Nerd Sniping

What are some of your favorite nerd snipes? I got my friend good today with this one.

Other urls found in this thread:

mathworld.wolfram.com/Circle-CircleIntersection.html
twitter.com/NSFWRedditGif

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

retard

Fuck you I have homework tonight

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

it could be the same proportion of A is covered as long as a

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

Test

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

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?

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.

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.

circle A and circle B are not pictured

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

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.

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

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

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

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

My first guess was (a-b)/2

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

mathworld.wolfram.com/Circle-CircleIntersection.html

fml

How about:

x = b - (2a - b)

That feels comfy.

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

Whoops, I mean:

x = b - (b - 2a)

That's right, right?

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

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

x=0

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

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

I think we have a winner.

user Presents: "Failing Math With Style"

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

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

He gave me the idea and it works.

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

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 user 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.

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)

So... x = -2a?

You messed up.

Yeah.

wheres b

Kek

[math]x=\phi[/math]

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

...

Tbought this was an anime face

Use Euclids second prop

Same. This site has ruined me.

No solution. Assume a=10^{-10^{10^{10}}} and b=1/a, then it is clear that even when x=0 Area A

Monte-Carlo anyone?

Nevermind misread the problem.

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.

This is what a Slavic cooking image board does to you

x=3.93 for r=4 and r=2

x=3.6 for b=4 and a=3

What program is this?

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.

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

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]

agar.io maths?

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.

post solution then you silly homosexual

? 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.

I ment "Decreasing"

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.

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

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.

Just give me some values faggot

(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.

mathworld.wolfram.com/Circle-CircleIntersection.html

How does addition effect prime factors?

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.

Human Rng, roll!

a = 5
b = 10
c = ?

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]

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.

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

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

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

this is a literal middle schooler, folks

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

mathworld.wolfram.com/Circle-CircleIntersection.html

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?

coefficients

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

*slow clap. That was pretty good.

kek

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.

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

yes you fucking normaltard

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

test

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

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.

>test

Hi Python script kiddie bot.

...

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

What's a nerd snipe?

it's this