Solve for the radius of the three cornered circles

Thanks to Umesh for the suggestion! If the large circle has a radius equal to 1, what is the radius of each small circle?
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[@pkmath12345]

Tangent points being collinear gives an important point to develop the work. Great job! Very neat and nice solution for sure

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@chandrahasreddy1729]

Aren't any two circles touching each other have collinear tangent points?

[@petey5009]

I love how these questions seem super complicated to solve but the actual math that goes into it is relatively simple

[@attsealevel]

which is why math is so important in our schools. it simply teaches students how to break down any problem, so that it's more easily solved. something the us has fallen woefully behind I'm afraid. as an engineer, who works and teaches in both boston and munich, i can attest to this fact.

[@salbahejim]

This is the fun part of math. Some amazing solutions can be derived from the simplest breakdown into common formulae. The challenge is figuring out how to do this, and that's where the fun lies!

[@TimoRutanen]

A good way to solve these kind of things is just starting to calculate all the stuff you can in the picture, and soon you will find a way to the solution. Or you don't, but often you do.

[@HWEWSWEW]

That is the difference between IQ and math knowledge. You can acquire a ton of math knowledge with a relatively average IQ, but no amount of studying math will ever make these types of questions easy, because it’s a totally different measurement.

[@quigonkenny]

And how a seemingly almost random placement of the circles gets you such a mathematically "clean" ratio. If that's one circle inscribed in the corner, the ratio is 1:(3-2√2), by comparison.

[@jimlocke9320]

At about 4:05, Presh eliminates the solution r = 1 by claiming that the small circles must have a smaller radius than the large circle. Another way to rule out r = 1 is to compute the value of 1 - 3r, the length of the bottom side of the right triangle. When r =1, that length turns out to be -2, which is invalid because triangle side lengths must be positive.

[@pieters286]

yes! this is a better test than 'claiming' r=1 is not a solution because same ss big circle.

[@pieters286]

the implicit limit is all sides >= 0

[@mmmmmratner]

I was thinking about the r = 1 solution, and it occurred to me that if you listen to the instructions in the video without watching the video, you may draw the fourth circle coincident with the first circle. In that case, following along the solution method yields a right triangle of base 3r - 1, height 1 - r, and hypotenuse r + 1, which has the same two solutions to the quadratic Pythagorean theorem, but this time r = 1 is the correct answer.

[@neoaureus]

True

[@VinyJones2]

@@mmmmmratner nice finding, thank you

[@FoxMcCloudV2]

You can also solve this problem by noticing, in the right triangle you mentioned, that the side lengths display an arithmetic progression. Since the only such right triangles are 3:4:5 right triangles (this is easily proven), we have (1 + r)/5 = (1 - r)/4 = (1 - 3r)/3, which is basically three equations in one. Solving any of these equations yields r = 1/9 as the only solution.

[@rahulvarma3248]

Wow!

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@jollyjoker6340]

If you call the smaller circle's radius 1 and the larger r and the triangle 6-8-10 instead of 3-4-5, you have r+1=10, r-1=8, r-3=6, making it pretty easy to see r=9.

[@highpath4776]

Now is it significant that three = number of small circles is in a ratio to 9 ( or 1/9th), so if you added more small circles to the bottom increasing height of them and distance out and infills, and then drew another circle (1) of unit 1, would the ratio (and r for smaller circles ?) remain at 1, or is it impossible to draw a larger circle still tangental to the external joining tangent of the small circles and be tangent to a small circle ? or (2) of unit less than 1 , would the ratio change (basically in proportion to the number of small circles - I suppose another way of thinking is to fit small circles in the small space between the edge of a square of unit 2 and a circle of radius unit 1, the more small circles you can fit in there, is the number of circles proportional to their radius ?

[@ultimatereality6846]

This guy has the strength, will and determination UNDEFEATABLE.... THANKS PRESH SIR FOR THESE VIDEOS

[@merchillio]

3:50 “this is easily factorable”. Yes, absolutely, super easy! I’m absolutely not foraging in my school notes from 20 years ago

[@sk4lman]

3:28 "Since we have a right triangle; Here we go again! " Made me laugh out loud for real 🤣

[@skybirdification]

I like this video because you spend most time on the important part of finding that triangle, and not most time on solving the resulting equation by moving around parts for 15 minutes. Thanks!

[@myonlineteacheracademy8207]

Great Math Teacher... A+ for explanations and video pace and transition of parts.

