Thanks

My friend: "What's the first thing you think of when you hear 'circles'?" Me: "Triangles." My friend: "Why triangles?" Me: "Triangles make circles less complex." My friend: "That doesn't make any sense!" ~2 weeks later, after a math contest~ My friend: "Triangles are the fundamentals of circles."

Solution made me think of formula for parallel resistor (caculated the replace resistor Rv): 1/Rv = 1/R1 + 1/R2 😁

I realize someone may have noted this earlier, but it was fun to see that if you use the quadratic equation that gives the first answer of 12-8√2, the "extraneous" solution of 12+8√2 is the radius of the circle that is externally tangent to the touching circles of radii 4 and 2 and the line on which they rest. A nice "efficient" result.

The equation 1/a + 1/b = 1/c is the "optic equation" (see Wikipedia) hence is connected to the heptagon and the "heptagonal triangle".

I usually tell my students "Whenever a circle touches anything, draw a line from the center of the circle to the point of tangency". Otherwise they will yell at me, "How the heck are we even supposed to *think* of making those right triangles?"

This is the first time i've answered one of this channel's math problems on my own *AND IT FREAKIN TOOK ME 3 HOURS*

I solved it using vectors: Naming C1 the center of the big circle, C2 the center of the medium one and C3 is the center of the small one. The first two coordinares can be placed. Then the vector C1C2 can be fully known and also the unit vector in that direction (ê1) Then, this vector C1C2 (lenght=6) can be expressed as this sum: 6 ê1 = (4+r) ê2 + (2+r) ê3 (#1) Where: ê1 is the unit vector from C1 to C2 ê2 is the unit vector from C1 to C3 ê3 is the unit vector from C3 to C2 Of course, we don’t know vectors ê2 and ê3 but we know that r+(r+4)sin(q)=4 (vertical distance from ground to C1) r+(r+2)sin(p)=2 (vertical distance from ground to C2) p is the angle that ê3 forms with an horizontal line and q is the angle that ê2 forms with the same horizontal. Solving for sin(q) and sin(p) can also give us cos(q) and cos(p): these values are: sin(p)=(2-r)/(2+r) cos(p)=2√(2r)/(2+r) sin(q)=(4-r)/(4+r) cos(q)=4√r/(4+r) Decomposing the ecuation (#1) gives you 0=0 in the vertical axis and for the horizontal: 4√2=4√r+2√(2r) Which results in r=12-8√2≈0,6863... Kudos to you if you read this!

For students at age 14 and 15. Even I'm 20 I can't solve this. 😭

This question came in JEE MAINS this year 😁!

Love the videos, hate the comment sections

Thanks for the puzzle, Presh! How I solved it: First, I'm interested in the line that goes thru both centers, where it intersects the external tangent line, and the angle at which it intersects it at. Call the point of intersection S. I notice that by using similar triangles, I could draw a sequence of similar circles inside that angle, all touching each other, all touching the line, all centers aligned, and each half the size of the previous circle. That's an infinite sequence that approaches point S. That gives me the distance from Blue Center to S: 4 + 2 + 1 + 1/2 ... etc = 8. That's the hypotenuse of a right triangle. I also know the height of that right triangle, it's the Blue radius, 4. So I have a 30/60/90 triangle, and the hypotenuse is a line of slope -1/2. Now I'm going to give the circle centers co-ordinates. Let's make the x axis the same as the line in the drawing, and put the y axis thru the center of the blue circle. So immediately we have that Blue center is at (0,4) and the blue and green centers are on the line y = 4 - x/2 Green center is therefore at (4,2). The distance from White center to Green center equals the sum of the radiii of the white and green circles; ditto White and Blue. So we have sqrt((wc_x - bc_x)^2 + (wc_y - bc_y)^2) = sqrt(4^2 + R^2) sqrt((wc_x - gc_x)^2 + (wc_y - gc_y)^2) = sqrt(2^2 + R^2) Substituting the values we know, sqrt((wc_y-4)^2+wc_x^2) = sqrt(R^2+16) sqrt((wc_y-2)^2+(wc_x-4)^2) = sqrt(R^2+4) We also know that the white circle is tangent to the given line, which is our x axis. So its y co-ordinate equals its radius: sqrt((wc_y-4)^2+wc_x^2) = sqrt(wc_y^2+16) sqrt((wc_y-2)^2+(wc_x-4)^2) = sqrt(wc_y^2+4) Squaring both sides and simplifying, wc_x^2-8*wc_y = 0 -4*wc_y+wc_x^2-8*wc_x+16 = 0 Solving the first for wc_y, wc_y = wc_x^2/8 Substituting into the second, wc_x^2/2-8*wc_x+16 = 0 Solving for wc_x, we get two answers. The second is about 13.7, much too large, and seems to correspond to a perverse answer with a huge white circle surrounding the other two. We'll focus on the first answer: wc_x = 8-2^(5/2) wc_x = 2^(5/2)+8 Substituting to find wc_y, wc_y = (8-2^(5/2))^2/8 [Edit: Originally made a big goof here. Solved for the distance from white center to the origin, which is completely irrelevant. wc_y is already the answer] Converting to decimal, R = wc_y = 0.6862915010152378 I thought I'd see some clever shortcut along the way, but nothing obvious presented itself. I expect that Presh has a much more elegant solution.

