Nerd Sniping

Is this unsolvable? Don't you have to know how a relates to b beyond being less than it?

>>8645994
retard

Fuck you I have homework tonight

>>8645994

It can be done with integrals, dunno if there's a more elegant way.

>>8645994
it could be the same proportion of A is covered as long as a<b

Just from eyeballing it, I think 2a might be close to the answer.

Test

>>8646074
>eyeballing it
>drawing made in MS Paint
cool, I bet you could get the exact answer by counting the pixels in between them

>>8646021

Scratch that. Tried it out to see what the answer is, but to find the value for x involves solving a transcendental equation.

x = b - a little bit

I (probably wrongly) assume you could make a triangle with two sides the length of a and one side the length of x.

So you just use the Pythagorean theorem to find the length of x.

a^2 + a^2 = x^2

Is that right?

>>8646187
If you're utterly sure about your answer, try plugging it back into the original situation and see if it makes sense at all. Assign a and b some values- doesn't matter what. Make it worst case scenario and say that a=1 and b=1000. Clearly the length of x in such a situation could not be anywhere close to sqrt(2), so this proves that this is not the correct answer. I'm feeling mean today, so I would like you to know that you are a Certified Angus Brainlet and I don't know what you're doing here.

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>>8646197
But if you placed a copy of circle A over the centre of circle B, just eyeballing it, you should be able to make a triangle with two sides the length of a and one side the length of x.

It just feels right. Going to go ahead and say I'm right on this.

I'd like to amend this by stating that πa^2 < 2πb^2, since this could work for any circle B that is at least half the size of A. I doubt that really changes anything, but it's more inclusive.

>>8645927
circle A and circle B are not pictured

>>8646207
Pythagorean Theorem only applies to right triangles. It doesn't apply except when a and b are two specific numbers.

>>8646218
Oh right, so you just do the sin/cos/tan meme to figure out the length of x.

You already know two sides of triangle have the length of a.

>>8646093
>tfw you were trying to do classical geometry and you end up in real analysis or abstract algebra again
that feel

>>8646221
And now think: did you use circle b at all in your "solution"?

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>>8646228
Kek, yeah it would probably better to use that triangle and just forget about circle A.

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>>8646241
I think I'm onto something here, but we need to restart the Apollo program to be sure.

>>8645927

My first guess was (a-b)/2

Pretty close.. plus or minus an awfully ugly bit..

http://mathworld.wolfram.com/Circle-CircleIntersection.html

>>8646276
fml

How about:

x = b - (2a - b)

That feels comfy.

>>8645927
The funny thing is, you haven't defined which is circle A and which is circle B.

>>8646288
Whoops, I mean:

x = b - (b - 2a)

That's right, right?

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>>8646016
That hurt, anon. I've been having a really shitty day and you just made me feel worse.

>>8645927

Well... it did not say that it had to be exactly 50%, so...

x=0

>>8646310
>it did not say that it had to be exactly 50%
Yes it did

>>8646316

It just said 50% had to be covered. Covering more than 50% still covers 50%.

>>8646324
I think we have a winner.

>>8646324
Anon Presents: "Failing Math With Style"

[math]x = 2 \sqrt{a^2 - \tfrac{1}{2}a^2}[/math]

>>8646355
nice try, Pythagorean Theorem guy, but your formula won't fool me.

>>8646363
He gave me the idea and it works.

>>8646351

Thank you! Thank you! I knew being a lazy anon would pay off one day. Thanks mom and dad for being such great role models!

>>8645927

OP. Another Snipe that you can do on anyone is:

1. Tell them to pay attention and prepare to decode a message to learn the number you will present.
2. Put 2 spoons on the table in a cryptic fashion.
3. Ask anon to tell you what number you are trying to show them.
4. With the other hand put 3 fingers on the table.
5. The number is 3 but they will guess 2.
6. Repeat with variety.

This will infurriate the person that figures out the trick last.

>>8646097
and that little bit is the parameter that sets the circle segment left of the vertical diameter equal to the two rounded triangles on the right side (although I wouldn't know their equations or how to arrive there quickly)

>>8646301
So... x = -2a?

>>8646387

You messed up.

>>8646389
Yeah.

>>8646355
wheres b

>>8646324
Kek

>>8645927
[math]x=\phi[/math]

>>8645927
is this even possible? because I got something of the form
[eqn] \sqrt{P(x)} = \cos^{-1}\left(Q(x) + \sqrt{R(x)}\right)[/eqn]
with P,Q,R polynomials.

So unless some things cancel out exactly, this cant be solved for x

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>>8646446

>>8646456
Tbought this was an anime face

>>8646528
Same. This site has ruined me.

>>8645927
No solution. Assume a=10^{-10^{10^{10}}} and b=1/a, then it is clear that even when x=0 Area A<<< .5(Area B).