[@badrunna-im]

I love these questions where there is a "degenerate" solution. Here it's r=1, which represents the case where the "big" circle coincides with the cornermost circle. It's still tangent to the two axes and two circles. That shows there's another way to construct the diagram following the instructions that satisfies the conditions, barring some of the arbitrary/intuitive things like scale.

[@maximilianosalvador9559]

If you took that solution the 1-r and 1-3r bits would go to 0 and -2 which doesn't make sense since we are working with positive lengths

[@arsenypogosov7206]

@@maximilianosalvador9559 Exactly.

[@shahriar1111]

I’m not a mathematician but I loved these comments

[@TheEternalVortex42]

@@maximilianosalvador9559 There are only (distinct) three circles in that degenerate case. The 0 makes sense because bottom left circle is coincident to the "big" circle, so the distance of their centers is 0. The -2 makes sense because the "big" circle is to the left of the rightmost circle.

[@mathmannix]

@@maximilianosalvador9559 It does make sense if you think about it correctly. Going from the center of the rightmost "smaller" circle to the center of the "big" circle (which coincides with the "smaller" circle closest to the origin): you go up 1-r = 0 units, and right 1-3 = -2 units (or left 2 units). And the distance (the "hypotenuse" of the degenerate "right triangle") is 1+r = 2 units (hypotenuse lengths are always positive.) It's a degenerate right triangle, with side lengths of 0, 2, and 2, and opposing angle measurements of 0 degrees, 90 degrees, and 90 degrees.

[@GlorifiedTruth]

I would never have dreamed of constructing that particular right triangle. Fantastic.

[@zlac]

Wow, it's actually much easier than it seems at first glance

[@rohangeorge712]

i like this problem because we actulaly have to thinkk of a solution and i was happy when i solved it talk about goood math problems

[@harishsathya9868]

Yes

[@nateisnotlate858]

I thought the same way

[@pierreabbat6157]

Here's how I did it. I offset the large circle, making a circle of radius 1+r. If you continue making a square lattice of small circles, the offset circle passes through a center directly below its center, a center directly left of its center, and the centers of the second and third large circle. Therefore, their coordinates relative to the center of the large circle are (0,-10r), (-6r,-8r), (-8r,-6r), and (-10r,0). So the offset circle has radius 10r, and the large circle has radius 9r. So 9r=1.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@jodfrut771]

Interesting way to solve.nice

[@ankurghosh2387]

@@AkashRaj-nq7qm whooshh

[@TheColtrane123]

Thanks for helping me recall the math I have lost for more than a decade.

I solved it as distance between the center of the large circle and second or the third small circle: (1-3r)^2+(1-r)^2=(1+r)^2 and it gives quadratic equation with solutions 1/9 and 1. Radius 1 of the smaller circle means that the first small circle is the large circle. Which is in some way may also be considered as a valid solution.

[@Ukrainian-22]

I like the way he says "and THAT'S the answer" 😊😊😊

[@kenweeraratne431]

He seems so pissed off when he says it like that

[@NoahUbf]

The eliminated solution 1 is also interesting. In this case we'll have 3 circles with r=1 and centers in (-1,-1), (-1,1) and (1,-1). The same way as our small circles they are tangent to each other, tangent to the X and Y axes and the later two of them are tangent to the "big" circle. The only their fault is that they lay in wrong quadrants.

[@markmolnar9901]

sorry, but no. the three other sircles would be at (1,3), (1,1) and (3,1) respectively as the x and y coordinates are R and 3R. they would touch the "big" circle from the top and from its right side.

[@thesimplekind8707]

@@markmolnar9901 circle (1,1) is the big circle , so it's not the big circle :D

[@thesimplekind8707]

There is no big circle , there are 4 circles (1,1) , (1,3) , (3,1) , (1,1) first and last coincide so the problem statement is not satisfied hence eliminated

[@mrb6309]

What a great way to start my morning. Thanks!

[@davidsalazar13]

Fantastic problem Umesh! And thanks Presh!

[@242math]

very well explained, thanks for sharing

[@ENTERTAININGGAMECHANNEL]

Awesome Teacher 🤩

[@chessmentor63]

Agreed, not sure why this showed up in my recommends, but I'm glad it did. Hadn't thought about math at this level in about 35 years, but this teacher does indeed do a fantastic job.