The graphic design on this was so satisfying that I almost forgot to pay attention to the math

"Did yiu figure that out?"..... Ehhhhh.. NO!!

I found this to be a nice looking answer: a + b - 2√(ab) r= --------------------- a/b +b/a -2 a=R1 b=R2 But I must admit your solution did have a nice simplicity to it.

Do you think you could make a video on how to approach challenging problems, to figure them out on your own?

Descartes' Theorem ("Kissing Circles"): Label circles 1, 2 and 3 where 1 is largest and 3 smallest. Let k(n) = 1/r(n) where r = radius. Then we have k(3) = k(1) + k(2) + 2.sqrt[(k(1).k(2)] = 1/2 + 1/4 +/- 2(sqrt(1/8)). The +/- alternatives give the radii of either a smaller or larger circle mutually tangent to circles (1) and (2); if you picture a larger circle to the right of the diagram which touches both circle 1 and 2, it has radius 23.31 (4sf) Enjoyed the proof a lot.

I used sine law and compound angle formula: (4+r) / sin(θ₁ + θ₂) = (2+r) / sin(θ₃ - θ₁) where: sinθ₁ = 1/3 cosθ₁ = √8/3 sinθ₂ = (2-r) / (2+r) cosθ₂ = √(8r) / (2+r) sinθ₃ = (4-r) / (4+r) cosθ₃ = 4√r / (4+r)

This equation is important in the formation of concrete when using circular materials, so spaces must be closed to obtain the greatest strength.The remaining spaces are filled by cement and sabale

I'm 15 , n was able to figure it out. Generally your problems are more difficult, but this one was quite easy, I did in the same way. Well, thanks a lot as you made me learn mathematics in a different and creative way. 😇😇😇😇😇

You have opened the door to genius and beautiful mathematics!

I had to think about it for a while, but the top of purple triangle for the general equation is as high as the radius of the large circle, R1. Since you can make a line from the corner of purple triangle, intersect with the center of the second circle, and go all the way down, the line drawn is equal to R1, and if you remove the radius of the second circle, you get R1-R2, which is the height of the purple triangle. Just leaving this here for peeps who might be wondering why the height is R1-R2.

The equation at 5:13 reminds a number of people of the formula for parallel resistors. Interestingly, the same idea is used to graphically find squared squares.

Fascinating problem. I got the correct answer. That equation at the end is sheer beauty. Note that if R1 = R2 then the smaller circle will have radius r equal to one fourth R1.

I'm a student of mechatronics engineer and I really like these problems. 👏🏾👏

Thanks sir for a new method, I was. Trying to solve this type of problem in a different manner , the method I had learnt is of common tangent, once again thank you

I used to be so good at math. But after high school I stoped doing it and forgot lots of stuff. It suck. I wanna relearn everything but don't know where to start or where to go to learn them.

