/sci/ - Science & Math

>>7227879
Define small.

>>7227879

>>7227879

Well it's 1/2 - (2tan(2arctan(1/2))^-1
but that's a very ugly answer IMO.

0.125

>>7227879
0,144935

>>7227979
>comma

>>7227980
old school

>>7227947
>mathfag comes up with some bullshit, doesn't even solve the problem
>>7227976
>engineer uses autocad for 10 seconds and solves the problem

>>7227980
non-murican comma

>>7227988
Engineeringfag assumes values.
Mathfag provides formula for all possible values.

Someone tell me why Im wrong, I feel retarded. Was it wrong to assume all 3 sides of the triangle are equal?

>>7227993
>The radius of the semicircle is 1.
also
>Well it's 1/2 - (2tan(2arctan(1/2))^-1
this evaluates to -1

>>7227998
nigger what are you doing, there is no such thing as a 1-1-2 triangle
you're thinking on 1-1-sqrt(2)

>>7228001
the red line in the middle should be sqrt 3

Anyways I realized why I'm retarded, there is a space left between the small circle and the top of the sqrt 3 line

>>7227999
Well no one said the formula was right, Godsarnit!

>>7227988
what so that top circle is the same size as the one below it?

>>7228019
What? The semicircle is defined as having radius 1. The circle inside of the semicircle has radius 0.5, you can see this geometrically. The small circle which we need to find the radius for, the one outside of the semicircle, has a radius of 0.125 when you solve it.

>>7227999
>this evaluates to -1
no, it evaluates to 1/8

Alright so the radius of the small circle is 1/8, why is this, intuitively?

>>7228069
finally someone asks the right question

>>7228069
The height of the triangle is $/frac{4}{3}$ the radius of the large circle. Now, if you imagine to add many circles over the two preexisting ones you see their diameters form a geometric progression. The first diaeter is 1, so the sum is $/frac{4}{3}=/frac{1}{1-x}$ where x is the ratio of the geometric (and, in this case, the diameter of the small circle).

It seems pretty complex, but it is really intuitive on a visual level.

Too bad I'm on my shitty phone and I can't draw.

>>7228082
I beg your pardon, I fucked up with the TeX. Let me rewrite with the right syntax.

The height of the triangle is <span class="math">/frac{4}{3}[/spoiler] the radius of the large circle. Now, if you imagine to add many circles over the two preexisting ones you see their diameters form a geometric progression. The first diaeter is 1, so the sum is <span class="math">/frac{4}{3}=/frac{1}{1-x}[/spoiler] where x is the ratio of the geometric (and, in this case, the diameter of the small circle).

It seems pretty complex, but it is really intuitive on a visual level.

Too bad I'm on my shitty phone and I can't draw.

>>7227998
sqrt(3)-1 is the distance between de medium circle and the top of the triangle, not the diameter of the small circle, so the radius must be less

>>7228094

FUCK BACKSLASH

>>7228069

The height of the triangle is <span class="math">\frac{4}{3}[/spoiler] the radius of the large circle. Now, if you imagine to add many circles over the two preexisting ones you see their diameters form a geometric progression. The first diaeter is 1, so the sum is <span class="math">\frac{4}{3}=\frac{1}{1-x}[/spoiler] where x is the ratio of the geometric (and, in this case, the diameter of the small circle).

It seems pretty complex, but it is really intuitive on a visual level.

Too bad I'm on my shitty phone and I can't draw.

>>7228097
I was wrong about that too, it't not an equalateral triangle

>>7228099
how do you know the triangle's height is 4/3's the radius?

(4/3-1/2) / (1/2) = (4/3-1-r) / (r)

5r/3 = 2/3-2r

(11/3)r = 2/3

r = 2/11

>>7228108
By a simple double-angle formula. Alternatively you can also use the bisector theorem, an even more elementary result - but that can take time.

I managed to draw a simple diagram. I'm sorry for the size of the pic.

>>7228131
Moron

>>7228105
How you know it? I'm confused then

What the fuck even is this shit? Are we assuming that the circle I'd tangential at all 3 intersections with the triangle? If so, this is basic trigonometry and the little radius is 1/6. But maybe I'm making too many assumptions.