[@arnavsarda2581]

He finally uses the pythagoras/gougu theorem after so many videos but I love how he just straight up avoids the argument lol

[@JuanMataCFC]

"here we go again" 😂

[@mehsansaeed]

Thank you for everything. May we know which programme you use to draw and animate your diagrams? Thanks

[@peters972]

I took ages and could not see how to do it but the explanation was very neat and simple, Ty. Next time I’ll try drawing it out instead of just staring.

[@monawarnaqvi1574]

Every word that was said flew over my head

[@afatsquirrel1853]

I love this problem and the way it solves

[@anil2296]

Got the answer quickly!!! I solved this problem during my JEE preparation.

[@richardflacid6038]

no you didnt

[@cool_sword]

I believe you Anil, and I'm proud of you

[@Under_Dog19]

@@richardflacid6038 Bro JEE is something what you can't even thing about to pass it. Second toughest exam of India. 1million students appears but only 5700 can pass.

[@Under_Dog19]

@@richardflacid6038 If a student fail to pass JEE then he/she can easily crack MIT.😂

[@richardflacid6038]

@@Under_Dog19 did you just mash your keyboard for the reply? readint that made me dummber.

[@saschahahn8994]

OMG. I solved it before I saw the video and made it so complicated, by spliting everything in 3 triangles. But I got the same equation at the end. Nice vid.

[@NestorAbad]

Nice problem, thanks for sharing! An interesting follow-up: As seen in the video, we can ignore the small circle on the top and just consider the two small circles tangent to the x-axis, because of the symmetry of the picture. We've just seen that for n=2 small circles, the solution radius is a rational number (r=1/9). Now, the question is: are there any other values of n that give a rational answer? If so, can you tell what must be the general expression for n?

[@sauravpradhan2349]

You comment is way before than this video was uploaded..

[@Peter-os8vl]

Should be: (2n+1-2*sqrt(2*n)) / (2n-1)² But only for sqrt(2*n) ∈ ℕ

[@JoseGomes-uh1gr]

@@sauravpradhan2349 That's because he must be one of Presh's patreons.

[@rogiervankoetsveld741]

You’ll get a rational solution for all n circles when n is element of the set of squared and then doubled natural numbers

[@rogiervankoetsveld741]

Fun fact: this is the number of electrons in subsequent electron shells

[@jtflypegasus]

That's very nice....beautiful solution. It's not obvious until it is obvious! Cheers.

[@ravishankargupta4858]

Love from India sir... I really appreciate your videos. It really helps me to think in different way absolutely very good thank you so much....even though I feel competitive examinors are taking questions from here ..

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@danielyu9144]

Cool problem, thanks for sharing

[@freakinccdevilleiv380]

Aweeeeeeesome, thanks. Perfectly explained 👍👍👍

[@nikolanikolov1800]

r=1 is also a valid solution. Notice that the first circle is also tangent to the lines and the other two. The word "big" does not mean "bigger" neither different.

[@user-oh2kt8lf6g]

No, it is not: at 3:13, we consider a right triangle. For r=1, the bottom side would have length of 1-3r= -2 and the right side would be 1-r=0 which is geometrically incorrect.

[@dougaltolan3017]

@@user-oh2kt8lf6g no, talk of triangles is only 1 route to a solution. r = 1 is a valid solution to the description. The meaning of this is that the description has an ambiguity that the drawing does not. Maths showed that ambiguity! Also maths dosent care if lengths are negative.

[@user-ym9rv4ur3i]

Another (similar but interesting) way to do it Big circle is (x-1)^2+(y-1)^2=1 Little circles can be represented by (x-r)^2+(y-r)^2=r^2, (x-3r)^2+(y-r)^2=r^2, and (x-r)^2+(y-3r)^2=r^2 Find the distance between the centers of bottom right circle (3r, r) and big circle (1, 1) in terms of r by using the distance formula/pythagorean theorem then subtracting the radius of the big circle. This (when simplified) gives you sqrt[10r^2-8r+2]-1 This is the radius in terms of the radius, so just set it equal to r and solve, which gives both r=1 and r=1/9. 1 can be dismissed although it fits the criteria; a circle (x-3r)^2+(y-r)^2=r^2 touches the circle (x-1)^2+(y-1)^2=1 when r=1 and r=1/9, but the former doesn't fit the problem at hand.

[@michaelbryson5870]

I solved it with a different triangle and got the same answer but I think his way is a little simpler. I made a triangle with vertices at the center of the bottom left circle, the center of the bottom right circle, and the big circle. I also knew the angle on the left was 45°. Then I used law of cosines to make an equation, simplified it to a binomial, and factored to get the solution.