1) Place x and y coordinate axis, with the line at the base being the y=0 line 2) Place the centers of the circles at c1 = (0, 4), c2 = (a, 2) and c3 = (x,y). You want to find y 3) Specify c2 by solving the trapezium with large base = 4, small base = 2, height = a and hypothenuse = 6 4) Find candidates for (x,y) by solving the distance(c1,c3)=4+y and distance(c2,c3)=2+y. 5) Use the fact that y>0 and y

I got the "connect the dots" bit. Adding the radii, all of that. I just had not thought of making my own "dots" to connect.

And we can thank Descartes again for inventing analytic geometry... The easiest way to solve this... by far (and for any radii).

Presh: "We can solve for big R2" Me: What about D2?

Or you can use Descartes' theorem, which for this case (with k3=0), simplifies to k4=k1+k2±2√k1k2, and plugging in k1=1/2 and k2=1/4, we get k4=(3±2√2)/4. Since we want the smaller circle, we choose the bigger value of k4, so k4=(3+2√2)/4, which means r=4/(3+2√2)=12-8√2.

The reciprocal of the radius of small circle is the sum of the reciprocals of the two larger circles' radii. Reminds me of resistance in parallel.

There's a direct formula for this. (1/√c) = (1√a) + (1/√b) Where, c = radius of the smallest circle a and b are the radii of the bigger circles.

Thank you sir for your creativity and sharing with students

This arrangement of circles has a beautiful application, in Number Theory, to "Ford Circles", which are closely related to "Farey Sequences". If you start with two touching circles, of equal size, tangent to a line, you can construct all possible fractions by inserting smaller circles tangent to previous circles and the line. Farey Sequences give you all fractions, automatically in reduced form, without any factoring, listing the fractions with the same ordering as the Ford Circles. An elegant bit of Mathematics!

This question was in my book and my teacher solved it.. The question was to prove the equation u gave for sooving..

Honestly I’d just draw the circles with the right measurements and just use a ruler to measure the radius of the smaller one 😂 we had a similar problem like this once and I got it right by drawing them and guessing the measurement, no rulers were allowed. It was fun 😂😂

Please can you tell which software you used to edit the video?

The formula for the solution is brilliant ! I don't expected that this is so easy.

Important intermediate result. When two circles are tangent to each other and some line is tangent to both circles, then the distance between the projections of centers of these circles onto this line is 2*sqrt(r1*r2), where r1 and r2 are radii of the circles. I found this result first, and only after that solved the given problem.

We can do this more simple by using tangent at the base.... 2√R1*R2=2√R1*r + 2√R2*r We get the same result

I draw these 3 circles in autocad and find the radius as 0.6863

No doubt that this is the best KZfaq channel ever, very very thank you sir for making such a great work. please go ahead like this

Again, super easy. I took a measuring tape and checked the little one against the other two. Only needed to do a little math in my head.

1/r=1/a+1/b reminds me of the resistance formula for parallel electric circuits

Kissing circles theorem

For 14 year olds!? Amazing. I think this was one of the hardest doable puzzles I've seen Pesh do.

Thank you Presh for your great videos. Math can be fun! It seems to me that this problem would be prettier if the result were an integer. For instance, with 225 and 100 for the radii of the two big circles you would obtain exactly 36 for the radius of the small one.

(*A Simplified version of your method*) Using Pythagorean theorem you can show that the "length" of the common tangent segment between two externally tangent circles of radii a & b is 2sqrt(ab). Now looking at the common tangent line: 2sqrt(R1.r)+2sqrt(R2.r)=2sqrt(R1.R2), By dividing by 2sqrt(R1.R2.r), this can easily be simplified to the formula you got at the end of the video.

This is in our FIITJEE package....in chapter circle Came in ntse stage 2

It's half past midnight. I just sat by, baffled, bewildered, and babbling. No inflated ego here. Not even by a long shot.

I was given the same problem in a high school exam recently except they helped you a bit by making the radius length labels be positioned so that they made the hypotenuse of the first right-angled triangle (bit of a visual clue).