>>8646089

Monte-Carlo anyone?

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>>8646939
Nevermind misread the problem.

>>8646942
brute force doesnt count, I have to review for a class but the answer should have to do with circle-circle intersection. You might not even need any calc.

You could probably get me with a molecular cloning question. Something that involves a tricky cloning project where you need to carefully consider restriction sites and insertion methods.

>>8646528
>>8646917
This is what a Slavic cooking image board does to you

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>>8646942
x=3.93 for r=4 and r=2

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>>8647739
x=3.6 for b=4 and a=3

>>8647739
>>8647742

What program is this?

>>8647768
Geometer's Sketchpad. 3.83 is not a precise answer though. I'd be shocked if the actual answer didn't have pi in it somewhere.

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>>8647779
I know, but maybe it spreads some light on someone. I can't solve it for myself anyways.

>>8647768
GeoGebra

x=15.96 for b=16 and a=2. Last pic.

>equation for the small circle with center at origin
y1=sqrt(a^2-x^2)
>equation for big circle, with x0 as the distance between the centers
y2=sqrt(b^2-(x-x0)^2)
>intersection point
y1=y2
a^2-x^2=b^2-x^2+2x*x0-x0^2
x=(a^2-b^2+x0^2)/2x0

The integral of y1-y2 from x=-a to x=(a^2-b^2+x0^2)/2x0 is set equal to a quarter circle (pi*a^2)/4. Plugging the integral into wolfram alpha and plugging in the limits of integration (as well as a bunch of algebra) yields:
[eqn] \pi a^2= \frac{a^2-b^2+x^2}{4x^2} \sqrt{-a^4+a^2b^2+3a^2x^2-b^4+b^2x^2-x^4} +a^2tan^{-1} \Big( \frac{a^2-b^2+x^2}{ \sqrt{-a^4+a^2b^2+3a^2x^2-b^4+b^2x^2-x^4}} \Big) [/eqn] [eqn] + \frac{-a^2+b^2-9x^2}{4x^2} \sqrt{-a^4+a^2b^2-3a^2x^2-b^4+7b^2x^2-9x^4} -b^2tan^{-1} \Big( \frac{a^2-b^2+3x^2}{ \sqrt{-a^4+a^2b^2-3a^2x^2-b^4+7b^2x^2-9x^4}} \Big) [/eqn] [eqn] +(x-a) \sqrt{b^2-x^2+2ax-a^2} -b^2tan^{-1} \Big( \frac{x-a}{ \sqrt{b^2-x^2+2ax-a^2}} \Big) [/eqn]
If anyone wants to continue solving this, I'm not sure it's possible to isolate x with all those inverse tangents.

One arrives to the solution for [math]x[/math] by solving this equation:

[eqn]\sqrt{b^2-\left(\frac{a^2-b^2+x^2}{2x}\right)^2}\sqrt{a^2-b^2+\left(\frac{a^2-b^2+x^2}{2x}\right)^2} + a^2 \arctan\left(\frac{\sqrt{b^2-\left(\frac{a^2-b^2+x^2}{2x}\right)^2}}{\sqrt{a^2-b^2+\left(\frac{a^2-b^2+x^2}{2x}\right)^2}}\right) = \\ \left(\frac{a^2-b^2-x^2}{2x}\right)\sqrt{a^2-\left(\frac{a^2-b^2-x^2}{2x}\right)^2}-a^2\arctan\left(\frac{\frac{x^2-a^2+b^2}{2x}}{\sqrt{a^2-\left(\frac{a^2-b^2-x^2}{2x}\right)^2}}\right)-\frac{a^2-b^2+x^2}{2x}\sqrt{b^2-\left(\frac{a^2-b^2+x^2}{2x}\right)^2}-b^2\arctan\left(\frac{\frac{a^2-b^2+x^2}{2x}}{\sqrt{b^2-\left(\frac{a^2-b^2+x^2}{2x}\right)^2}}\right)-\frac{\pi}{2}(a^2+b^2)[/eqn]

[Left as exercise for the reader]

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>>8645927
agar.io maths?

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>>8645994

Yeah, exactly. There are not enough constraints for the solution to be very useful.

I guess if someone did try to create an equation, we would have a cool way of graphically representing bb-8 from star wars though.

>>8646016
post solution then you silly homosexual

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>>8647958
? Of course this problem has a unique solution. [eqn]f(h)=Area(A(h)\cap B)/Area(A(h))[/eqn] is a strictly monotonically increasing continuous function from [eqn][0,a+b]\to [0,1][/eqn] with [eqn]f(0)=1[/eqn] and [eqn]f(a+b)=0 [/eqn]. Thus there exists a unique solution.