[@rajdipdeka10k]

Good good

[@tokoloshgolem]

A master case in the unnecessarily jargonised verbiage that causes so many people to turn away from trying to understand geometry and mathematics. Such a simple concept made to sound as difficult as possible

[@afwaller]

Short and sweet - love it

[@roriedeweber]

How I would have done this on a multiple choice test because I’m lazy: - make the radius of the big circle (1) go down and to the left (making a “square” in in bottom left corner [mathematically, it’s probably not an actual square]) - you can now see that the two smaller circles are about half of the distance to the radius (1) of the big circle. *1/2= .5 - then divide that in half again because there are two circles *.5/2= .25 - then divide in half again (because a radius is half of a diameter) *.25/2= .125 - then pick the closest answer to mine - would probably be the actual answer because that’s pretty dang close That’s how I did it before I watched. I knew that’s not how they were going to do it in the video, but that’s pretty much how I’m passing my math classes at the moment.

[@derekirish5121]

I did the same thing. There would be 4.5 small circles in the radius of the big circle. 1 divided by 4.5 will give you the diameter of each small circle, then divide by half to get the radius (.11111 same answer in the video) Lazy brilliance.

[@canbalcioglu4229]

solved that one too:) took me some time but finally did it. pls put more geometry questions like these. these are like brain candy🤩

[@tobiasL1991]

Very elegant solution.

[@Shareholder80]

A very very good problem. A high school concept solved a beautiful question.

[@mzatzk5135]

Can’t 9r^2 - 10r + 1 simply become (9r-1)(r-1) ? Your way gives the same answer, so it all works out. But I feel like my way is the more intuitive way

[@rudrodeepchatterjee]

Well, you can do both ways, and both ways will give you the same answer.

[@omurice8534]

there are two ways to solve quadratic equations right?? both are fast

[@robertveith6383]

@@omurice8534 Wrong. There are more than two ways to solve quadratic equations.

[@chemistry4life]

i also factorised to (9r - 1)(r-1). i think this way is simpler as well as it doesnt involve fractions in the factorisation process

[@study69696]

mid term splitting is our all favorite

[@wjodf8067]

got me flatfooted on that one

[@gabor6259]

I figured this out in a much more convoluted way but I enjoyed it. Nice problem.

[@brunoterlingen2203]

Brilliantly explained.

[@benheideveld4617]

Beautiful application of the Baudhayana theorem!

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@jormdeworm]

I had a similar question on my exam, except they asked the entire surfice between the circle and the two lines, and not three circles in that surface. I must say it was also much easier than this😅😂

[@khatharrmalkavian3306]

I came up with this prior to watching the solution. I'm not very confident in it because it came down to the small radius being either 1 or 1/9, but my process was as follows: The centerpoint of the topmost small circle is (r, 3r) because the center is r above the bottom and the diameter of the circle underneath is 2r, and the centerpoint is x=r away from the axis. The centerpoint of the large circle is (1,1). If we construct a right triangle with the hypotenuse running between the centerpoints of the large circle and the topmost small circle, then it will have a leg of length 1-r running along the x axis, and a leg of length 1-3r running along the y axis. We also know that the two circles are tangential, and thus can deduce that the length of the hypotenuse is the sum of the radii of the circles, or 1+r. From this we can deduce via Pythagoras that (1+r)^2 = (1-r)^2 + (1-3r)^2. r^2+2r+1 = r^2-2r+1 + 9r^2-6r+1 r^2+2r+1 = 10r^2-8r+2 2r+1 = 9r^2-8r+2 1 = 9r^2-10r+2 0 = 9r^2-10r+1 0 = (r-1)(9r-1) r = 1 or 1/9

[@khatharrmalkavian3306]

Ayy, I got it.

[@alittax]

Beautiful, thank you!

[@sillymel]

I made a triangle from the centers of the large circle and two adjacent small circles, calculated the side lengths to be r+1, 2r, and (1-r)sqrt(2), then determined that the angle opposite the side with length r+1 was 45 degrees and used the Law of Cosines to get an equation I could solve for r. Your solution was definitely more elegant, but mine also worked... eventually.

[@RandomGuy-yf4wf]

How 2r and why 45°?

[@sillymel]

@@RandomGuy-yf4wf I think I may have been a bit ambiguous about the placement of the smaller circles. They're supposed to be adjacent to _each other,_ not necessarily the larger circle. (Only one will be adjacent to the larger circle.) If you still don't see it, I can try to explai further, but it'll be a bit hard to do with just text.