Presh, could you please explain that last rationalization step at 3:20 that results in 12 - 8√2. How to get from 16 / (12 + 8√2) to 12 - 8√2? Thanks!

JEE Mains 2019 ka question h

Using direct common tangent 2√(4*2)=2√(4r) + 2√(2r) Square both side

You can also first figure out that the overall horizontal distance (base of the purple triangle) is 4-radical-2. With the blue triangle, you can then set up the Pythagorean equation using x as the horizontal side, and for the green trial use "4-radical-2 minus x" as that horizontal side. Two equations, two variables. The downside is that the algebra gets messy and requires the quadratic formula; the upside is that you avoid the trickiness in 3:10.

I liked deriving the general solution. It ended with a very elegant expression. Coincidently, this has a direct relationship to Farey Sequence & Ford Circles. It's worth looking these up.

I am a simple Indonesian. I see "Indonesia" in title, I upvote..

Hello there, I am 14 and I completed it half (you taught me the other one) thank you!

Got this one, did it in a similar way. I have to admit that i thought it was going to be easy but took me some good 15 min to solve

nice formula at the end

I'm so annoyed at myself. I found the two right-angle triangles from this proof pretty much immediately, but then I gave their small angles variable names and started using trigonometry to find their bases because I FORGOT THAT THE PYTHAGOREAN THEOREM EXISTED. I gave up and just watched the video, and the second Presh said "... we can solve using the Pythagorean Theorem..." I went numb with the shock that I could have forgotten such a basic thing.

A slightly different approach. Set a medium circle center at (0,r1), locate second large circle at (2(r1r2)^.5,r2-r1), set third small circle at (Xs,Ys). Then set distance between centers equal to r1+Ys and r2+Ys but change Ys to small circle radius. Solve resulting eqns and you get the same results as above. 🙂

0:16 The small circle's raidus seems to be 1. 4 to 2 to 1. It seems like each circle's diameter is complete with it's own radius and the other circle's diameter. Nth circle has a diameter of 8. It's radius is 4. The circle smaller than the nth one has a diameter of 4. Dia. of N-1= radius of N

I like how he says keep watching the video for "a" solution

Wonderful wayof proceeding thanks. Dr.JPP

I solved it using the fact that the locus of points that is equidistant between a circle and a line is a parabola. I constructed 2 parabolas, equidistant from each given circle and the x-axis on the cartesian plane whose intersection point is the center of the circle in question. From there, it's easy. I've done similar problems with inscribed circles in college geometry.

Nobody: Random dude with a huge ego: *_lOl tHaT waAs EeeEEEeZY 😂😂😂_*

This question came in ssc cgl 2017

I cannot express my gratitude for such a wonderful problem and clear proofs.

I got ✓2-1 by creating a right angle trapezoid and getting the length between the middle of both the circles (which is 6) and the sides being the radi of the circles beeing 4 and 2 then creating a right triangle which the side is paralel to the base of the trapezoid( 6^2 - 2^2 = 4✓2) then 4✓r + 2✓r *✓2 and getting r+1 = √2 and the final awnser being r = √2 - 1

Mind your decision: Challenging problem from indonesia Jee aspirants:Hold my beer

This is...a multiple choice problem. Usually within 60 questions that has to be solved in 60 minutes. This question either appeared on high school graduation test or university enterance test....

this question has been asked in so many competitive exam in india..2 or 3 years ago.. right now I'm obsessed

In general, if the radii of the two circles are a and b, where a>b, then the radius of the small circle is: c=ab[a+b-2(ab)^(1/2)]/(a-b)^2. Here a=4 and b=2, so c=4*2[4+2-2(4*2)^(1/2)]/(4-2)^2=8[6-2*8^(1/2)]/4=2[6-2*2*2^(1/2)]=2[6-4*2^(1/2)]=12-8*2^(1/2).

at 4:50: i don´t understand how to get to the term on the bottom right?

I work out this problem with my students in Mexico

Creative ideas...thanks to you, I Love Mathematics as much as I love my Life.

I get this problem from a math competition not too long ago, i too am from indonesia, i cant solve it. Thank you, now i know the answer, been searching since!