>>8647993
I ment "Decreasing"

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>>8645927
Would you be so kind to give us some values for a and b?

I have the symbolic solution but it's too complicated for mathematica.

>>8648019

No, that's the point. Without knowing a and b, there aren't enough constraints for a meaningful solution.

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>>8648019
Let me explain some of the variables and the process.

a, b are as stated on the diagram, c is x, but I used c because x is the integration variable.

Assume the center of circle a is (0, 0), then b is centered at (x, 0). f1 is the equation for the top half of circle a, f2 is the equation for b. mid is the x-coordinate where the two circles intersect.

To calculate the area, we need to integrate f1 from c - b, the leftmost point of circle b, to mid, and then integrate f2 from mid to a. Adding those two integrals, setting them to a^2/2, and solving for c should get the answer.

Also, made a little typo in the last one for f2. It's fixed now.

>>8648032
Just give me some values faggot

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>>8647963

(Area of Small Circle - Area of Lens)/(Area of Small Circle) = 0.5 where a < b

The equation depends on a, b, and x. That's 3 variables and 1 equation.

There needs to be more constraints for the solution to be very useful. Otherwise, you could just change the size of one of the circles, and get a new x.

http://mathworld.wolfram.com/Circle-CircleIntersection.html

How does addition effect prime factors?

>>8648047
Think the point is to solve for x in terms of a and b. Under your logic formulas like mv^2 / 2 are useless because it's 2 variables and 1 equation.

>>8648046

Human Rng, roll!

a = 5
b = 10
c = ?

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>>8648052
Mathematica shit confirmed.

I'm going to bed. If anyone else wants to try here's my code:

a = 5
b = 10
f1 = (b * b - (x - c) ** 2) ** 0.5
f2 = (a *a - x * x) ** 0.5
mid = (c * c + a * a - b * b) / 2 / c
i1 = Integrate[f1, {x, c - b, mid}]
i2 = Integrate[f2, {x, mid, a}]
Solve[i1 + i2 == Pi * a * a / 2, c]

>>8648050

The point is that this thread is called "Nerd Sniping" because nerds regurgitate what they have been taught instead of using critical thinking.

I already conceded that you will have a cool equation for bb-8. Now leave me alone.

Kudos for finding the bb-8 eq.

>>8648047
1 variable 1 equation. He told us that a and b are constant.

>>8648057

Arbitrary constants... that change... depending on what you choose... sometimes called variables...

With your logic, x is a constant too ;-P.

>>8648059

this is a literal middle schooler, folks

>>8648059
you clearly need to solve for x in an algebraic closure of R(a,b) you idiot

http://mathworld.wolfram.com/Circle-CircleIntersection.html

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>>8648066

You are just butthurt that you took all night/day solving this stupid equation when it took me 10 minutes.

Let me ask you, what do you think a, b, and c are in the quadratic equation?

What word would you use to describe these unknown values?

>>8648079
coefficients

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>>8648070

Okay, I am done talking to a baby. When you get past Algebra 1, give this problem another try.

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>>8648085

*slow clap. That was pretty good.

>>8646528
kek

>>8648019
Try doing it for a=b=1. Then try for b=2,5,10,100, etc. I'm curious to what this function looks like. As b approaches infinity, x should approach b.

>>8645927
so to be a nerd I have to study euclidean geometry I will never use on top of doing my regular coursework?

>>8648160

yes you fucking normaltard

>>8648160
>to be a nerd I have to study euclidean geometry
so to be a nerd I *get* to study euclidean geometry
ftfy

test

>>8648170
>>8648179
I mean, I read euclids elements in highschool up to like prop 30 or something then considered it a waste of time. It's interesting, I'd like to though descartes' "geometry" one day

>>8648056
Actually, nerd sniping is a term coined by Randall Munroe for introducing a nerd to a problem like this and watching them drop everthing to try (and usually fail) to solve it. In fact, OP nerd sniped nearly everyone here by making them debate this problem rather than discussing nerdsniping.

>>8648180
>test

Hi Python script kiddie bot.

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>>8646258

>>8648184
wow you decided something wasnt worth your time so that you could seem smarter? Thats fucking awesome!

>>8645927
What's a nerd snipe?

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>>8648379
it's this

>>8645927
x increases as a and b increases, there is not a actual relationship between them. Problem solved.

>>8648387
Oh damn kek
>xkcd
Still kinda funny tho

>>8648387

Eh, just realized this applies to me, fuck

>>8645927

Oh you got me good OP

>>8645927
In the limit b>>a it is clearly: x = b-0.5a

If the limiting case for a=b is solved, then the derivitaves of both limits can be found, and extrapolated from there.

>>8648630
If a=b then
x = a * 0.807945506599034418637923480132630885804471929148196844500195203467741099942590707002486780330445457418982...

Where are circles A and B?