[@ProfessorDBehrman]

Construct the right triangle with the right angle at the midpoint of the centers of the two small circles. The sides are r+1 , rsqrt(2) and d. The distance from the center of the big circle to the corner is sqrt(2). Therefore, sqrt(2) = d + 2rsqrt(2). With these two equations and the Pythagorean theorem, you can solve for r. No need for the angle.

[@sillymel]

@@ProfessorDBehrman So, first, this is exactly the sort of ambiguity I was talking about. I meant that the small circles were adjacent to _each other,_ not that they were both adjacent to the large circle. Other than that, nice derivation, although it took me a bit to figure out how you got the “2r sqrt(2)” figure.

[@nirajkumarverma5299]

Smart application of tangency criteria of two circles to create an interesting problem.👍👍👍

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@i_am_keerthana4217]

Wow, exceptional perspective

[@englishlife5838]

Love your solution

[@UttamKumar-gf4hd]

Very interesting. I love it alot! 😍

[@chrisj5443]

How about that. He solved it the same way I did! To extend this problem, one could develop an equation to determine the small-circle radius for a variable number of stacked small circles which fit under the large circle. The quantity of small circles would be 1, 3(as here), 5, 7, 9, and so on. Note that the product of the small-circle radius and the number of small circles will approach a limit of 1 as the number of small circles approaches infinity (for 3 small circles, it's 3x0.111=0.333).

[@HankGD-oj3sl]

I have a really cool problem I’d love to suggest, where could I do so? I don’t see an email address in the about section on my phone

[@PenandPaperScience]

Beautiful solution! :)

[@Chrisoikmath_]

I am looking forward to being able to solve these difficult problems!

[@juliocesarsalazargarcia6872]

With the right training and coaching you can improve more than you would expect. If you need help I can coach you, I teach online.

[@MrUncleKurt]

Kindly remind me why I can be confident in the possibility of construction of the large circle, tangent to the two smaller ones AND the line segments. Where might I look to find a proof of that construction? Thanks.

[@user-pz8be1kf1p]

Thank you sir ❤🙏

[@jimmykitty]

Brother! I can remember your profile picture.. 😥

[@amclauzeo4373]

Thanks man for this great question it challenged my mind and after this video i subscribed your channel

[@SQRTime]

If you are interested in geometry questions, checkout our playlist Geometry: kzbin.info/aero/PLgwyp12I7OOCXinAb1ZgUg-T5qdruyfKk Hope you enjoy

[@cipher3966]

My theory was to draw a triangle at the base containing the three circles. Finding the lengths of each side will give me the area. With that I could work out the maximum radius three circles can have to fit inside. Does this count as a pass?

[@zachzero7167]

Great question thank you for providing us such a wonderful video

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@winteranfang2773]

You can easily solve these kind of problems on a checkered notebook. You can place 4.5 circles which has r radius in the big circles radius length. So, radius of the small circles equal of the 1/9 of the radius of the big circle.

[@wwoods66]

That relationship of the squares of the sides of a right triangle looks like it will be useful. Maybe we should give it a name? The Preshtalwalkean Theorem, or something?

[@Grizzly01]

It might crop up a couple of times, but it's very obscure. Hardly anyone knows about it.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@Tiqerboy]

Wow, 🐯 solved it exactly as Presh did. Had to think for a moment to set up the solution as described but it wasn't too hard. Instead of saying "here we go again", you can say Pythagorean Theorem, Presh, no one is going to bite 😂

[@Merajmohdkhan]

Old days memories 😍

[@dr.sabdaramanchowdhury7102]

Wow! It's just fantastic.

[@granhermon2]

Thank you! What's the name of your theme song?

[@nirajpradhan9734]

what do you use to make your videos? i love them.

[@shubhambisht9311]

Finally a question I was about to solve

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@joeyfalcon3278]

Drew a vertical radius down to the x-axis. From there, I saw that the two circles (one on top of the other) were about 1/2 the length of the bigger radius. The center line going through the two small circles would have a length of 1/2. Which would then mean the diameter of each small circle would be 1/4. Since we want the radius, we would then divide again, making the small circle have a radius of 1/8. This was a visual way that I saw it, obviously it isn't correct but solving this problem numerically wasn't too bad either after the explanation.

[@btwin93]

estimated the small r exactly the way you did just by looking at the pic :D took 5s but as we know now it's not correct haha cheers!

