/sci/ - Science & Math » Thread #12472938

31KiB, 400x400, radius.png

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Anonymous Wed 16 Dec 17:57:17 2020 No.12472938 View Reply Original Report

Quoted By: >>12472948 >>12473010 >>12473096 >>12473122 >>12473256 >>12474014

Looking for puzzles like this if anyone has any. Or if you guys have anything similar. Riddles work also.

ALSO FEEL FREE TO TRY AND SOVE IT!

Anonymous

View Same Google iqdb SauceNAO Inner radius.jpg, 13KiB, 300x299

Anonymous Wed 16 Dec 2020 18:01:02 No.12472948 Report

Quoted By: >>12473057 >>12473096 >>12474573

>>12472938
Here is another to try and keep y'all stimulated during the holidays.

Find the radius of the inner circle.

Anonymous

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Anonymous Wed 16 Dec 2020 18:23:46 No.12473010 Report

Quoted By: >>12473036 >>12473096 >>12473164

>>12472938
let me guess, it has something to do with this, right?

Anonymous

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Anonymous Wed 16 Dec 2020 18:30:48 No.12473036 Report

Quoted By:

>>12473010
I don't.... How....

Anonymous

Anonymous Wed 16 Dec 2020 18:37:44 No.12473057 Report

Quoted By: >>12473164

>>12472948
That should be the extra length of the hypotenuse, right? So if the big circles are 1, the little circle is ?2-1?

Anonymous

Anonymous Wed 16 Dec 2020 18:49:21 No.12473096 Report

Quoted By: >>12473111

>>12472938
>>12472948
>>12473010
how is it possible to solve this at all with no numerical data?

Anonymous

Anonymous Wed 16 Dec 2020 18:55:01 No.12473111 Report

Quoted By:

>>12473096
just assume that radius or side is "a" or equal to one. this about finding a formula to describe something out of something else, in this case the inner circle by the side of a square

Anonymous

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Anonymous Wed 16 Dec 2020 18:58:22 No.12473122 Report

Quoted By:

>>12472938

Anonymous

Anonymous Wed 16 Dec 2020 19:09:19 No.12473164 Report

Quoted By: >>12473190

>>12473010
nice
>>12473057
you want the radius of the little circle in terms of the size of the square
I think its
$\dfrac{2-\sqrt{2}}{4\sqrt{2}}$

Anonymous

Anonymous Wed 16 Dec 2020 19:16:13 No.12473190 Report

Quoted By:

>>12473164
If the big circle radius is 1 then the square side is 4 so (?2-1)/4 which is the same value that you wrote

Anonymous

Anonymous Wed 16 Dec 2020 19:33:10 No.12473256 Report

Quoted By: >>12474272

>>12472938
cant you just scale this up or down, and r could be anything?

Anonymous

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Anonymous Wed 16 Dec 2020 22:47:55 No.12474014 Report

Quoted By: >>12474272 >>12474609 >>12474657 >>12474661 >>12476364

>>12472938
Does this count?

Anonymous

Anonymous Thu 17 Dec 2020 00:21:53 No.12474272 Report

Quoted By:

>>12473256
Then you set the side length as a function r(a).

>>12474014
Yes please

Anonymous

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Anonymous Thu 17 Dec 2020 01:55:32 No.12474573 Report

Quoted By:

>>12472948
$2r + \frac{s}{2}= \sqrt{2}\frac{s}{2}$
$2r = \sqrt{2}\frac{s}{2} - \frac{s}{2}$
$2r = \frac{s}{2} (\sqrt{2}- 1)$
$r = s \left(\dfrac{\sqrt{2}- 1}{4}\right)$

Anonymous

Anonymous Thu 17 Dec 2020 02:12:54 No.12474609 Report

Quoted By: >>12474674

>>12474014
can you do it for the general case where the height of the cone is a multiple of it's base radius?

Anonymous

Anonymous Thu 17 Dec 2020 02:33:23 No.12474657 Report

Quoted By: >>12474674

>>12474014
goddamn there's some algebra (too much!) for this one if im not mistaken

Anonymous

Anonymous Thu 17 Dec 2020 02:36:55 No.12474661 Report

Quoted By: >>12474674

>>12474014
then after all the algebra you need to solve a 4th order polynomial! lol
THere's gotta be an easier way!

Anonymous

Anonymous Thu 17 Dec 2020 02:43:35 No.12474674 Report

Quoted By:

>>12474609

I am not aware of the general case sorry anon.

>>12474657
>>12474661

It has been a while since I've solved this problem. From what I remember, there's not a terrible amount of algebra involved and I definitely didn't have to solve a fourth order polynomial. All I can really recommend is to keep it as simple as possible. For me I took more time thinking about how to approach the problem than actually computing the answer.

Anonymous

Anonymous Thu 17 Dec 2020 16:16:28 No.12476364 Report

Quoted By:

>>12474014
The trick here is that the circle's diameter lines up with the square's diagonal, not its edge. So instead of the intuitive, rational number ratio 1:3::1-x/2:x -> x=6/5, we get the fake and edgy "metanumber" ratio 1:3::1-x/?2:x -> x=(9?2-6)/7

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