/sci/ - Science & Math

There is no x dumbass

Alternatively, find the *real* Pac Man.

>>2966948
There's one in OP's post, you idiot.

i found it

>>2966926
>find x

>x

/thread igentog nigger

here it is: x

It's easy, it's one.

We know that on the big circle, r squared is 6, and on the circle with 2 in radius, radius squared is 4. So what's left in the little circle with r (which I think is your x) is 2, so the r is 1.

>>2966926
OP is obvious woman.

x is in OP's post
r is approximately .85?

>>2967153
Leave

3^2=6?

In any case, r=1 can't be true since r is definitively smaller than the radius of the third circle which is already 1.

The problem is easy if we know that the triangle connecting three of the centers of the circle has an orthogonal angle. but I don't know.

I'll try to solve it without using that assumption.

there was a facinating thread about this ages ago but i cant find the god-damn picture! i am so pissed off right now. ive just searched through my old notes and ive no fucking idea what i did, but i think the answer may be <span class="math"> \frac{\sqrt13 - 1}{3} [/spoiler]
= 0.87

can anyone confirm?
also, i forget the method, i'll try and work it out now.

>>2967180

>>2967203
>Be stupid
>Claim to be trolling
>Every day

>>2967204
You so mad, I don't even know why I asked the question.

>>2967216
>Mad
Nope.

>>2967183
>orthogonal

Good god, just say "right triangle."

>>2967185
Hints inc.

lets call the four circles S3, S2, S1 and Sr, where the number is the radius, obviously.
I'll flip the coordinate system and shift it to the centre of S2. And I'll call the unknown quantity x for OPs sake.

S3 viewed from the S3-centre coordinate system is implicitly parameterized by
X^2 + Y^2 = 3^2

from the centre of S2 we have that S3 is therefore
X^2 + (Y+1)^2 = 3^2 (shifted by 1 in y direction)
so
X^2 + Y^2 + 2Y = 8

set
X=s*sin(t)
Y=-s*cos(t)
(the coordinates in the picture)

then
X^2 + Y^2 + 2Y = 8
becomes
s^2 - 2 s cos(t) = 8
solving that...
s of the big circle viewed from the centre of S2 is
s(t)=cos(t)+squrt(8+cos(t)^2)

then x(t) is that minus 2 (the radius of S2) divided by 2.

so the triangle has lenghts:
a=2+x(t)
b=3
c=1+x(t)

now the Law of cosines holds

c^2 = a^2 + b^2 - 2ab cos(t)

plug in a, b and c as functions of t.

give it to wolfram alpha.

???

Profit.

>>2967249
Find a relation between theta and k (equation of a half-circle)
>>2967254
Find a relation between theta and r (law of cosines)
>>2967262
The relation between theta and r

three equations, three variables.

oops. I mean this pic, not emma

>>2967273
>The relation between theta and r
K and r, rather.

ITT: Say a number that means nothing as a solution to a problem that you created yourself.

well I thought the centre of the circle with x lies on half the distance between the big and the second big circle, but that turns out to be false - it's more complicated.
anyway here the approximation as seen in the pic

r = 0.86

mathematica-code:

x[s_, t_] := s Cos[t]
y[s_, t_] := s Sin[t]
s /. Solve[x[s, t]^2 + (y[s, t] + 1)^2 == 3^2, s] // Simplify

s[t_] := -Sin[t] + Sqrt[8 Cos[t]^2 + 9 Sin[t]^2]
r[t_] := (s[t] - 2)/2
a[t_] := 2 + r[t]
b[t_] := 3
c[t_] := 1 + r[t]

t /. NSolve[
c[t]^2 == a[t]^2 + b[t]^2 - 2 a[t] b[t] Cos[t - 3 \[Pi]/2], t]

(* angle T in \[Degree] and r[T] *)
T = -0.924;

Show[ContourPlot[{x^2 + y^2 == 2^2, x^2 + (y + 1)^2 == 3^2,
x^2 + (y + 3)^2 ==
1^2, (x - a[T] Cos[T])^2 + (y + Abs[a[T] Sin[T]])^2 == r[T]^2}
, {x, -3, 3}, {y, -4, 2}],
Graphics[{Red, Opacity[.5],
Polygon[{{0, 0}, {a[T] Cos[T], a[T] Sin[T]}, {0, -3}}]}],
PolarPlot[{a[t]}, {t, 0, 2 Pi}]]
r[T]