(4/3-1/2) / (1/2) = (4/3-1-r) / r

5r/3 = 1/3-r

(8/3)r = 1/3

r = 1/8

>>7228142

>>7228142
Fuck I accidentally disregarded the space at the top. 1/8...

>>7227947
>>7227988
tan(2arctan(1/2))=2*(1/2)/(1-1/4)=4/3
1/2-(2*4/3)^-1=1/2-3/8=1/8

>>7228142
>>7228172
Why do you use the word "fuck" so much?

How to understand the arctan solution intuituvely?

>>7227988
>mathfag comes up with some bullshit
>engineer uses autocad while sucking on two dicks
ftfy

>>7228643
>2 hours later mathfags are still trying to solve the problem

I saved this from a thread on /sci/ back when it was first introduced. I was in high school calculus then, and couldn't figure it out. 5 years later, I still have no fucking idea how to solve this.

>find the total area of the grey circles, assuming the large semicircle has radius 1.

>>7228661
I have no idea if this is correct, and I have no idea if this is how you latex

\pi r^2 \sum_{n = 1}^{\infty} \frac{1}{2^n}

= (pi*r^2) * 1 / (1 - 0.5)
= 2pi*r^2
= pi/2

>>7228732
>I have no idea if this is correct, and I have no idea if this is how you latex

Last try at latex

<span class="math">
\pi r^{2} \sum_{n = 1}^{\infty} \frac{1}{2^n}
= (\pi*r^{2}) * 1 / (1 - 0.5)
= 2 \pi*r^{2}
= \frac{\pi}{2}
[/spoiler]

>>7228737
how did you arrive at that sum? The factor in the sum should represent the relative sizes of the grey circles; why is it 1/(2^n)?

I had no idea what I was doing before. I think I got it now. I still have no idea how to prove the largest grey circle has a radius of 1/4...

<span class="math"> \pi \sum_{n = 0}^{\infty} \left( \frac{1}{8} \right)^{n} [\math]

If a geometric series doesn't start at 0, will it suffice to use the a/(1-r) formula and then just subtract S_0?

If so, the answer is (8pi - 7)/7[/spoiler]

>>7228812

<span class="math"> \pi \sum_{n = 0}^{\infty} \left( \frac{1}{8} \right)^{n} [/spoiler]

= pi / (1-(1/8)) - 1
= (8pi - 7)/7

Pretty sure im wrong still

>>7228062
Yeah which is the answer that >>7227976 got using autocad. And like they teach you in physics, if two people get the same answer, that must be the correct answer.

>>7227879
Assuming that the point where the small circle and medium circle intersect is the point on the semi circumference such that the shortest possible path from the point to some point on the diameter of the semi circle is longer than the path of any other point on the semicircle's circumference, then the tangent line going through that point must be parallel to the semi circle's diameter (as the longest possible chord which is perpendicular to the semicircle's diameter is the radius), which is also parallel to the tangent lines going through the same point but corresponding to the medium circle and small circle, giving us that the diameter of the medium circle is also the radius of the semicircle (as the tangent lines passing through two different points in a circle are parallel iff those points are endpoints of the circle's diameter) and that the lowest point of the small circle's tangent line is parallel to the semicircle's diameter. Assuming this, the small radius is 1/5.

>>7227980
I hate this comment so much. Presumably an American is dissatisfied with the common European tradition of using the comma and expresses this with an annoying reaction image.
I just hate this so much.

>>7229429
bienvenue a la internette

>>7228636
Pythagoras.
Pythagoras.
Pythagoras.

>>7228812
Which problem are you talking about?

Anyway, no greatest gray circle ITT has radius 1/4

>>7229471
ah, j'ai ri

>>7228661
First step:

The radius of the largest gray circle is <span class="math">\frac{4}{9}R[/spoiler].

The rest of the problem is left to the reader.

>>7229644
Why is it 4/9 R?
How should a problem like this be treated?

>>7227879

>>7229731
Let <span class="math">x[/spoiler] be the length of the unknown radius.

<span class="math">OC=\frac{R}{2};
CK=\frac{R}{2}+x;
OK=R+x
KH=x [/spoiler]

Now we can express the area of the traingle OCK in two ways: by Heron's formua (LHS) and by the usual formula (RHS)

<span class="math">\sqrt{R\cdotx\cdot\left(\frac{R}{2}-x\right)\cdot\frac{R}{2}}=\left(\frac{R}{2}x\right)\cdot\frac{1}{2}[/spoiler]

Solving this simple equation yelds the desired result.

