UJRefugees.net

Posted: **Sat Oct 03, 2020 4:39 pm**

(or "maths" to AH...)

Draw a circle of radius 1 between the line of the x axis and the line formed by the function y=x, with the circle tangent to both lines. Then draw a smaller circle tangent both to each line as well as the first circle. Continue drawing smaller circles in the same fashion down to infinity. Sum up the circumferences of all the circles (C), and then sum up the areas of all the circles (A). Find the answer in the form C squared divided by A.

hints: the radius of each subsequent smaller circle is directly proportional to the radius of the next larger circle in a very simple way. The center points of all the circles are colinear.

Just the answer is good enough, you don't necessarily need to show your work. But don't google the solution!

Posted: **Sun Oct 04, 2020 12:32 pm**

in·ter·est·ing
/ˈint(ə)rəstiNG/
adjective
arousing curiosity or interest; holding or catching the attention.

Sorry, QD. Not for me.

In one of the Tier 1 shops I used to supervise we had this guy we would call in when we had computer/programming issues with our robots if our skilled trade guys couldnt figure out. He was CRAZY expensive but he solved our problems 100% of the time. While shooting the shit with him one day he mentioned that he likes to solve math problems for "fun".

Posted: **Wed Oct 14, 2020 1:31 am**

if no one has solved this, i may give it a go. I haven't read any answers.

Posted: **Wed Oct 14, 2020 1:33 am**

Here is where I punched out:

“line of the x axis...”

Posted: **Wed Oct 14, 2020 1:36 am**

HighNDry wrote: ↑Wed Oct 14, 2020 1:33 am Here is where I punched out:

“line of the x axis...”

well, i have the first circle plotted. Now i'm trying to figure out how the ratio of the first circle to the 2nd circle. If I can figure that out, i think the solution will be simple.

Posted: **Wed Oct 14, 2020 1:39 am**

Flumper wrote: ↑Wed Oct 14, 2020 1:36 am
HighNDry wrote: ↑Wed Oct 14, 2020 1:33 am Here is where I punched out:

“line of the x axis...”
well, i have the first circle plotted. Now i'm trying to figure out how the ratio of the first circle to the 2nd circle. If I can figure that out, i think the solution will be simple.

You’re a better man than I, Good Will Flumping.

Posted: **Wed Oct 14, 2020 4:19 pm**

Okay, so it looks to me that each circle becomes smaller by a proportion of 0.446463 (I could give the formula, but its complicated).

So now we have a series of C and A to calculate. In order of larger circle to smaller (infinity)

And then all we have to do is add up the C and A. Then Square C and divide by A.

I get 32.837508895265000000

Posted: **Wed Oct 14, 2020 5:12 pm**

HighNDry wrote: ↑Wed Oct 14, 2020 1:39 am Good Will Flumping.

I think he is more like the retarded blind kid from India but Texan.

Posted: **Wed Oct 14, 2020 8:07 pm**

While, we wait on QD, I will post the forumula that I got for comparing the circles as they get proportionately smaller and smaller.

[ (sin 33.75) (cos 56.25) (2) (r) ] / [ (sin 33.75) (cos 56.25)(2)(r) + (sin22.5)(2)(r) ]

that will reduce to 0.617317 / 1.382683 = 0.446463

admittedly, my biggest problem is i get in a huge hurry and generally make a glaring mistake.

Posted: **Wed Oct 14, 2020 9:46 pm**

Algebra and Calculus, I can make logic of.
Trig, not so much.

Posted: **Wed Oct 14, 2020 9:52 pm**

DiverTexas wrote: ↑Wed Oct 14, 2020 9:46 pm Algebra and Calculus, I can make logic of.
Trig, not so much.

I could have it wrong, but i'm sure i made it harder than it has to be.

Posted: **Wed Oct 14, 2020 10:19 pm**

HighNDry wrote: ↑Wed Oct 14, 2020 1:39 am
Flumper wrote: ↑Wed Oct 14, 2020 1:36 am
HighNDry wrote: ↑Wed Oct 14, 2020 1:33 am Here is where I punched out:

“line of the x axis...”
well, i have the first circle plotted. Now i'm trying to figure out how the ratio of the first circle to the 2nd circle. If I can figure that out, i think the solution will be simple.
You’re a better man than I, Good Will Flumping.

Posted: **Thu Oct 15, 2020 2:50 am**

Flumper, thanks for working on it. And yea, you're making it a little harder than you need to. First, don't worry about using a calculator, we don't need a "number". This isn't a multiple choice chemistry exam. We only need the answer in a calculable form.

But I am happy to tell you your answer as a numerical solution is remarkably close, to the first decimal place.

Another hint, and you've probably already seen it, is that radius r of the first drawn circle beside the given original circle has a radius less than 1. Due to the linear equation to give the other line that all the circles touch, and how it cuts through the graph, shows a simple symmetry. Once you know r-sub1, you know all the radius of every smaller circle. There is a simple symmetrical relationship between the radius of a smaller circle and the radius of the next larger circle. So you only have to figure out that first radius in relation to the radius of the given circle. Which is unit. A lot of stuff drops out, leaving a relatively simple formula in calculable form. QED and shit.

