Anonymous No.10565454]

>>10561175
If the bug, I think it means a failsafe in a feature, is not used to deride a point in it, as a way to say that the bug can use an argument you have available to make it to another point and then use that to make itself more useful to the box as by showing it a new path itself, and thus teach it something new, that box is not a feature and they cannot complement each other, thus not B, but then if it can actually argue for a point and the movement is away from the box and suddenly the box can urge a point to it in time, the shortest path would be into the box across the thing and then the deferential of it should be a common number to from 0 to 1 and then use the issue that the bug is a feature to allow that more common numbers can exist in it as for floating point math. But the derision here is that that means you are being fed a problem and the solution is in the working parts of it being filtered to allow a civil discussion so you just plot to point out all the features and choose the shortest one to a ratio that isn't displayed. So that you can engage in the bug to feature dispute as the only non aryan in the room. This is Von Neuman, btw, not a simple physics equation.

[Anonymous No.10566261]

I could see from the diagrams that OP has given, the maximum distance lies somewhere in the neighborhood point P. Which lies on the top plane. We can label points on the top plane by the coordinates (x,y); with the point (0,0) right above point A.

The distance [math] D(P) [/math], which is the distance (length of the geodesic) from point A to point P, can be explicitly written as:
[math] D(P) = D(x,y) = \text{min} \left { \sqrt{(2+y)^2 + x^2}, \sqrt{(1+y)^2 + (3-x)^2} \right } [/math]

To find the point (x,y) that maximize the distance [math]D(x,y)[/math] on the domain [math] [0,1] \times [0,1] [/math], we take the partial derivatives and set them equal to zero:
[math] \frac{\partial D}{\partial x} = 0 [/math]
and
[math] \frac{\partial D}{\partial y} = 0 [/math].

The thing is I don't know how to take the derivatives of D(x,y). Once I figure out how to do this, We can solve for the exact location of point P on the top plane.

[Anonymous No.10566272]

I could see from the diagrams that OP has given, the maximum distance lies somewhere in the neighborhood point P. Which lies on the top plane. We can label points on the top plane by the coordinates (x,y); with the point (0,0) right above point A.

The distance [math] D(P) [/math], which is the distance (length of the geodesic) from point A to point P, can be explicitly written as:
[math] D(P) = D(x,y) = \text{min} \left \{ }\sqrt{(2+y)^2 + x^2}, \sqrt{(1+y)^2 + (3-x)^2} \right \} [/math]

To find the point (x,y) that maximize the distance [math]D(x,y)[/math] on the domain [math] [0,1] \times [0,1] [/math], we take the partial derivatives and set them equal to zero:
[math] \frac{\partial D}{\partial x} = 0 [/math]
and
[math] \frac{\partial D}{\partial y} = 0 [/math].

The thing is I don't know how to take the derivatives of D(x,y). Once I figure out how to do this, We can solve for the exact location of point P on the top plane.

[Anonymous No.10566276]

I could see from the diagrams that OP has given, the maximum distance lies somewhere in the neighborhood point P. Which lies on the top plane. We can label points on the top plane by the coordinates (x,y); with the point (0,0) right above point A.

The distance [math] D(P) [/math], which is the distance (length of the geodesic) from point A to point P, can be explicitly written as:
[math]D(P) = D(x,y) = \text{min} \left \{ \sqrt{(2+y)^2 + x^2}, \sqrt{(1+y)^2 + (3-x)^2} \right \}[/math]

To find the point (x,y) that maximize the distance [math]D(x,y)[/math] on the domain [math] [0,1] \times [0,1] [/math], we take the partial derivatives and set them equal to zero:
[math] \frac{\partial D}{\partial x} = 0 [/math]
and
[math] \frac{\partial D}{\partial y} = 0 [/math].

The thing is I don't know how to take the derivatives of D(x,y). Once I figure out how to do this, We can solve for the exact location of point P on the top plane.

[Anonymous No.10566533]

>>10561175
It's a time cube, it reaches ALL FOUR POINTS SIMULTANEOUSLY.

[Anonymous No.10566575]

Thus shit is not hard. There is controversy because half of the faggots here are assuming that the bug can fly and that the cuboid is empty and the other half asume that the the cuboid is solid and the bug must walk on its surface. OP's fault for not stating all the conditions at the beginning. Not a math problem but an interpretation one...

[Anonymous No.10567180]

>>10566276
I finally solved it.

If the top side of the box is the x - y plane.

The point (x,y) on the plane that maximizes [math]D(x,y)[/math] on the domain of [math] [0,1] \times [0,1] [/math], then that maximum is at x = 3/4 and y = 3/4.

I found the maximum using this approach:
1. set x = y = t, because the box is symmetric. down the line y=x.
2. set [math] (2+t)^2 + t^2 = (1+t)^2 + (3-t)^2 [/math], because I needed to see on which interval of t the first term the set dominates the second. And find where t is maximum on that domain. This turns out to be at t = 3/4.
3. Since t = 3/4 it follows that x= 3/4 and y = 3/4.

[Anonymous No.10567221]

>>10561175
Assuming A really is in the bottom corner and the drawing is just really bad (how difficult is it to draw a dot in the actual corner btw), then the furthest you can travel is 1*2, so any of the corners above and adjacent to or opposite A. Surely. And assuming the bug can't fly...

[Anonymous No.10567236]

[Anonymous No.10567239]

>>10567236

[Anonymous No.10567352]

>>10561175
The answer is A

[Anonymous No.10568733]

>>10561175
The answer is B because of the black hole centered at B which prevents the bug from ever leaving B. Since it's a black hole the bug will go to B before it reach's A, therefore there is an infinite set of greatest distances to A because all movement is futile and a net waste.

[Anonymous No.10570381]

>>10561175
Wait. If the bug is inside is it crawling or flying?

[Anonymous No.10571168]

the answer is to heckle the bug until his self esteem is destroyed, ruining his will to move.

[Anonymous No.10571249]

>>10561380
wrong