An important thing to consider in order to solve the whole problem is that the centres of the gray circles lay on a circle with centre halfway between O and C. This also is a simple result.

I'm sorry again for the size of the pic. Wish I could use my PC.

>>7229746

>>7229759
>>7229731
Forgot a semicolon between x and K.

Also

<span class="math">\sqrt{Rx\left(\frac{R}{2}-x\right)\frac{R}{2}}=\left(\frac{R}{2}x\right)\frac{1}{2}[/spoiler]

I'm a shame.

>>7229766
>>7229731

<span class="math">\sqrt{R x \left(\frac{R}{2} - x \right)\frac{R}{2}} = \left(\frac{R}{2} x \right)\frac{1}{2}[/spoiler]

>>7229759
>the centres of the gray circles lay on a circle with centre halfway between O and C

>>7229759
<span class="math">OK=R - x[/spoiler]

>>7229773
Are you a retard?

>>7229759

>An important thing to consider in order to solve the whole problem is that the centres of the gray circles lay on a circle with centre halfway between O and C.

It's an ellipse. The center is still between c and o tho.

>>7229800
Are YOU a retard?

>>7229816
Ok I was just testing whether you were an actual engineer. Turns out you are.

>>7229813
an ellipse with the centre halfway between o and c would intersect one of the circles somewhere in the left part of the figure

>>7229822
It intersecs both in their intersection.

>>7229821
I know cad is not the best way to prove the statement is wrong, but do you really believe it isn't wrong?

>>7229823
that would be a circle :^)

and as >>7229773 shows, such circle doesn't touch the grey circles' centres

>>7228661
doesn't it approach infinity?

the smaller grey circles seem to continue on, as they continue to decrease in size, so there are infinitely many smaller grey circles

or is it impossible for it to be infinite because of the finite area (the large semi circle) in which the grey circles are contained?

>>7229828
>doesn't it approach infinity?
please kill yourself

>>7229830
reported :^)

>>7229822
Think of the centre of the largest circle (the semicircle in pic) as the origin of a system of Cartesian axes. Then the equation of the ellipse is

<span class="math">\frac{\left( x + \frac{R}{4}\right)^2}{\left(\frac{3}{4}\right)^2} + \frac{y^2}{\left(\frac{2}{3}\right)^2} = 1[/spoiler]

>>7229840
Forgot
a and b are actually 3R/4 and 2R/3 respectively

>>7229824
No, I said I was kidding. It is true, the shape is an ellipse. The point is I did not use that fact in my solution, so I was mistaken like a fag.

>>7229840
>>7229844
that ellipse intersects the left circle (the one with the center in c), and still misses the rightmost grey circle's center

>>7229827

>that would be a circle

That is clearly false. It is an ellipse. The centre is halfway between O and C and the foci are each <span class="math">\frac{\sqrt{17}}{12}[/spoiler] from the centre, obviously on the greatest diameter. The intersection is then clear.

(Actually, you don't even need the characteristics of the ellipse to see why that is true)

>>7229765
how do you find PN= CN/CM times OM ?

and you write arcsin and sin ?

how do you find the formula for CM ?

>>7229859
y=0
<span class="math">\frac{\left( x + \frac{R}{4}\right)^2}{\left(\frac{3}{4} R \right)^2} = 1[/spoiler]

then x=-R or x=R/2

>>7229864

>how do you find PN= CN/CM times OM ?
because the small triangle and the large triangle are similar
PN : CN = OM : CM

>how do you find the formula for CM ?
i applied the law of sines
CM : sin CAB = AM : sin ACM

>>7229871
so what? you found the intersections between the ellipse and the x axis, not between the ellipse and the "c circle"

>>7229889
Well it turns out that X=-R is both on the C circle and on the O circle. Remember >>7229840 , O is centred in the origin

>>7229877
>because the small triangle and the large triangle are similar
can you name the two triangles ? ( I do not see them)

>>7229759
>>7229813
>>7229823
>>7229840
>>7229844
>>7229848
>>7229860
>>7229871
Protip: it's not an ellipse

>>7229902
true, but there is another intersection you forgot, try making a system with your ellipse and the left circle's equation

>>7229911
the one with CN as height and the one with CM as height

>>7229918
okay thanks

>>7229915
http://jwilson.coe.uga.edu/EMAT6680Fa11/Lee/asnmt07hylee/asnmt07hylee.html

>>7229917
What's the other interception? I see no such thing. You have the equations. Show me.