Posted: **Thu Oct 15, 2020 2:55 am**

QillerDaemon wrote: ↑Thu Oct 15, 2020 2:50 am Flumper, thanks for working on it. And yea, you're making it a little harder than you need to. First, don't worry about using a calculator, we don't need a "number". This isn't a multiple choice chemistry exam. We only need the answer in a calculable form.

But I am happy to tell you your answer as a numerical solution is remarkably close, to the first decimal place.

Another hint, and you've probably already seen it, is that radius r of the first drawn circle beside the given original circle has a radius less than 1. Due to the linear equation to give the other line that all the circles touch, and how it cuts through the graph, shows a simple symmetry. Once you know r-sub1, you know all the radius of every smaller circle. There is a simple symmetrical relationship between the radius of a smaller circle and the radius of the next larger circle. So you only have to figure out that first radius in relation to the radius of the given circle. Which is unit. A lot of stuff drops out, leaving a relatively simple formula in calculable form. QED and shit.

yeah, i know how the graph of it all works. basically you have a 45 degree angle with a 2 inch diameter circle rolled into it. then on the other side, smaller and smaller circles that fit down to infinity. I get the construction of it all. But, I think my answer is right. although, it didn't reduce to a simple equation. But, I only gave my answer to 2 decimal places. I have it out to 7 or 8 decimal places.

Posted: **Thu Oct 15, 2020 3:01 am**

another way to get to the "ratio" of how the circles get smaller and smaller compared to each other is to:

[1 - sin(22.5) ] / [1 + sin(22.5) ] = 0.446463. At least I think that's right (going from memory).

So first circle has radius of 1.
2nd circle has radius of 1 x 0.446463
3rd circle has a radius of 1 x 0.446463 x 0.446463.
4th circle has a radius of 1 x 0.446463 x 0.446463 x 0.446463

nth circle has a radius of 1 x (0.446463)^(n-1)

Posted: **Thu Oct 15, 2020 3:06 am**

I don't find any of this interesting... What's wrong with you nerds

Posted: **Fri Oct 16, 2020 1:55 am**

I did invest a little bit of time into this thread, so I would appreciate at least some response.

I still don't understand what I have wrong. The way you worded the question, concerns me a bit. You said:

Sum up the circumferences of all the circles (C), and then sum up the areas of all the circles (A). Find the answer in the form C squared divided by A.

So, you aren't asking to square the circumference of each circle and add those numbers. You are asking to add up all of the circumferences and then only square the sum of them all. That's what I did. And then divided by the sum of all of the Areas.

Posted: **Fri Oct 16, 2020 2:04 am**

Posted: **Fri Oct 16, 2020 2:13 am**

You lost me at math

Posted: **Fri Oct 16, 2020 2:31 am**

Stapes wrote: ↑Fri Oct 16, 2020 2:13 am You lost me at math

LOL... I’m with you 100%. This thread’s giving me a headache.

Posted: **Fri Oct 16, 2020 5:25 am**

One solution...

Posted: **Fri Oct 16, 2020 2:14 pm**

That is exactly what I got.

32.84.

4pi * [sqrt ( 4 + 2 * sqrt(2) ) ] = 32.84.

Actually, my answer was 32.8375089, I just didn't post all of the digits.

Posted: **Fri Oct 16, 2020 2:20 pm**

Flumper wrote: ↑Fri Oct 16, 2020 2:14 pm That is exactly what I got.

32.84.

4pi * [sqrt ( 4 + 2 * sqrt(2) ) ] = 32.84.

Actually, my answer was 32.8375089, I just didn't post all of the digits.

I think you answered him before his next reply. He was asking for the equation in that last response, and you gave it to him but he didn't respond to that yet so you may have gotten what he was looking for. Because I believe you first just posted the answer of 32.84, but I think he was looking for the equation. That's why he said you got the 32.8 right.
I'm thinking, but I could be wrong

Posted: **Fri Oct 16, 2020 2:28 pm**

necronomous wrote: ↑Fri Oct 16, 2020 2:20 pm
Flumper wrote: ↑Fri Oct 16, 2020 2:14 pm That is exactly what I got.

32.84.

4pi * [sqrt ( 4 + 2 * sqrt(2) ) ] = 32.84.

Actually, my answer was 32.8375089, I just didn't post all of the digits.
I think you answered him before his next reply. He was asking for the equation in that last response, and you gave it to him but he didn't respond to that yet so you may have gotten what he was looking for. Because I believe you first just posted the answer of 32.84, but I think he was looking for the equation. That's why he said you got the 32.8 right.
I'm thinking, but I could be wrong

Well, I don't have a way to post pictures and making mathematical signs and symbols and equations is impossible with just a keyboard.

and, I took him at his original post where he said

Just the answer is good enough, you don't necessarily need to show your work.

Posted: **Fri Oct 16, 2020 3:08 pm**

i got lost right after draw a circle.

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