[@justanotherguy469]

Brilliant! I'm so excited!

[@HoSza1]

Is it possible to pack smaller circles in the corner using a rectangular grid (in rows and columns) so that the big circle touches the topmost circle in each column and the rightmost circle in each row? n=3 is an example for such a grid, but are there any more values? And what are they if they exist?

[@markmolnar9901]

if you have N (0 < N) small circles in a column, their radii (0 < R < 1) would be given by the formula R = (2N + 1 - sqrt(8N)) / (2N-1)^2. another solution of the pythagoras-equation [R' = (2N + 1 + sqrt(8N)) / (2N-1)^2] belongs to somewhat greater circles in the x

[@MinhNguyen-nu4nx]

that's was amazing.i,m from VietNam and i love your content so much!

[@nothing-uw4kk]

Cool exercise with cool channel 🔥🔥🔥

[@timothymchugh6232]

Cool, now when I see this in the real world I’ll know exactly what to do. Nice challenge

[@BELKOchannel]

this was good as well. thanks

[@okawashi]

This was easier than I expected

[@white8993]

This movie is quit easy to understand. Thanks!

[@WhiteGandalfs]

This was a nice problem. Solvable but one had to construct a helpful mesh from the given bits. That's in contrast to the previous few ones, where the wit of the problem was based on arbitrary withholding of definitions without which the problems were either unsolvable or straight nonsense.

[@moobles2998]

Just looking at the proportions of the circles compared to one-another before doing any math, I noticed you can fit roughly 4 or 5 of the small circles within the radius of the larger circle. As such you can intuit it should be twice the amount of radii as circles, so 2*4^-1 or 2*5^-1, so between 1/8th and 1/10th. Knowing these kinds of problems tend towards easily written answers, and intuiting that when the number 3 is involved in any way, you tend to get multiples of 3 somewhere. 1/9th was a quick choice. Lovely to see an initial estimate be proven mathematically though! As fun as estimation is, it will never beat absolute mathematical proof.

[@mathswallah3831]

Ohh my god .... Thankyou for solved my problem.

[@birdgincrit]

I'm not skilled at math but i like it, especially 'puzzles' like these. Geometry stuff. I just tried to determine every measure i could think of: the surface of the four empty corners, the diagonal line of the four imaginary squares, deviding the circle in eight pieces and see how the lines would interact/intersect with the smaller circles, etc. I was hoping that if i do all that eventually i would stumble on a number/measure or insight that i needed to calculate the radius but i just couldn't make it happen. Eventually i thought well.. it looks a bit like you can put four circles on top of each other and that distance would equal 1 so i came up with r=0.0625 (1/8). But i knew that was wrong and obviously that isn't the way to approach these things. Anyway, fun stuff!

[@specificgravity-thedancing9700]

This was fun!! Thanks!

[@mf0u3021]

Mindblowing how much ‘basic’ maths I’ve forgotten since my degree. Haha

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@anvaykhedulkar339]

cant we just substract the area of sector from 1 by 1 square and divide the area by 3Pi(r)^2 and determine the value of r from there ? ofc considering these are the largest circles that can be formed in that area.

[@prosp3421]

Big fan of ur work, sir

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@johnnewson939]

great solution

[@PhysiKarlz]

I'm impressed that you have three blue balls. The normal distribution is around 2.

[@rameshks8449]

Thanks And very interesting replies

[@dankierson]

More obvious to look at the line from the origin to the big circle center. Sqrt(2) = Sqrt(r) + Sqrt(r) + Sqrt{ Sqr(1+r) - Sqr(Sqrt(2)r)}

[@theovolz3073]

That is so clever!

[@egregiouss1865]

Wow!!! Very nice task!!!

[@ginabegier4521]

Thank you for stimulating my old brain.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/gorVZYWMm5qoe9k

[@NorkFN]

Can you make a video about why there are false roots in some problems, like r = 1 in this one. I do not understand where it comes from, would help a lot.

[@tusgyu851]

So we have distance from center of circle to tangent line or (right angle tringle) =1 and if we see those small circles and imagine upto center of the circle then there will be 5 small circle.and if we divide 1/5=0.2 so each small circle will have Approximately 0.2/2= 0.1 radius. ....

[@darthm985]

QUESTION: will 3 circles draw in the bottom left corner of the 1st small circle as were the 1st three small circles have r=1/81th ? will someone want to bench on it?

[@chotana]

It sounds so easy when you explain ... :D