And still nobody solved the second problem. It's even a historical one.

>>7228005
No, you're thinking of a 30/60/90 triangle. This is a 45/45/90 triangle. Different rules.

>>7228005
>>7229984
My bad, I looked at it wrong. You're right.

>>7229968
what second problem? i only see one question in the OP

ellipse solved:
A = 3 / 4
B = √2 / 2

Circles are totally interesting.

>>7229352
tl;dr what the fuck is wrong with you to write a wall of text in response to a geometry problem, and your answer is incorrect, it's easy to check that in mspaint or with a ruler or someshit

>>7230028
>>7228661

>>7227879
Were can I get more problems like this?

>>7230231
http://www.mfdabbs.pwp.blueyonder.co.uk/Maths_Pages/SketchPad_Files/Japanese_Temple_Geometry_Problems/Japanese_Temple_Geometry.html

>>7227988
>>7227993
>>7228643
The real winner here is codingfag. Best of both worlds in this context.

>>7229429

>>7229828
Back to high school, retard.

>>7230266
Niiice, thank you so much

>>7230274
What?

>>7229429
Mericans gon be mericans

>>7229828
>>7230287
sniggered

>>7228661
the radius of the nth largest grey circle is
<span class="math">r_n=\frac{4}{(2n-1)^2+8}[/spoiler]

>>7230615
how do you figure that out though

>>7230615
So beautiful and elusive. I just want to understand the ways of these circles.

>>7227900
Underrated

>>7230645
>>7230671
Use the fact that the nth circle touches 3 other circles tangentially:
>The circle with radius 1
>The green circle
>And the (n-1)th circle (for n > 1)

Write those 3 statements using Cartesian coordinates (the distance bw the centres of two circles which touch tangentially is equal to the sum or difference of their radii, depending on how they make contact)

Do some simplifications and you'll get a long-ass relation bw r(n) and r(n-1). Using that and the fact that r(1) is 4/9, let Mathematica do the rest.
>inb4 matlab
No

>>7230795
I used Möbius transformation on the complex plane

Anyway, the historical problem I was referring here >>7229968 was in fact Descartes' fourth circles problem (Descartes was the one who solved it), which you solved as a part of your solution.

Solution in general for a semicircle of radius R and triangle having gradient m, then radius of small circle is

r = R (m-1)/(1+sqrt(1+m^2))

For special case of R=1 and for an equilateral triangle m=sqrt(3) then r=(sqrt(3)-1)/3

>>7230855
But the poin of the problem is that the triangle is tangent to the bottom circle, the one with radius 1/2

It's up to you to find the gradient.

(which is 4/3)

>>7230859
yes I see. You are correct. Thank you for pointing this out. Insisting that the discriminant of the quadratic equation is zero for common roots(tangent point) leads naturally to a gradient m=4/3

Thanks ^__^

>>7230681
Overrated now that you said that.

>>7230906
A-are you solving that with analytic geometry?

final answer is 1/8.
-second degree equation stuff to get the height of the triangle (4/3)
-homothetic figures to get small circle radius to middle circle radius ratio

>>7227879
semicircle has a radius of 1 , the bigger circle has a radius of 0.5. To get the diameter of the smaller circle you would have to subtract the diameter of the bigger circle from the height of the equilateral triangle, so ((rad3)/2) ×2)-(1))/2 is the radius of the small circle

>>7232159
Can someone confirm this?

>>7232179
Yeah I can confirm it's wrong. The diameter of the small circle is NOT equivalent to the height of the triangle minus the diameter of the large circle, and there is no equilateral triangle.

See >>7228148

>>7232190
shit, wasn't an equilateral traingle after all.

>>7227879
Circumference of small circle = r of big circle ???

>>7232298
No???

>>7227879
Math is dumb.

>>7232338
1/16th* typos are dumb

>>7229828
How the fuck can a finite area contain infinite area you imbecile

>>7232668
Why is /sci/ so aggressive to those who don't know their maths?