There is a direct formula for this 1/√c = 1/√a + 1/√b Where c = radius of smallest circle b = radius of bigger circle a = radius of biggest circle

can you teach me how r = 32 -(4 + 2√2)^2 became r = 12 - 8√2?

Quite straightforward with the apparatus of cartesian geometry

I love the way you say Devan's school! XD Love from Indonesia. 🇮🇩

Just when I solved it, you one up me with the general solution. Ahhhhhh.

I had to put the purple triangle up side down to understand why its hight is R1-R2 😅

I noticed that the horizontal distance between two circles is four times the multiplication of the radii of the two circles. If we name the radii of the circles a, b, c. Where c is the circle in the middle. The horizontal distance between the right and left circles equals 4c(a+b). I was thinking how the horizontal distance between right and left circles increases as c increases. If change is dc. Horizontal distance increases 4dc(a+b). This is useful if we need to know the radius of the circle needed to keep a specific distance between two other.

I cannot emphasize this enough: *If you see a tricky circle problem, always connect their centers or draw radii. Then you can simply use the Pythagorean theorem.*

1/✓c=1✓a+ 1✓b Where c is the radius of smaller circle and a,b the radius of other 2 circles

I think you haven't seen the questions of an Indian extrance exam CAT 😂

I found something interesting while fiddling with the general solution a bit. I tried inserting Euler's number (e = 2.71828...) and 1 as R1 and R2 with a hypothesis that the small circle would be 1/e. Sadly, it was close but not quite. However, while calculating it, I noticed another thing. It looked like the Golden Ratio (1.618...) would instead work when squared. And then it dawned upon me, the generalized formula just straight up turns into a modified version of the definition of GR itself when you insert R1 = GR^2 and R2 = 1, with a result of r = 1/GR^2: 1/sqrt(1/GR^2) = 1/sqrt(GR^2) + 1/sqrt(1) => 1/(1/GR) = 1/GR + 1 => GR = 1/GR + 1 In other words, this creates an infinitely repeating pattern where adding a larger circle so that it touches the previous two largest circles and the "ground" level would always result in a circle that has a radius of GR^2 times the radius of the previously largest circle, and adding smaller ones under the two previously smallest ones results in a circle that has a radius of 1/GR^2 times the radius of the previously smallest circle! Math can sometimes be so unbelievably beautiful.

This model question was recently asked in jee main 2019 January. Radius were given a,b and c and we have to find relation between a,b and c.

Me gustó la generalización del problema

I don’t understand how you knew the hypotenuse of the first triangle was 4 + r

eyy I think this is the first video where I actually got the right answer in almost the exact same way as the video! the only thing i did differently was i calculated the actual length of the last leg between the 2 larger circles (4*sqrt(2)), set that value equal to the sum of the 2 other triangle legs with the variable, and turned it into a quadratic. just had to give myself a quick refresher on factoring polynomials and then i was all set!

you can also use descartes' 4 circles theorem (search it up, use wikipedia if you havent heard of it) its the special case where one of the circles is a line curvatures are: 1/4, 1/2, 0, and 1/r so 1/r = 1/4 + 1/2 + 2*sqrt(1/4*1/2) the coefficient to the square root is positive because the circle is the smaller one (refer to wikipedia for more info) r = 1/(1/4 + 1/2 + 2*sqrt(1/4*1/2)) r = 1/(3/4 + 2*sqrt(1/8)) use a calculator because doing sqrt(1/8) without one is ridiculous and you get 0.686 edit: all the math stuff

Well, bro, I got this in my test and surprisingly i cracked it :)

Descartes Theorem for tangent circles! a line is a circle of infinite radius!

I like PreMath because he goes slowly. And other guys who use writing boards in their vids which is the best approach. And, I don't want to replay and pause because it takes more of my time.

This solution reminds me of the method for calculating the equivalent value of 2 resistors in parallel IE sqrt(r)=(sqrt(R1)*sqrt(R2))/((sqrt(R1)+(sqrt(R2)) so r=((sqrt(R1)*sqrt(R2))/((sqrt(R1)+(sqrt(R2))) squared and it works :) great channel btw