>>8645927
My very first instinct is to try some sort of convolution integral here but I have partial diff. and econometrics finals to study for

>>8648632

how can a=b if a<b

>>8648667
That's a good question! I don't know how OP expects anyone to infer what circle A and B are. I bet OP is a jew and didn't want to pay for circle A and B, and thus this whole problem is unsolvable. Case dismissed!

>>8649601

kek

>>8648698
They're talking about limits, but without a proper notation.

>>8649658

how hard is it to write as a->b
why the fuck would you write a=b when you mean a->b
fucking hell

set the area integral for the intersection equal to .5 * pi * a**2 and solve for x, right?

>>8649658
You don't even have to take limits here.
The problem still has unique solutions for [math] b \leq a < \sqrt{2} \, b [/math]

>>8649707
With a being bigger than b it's still just as hard as solving when a is smaller than b. It only helps when a=b. For that you'd just have to find the secant line across their intersection and find when the sector that it makes equals π/4a^2

>>8649707
Please, can you solve it?

>>8649732
x is the solution of

[eqn] \frac{1}{2} \pi a^2 = a^2 \cos^{-1}\left( \frac{x^2 + a^2 - b^2}{2 x a} \right) + b^2 \cos^{-1}\left( \frac{x^2 + b^2 - a^2}{2 x b} \right) - \frac{1}{2} \sqrt{(-x+a+b) (x+a-b) (x-a+b) (x+a+b)} [/eqn]


The only way to solve this equation is numerically.

>>8649797
>The only way to solve this equation is numerically.
Brainlet alert

as b approaches infinity and/or as a approaches zero, x approaches b

>>8650571
good test for solutions.

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Heres a numerical solution

>>8649797
Devide a,b,x by b then in terms of r = a/b its
[eqn]\frac{1}{2} \pi r^2 = r^2 \cos^{-1}\left( \frac{x^2 + r^2 - 1}{2 x r} \right) + \cos^{-1}\left( \frac{x^2 + 1 - r^2}{2 x} \right) - \frac{1}{2} \sqrt{(-x+r+1) (x+r-1) (x-r+1) (x+r+1)} [/eqn]

then you can solve for x(r)

>>8648593
feels good man

another step of simplification get you:
[eqn]r^2 \sin^{-1}\left( \frac{x^2 + r^2 - 1}{2 x r} \right) = \cos^{-1}\left( \frac{x^2 + 1 - r^2}{2 x} \right) - \frac{1}{2} \sqrt{(-x+r+1) (x+r-1) (x-r+1) (x+r+1)} [/eqn]

>>8651019
What program would I use to view this in equation form?

>>8650767
If you plug in r=a/b those two equations dont match up though, at least not in the cosine parts. Is it even possible to solve a function like this?

>>8645927
tfw we're all getting nerd-sniped by OP

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>>8645927
Suppose you have a number of congruent circles in a plane. If you have a direct line of sight between two circles then the area that you can see from one circle is the half facing the other. Prove that no matter how many circles there are the total amount of area that cannot be seen from any circle is a constant.

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>>8652565
Is this not just a simple extremal principle question? Ie you just need to consider a "perimeter" traced by the outermost circles in each direction, as all of those inside are visible from all angles, then consider the region of each of these "outer" circles which cannot be seen by any other.

Intuitively it should be a whole circle, so maybe you could introduce a large circle which contains all of the smaller circles, and consider the regions of the circumference of these circles which will first be intersected by a radial line drawn from the circumference of the larger circle to its centre.

If that doesn't prove fruitful, I'd construct an induction on the number of circles: if a newly added circles lies within our old perimeter, then it is entirely visible, and our total is unchanged. Else, we can consider the two circles in the perimeter between which it will be placed. Simple Euclidean geometry should show the total visible area is unchanged.

Certainly the cases for n=1,2,3 seem fairly simple.

Anyway, I'm meant to be writing my dissertation not shitposting on /sci/, but your nerd snipe has proven quite effective. I'll leave the details to someone with more time on their hands

>>8652565
this is just basically a restatement of the polygon exterior angle theorem

>>8652831
>>8652837
This is a good point the "extremal" set of circles will naturally define the certifies of a convex polygon, to which the theorem regarding the angles therein can be applied

>>8652565
For n circles
Total unseen angle = 2[pi]n - [pi](n-2) - [pi]n = 2[pi]

>>8652860
Nice proof you've got there

>>8652863
I left it as an exercise for the reader.

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>>8652872
>>8652872

>>8652390
you plug in r=a/b those two equations dont match up though

i re-scaled x as well and forgot / didnt want to use another variable, replace r -> a/b and x -> x/b and it matches up.

>Is it even possible to solve a function like this?
I dont think so, not even a simplified one could be solved by mathematica.