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."
@mehujmehuj22295 жыл бұрын
Triangles are the basics of all geometry, just continue drawing them until you draw the right ones :D
@sanjeevgoel88655 жыл бұрын
😂
@pablomalaga46764 жыл бұрын
@@mehujmehuj2229 I think also of sailing and navigation
@vivekkedia21404 жыл бұрын
Thanks to let me know i will try to use it in future
@DarthAlphaTheGreat3 жыл бұрын
Triangles make everything easier.
@Guust_Flater5 жыл бұрын
Solution made me think of formula for parallel resistor (caculated the replace resistor Rv): 1/Rv = 1/R1 + 1/R2 😁
@user-us5cw3eq8y5 жыл бұрын
This is a wonderful idea to use the geometry of circles to analyze electrical circuits🤔
@chaos.corner4 жыл бұрын
@@user-us5cw3eq8y You actually kinda do with some advanced analysis. You can treat an AC current as the real part of the complex function I=Ae^iwt
@betailleuh4 жыл бұрын
I thought about it too
@arielfuxman88684 жыл бұрын
Made me think of optical lenses.
@Walczyk4 жыл бұрын
@@user-us5cw3eq8y except not because the equation uses square root radii which is not obviously geometrizable. It appears that it would involve inversion of some kind, points to lines and lines to points.
@kevinstewart25724 жыл бұрын
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.
@TitoTheThird5 жыл бұрын
The equation 1/a + 1/b = 1/c is the "optic equation" (see Wikipedia) hence is connected to the heptagon and the "heptagonal triangle".
@victorgorelik738311 ай бұрын
“Solve without pen. 419”
@taufiqutomo4 жыл бұрын
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?"
@hassanakhtar78744 жыл бұрын
Good job! Be sure to tell them circles are by definition shapes where every point is equidistant from the center. And in order to prove things or come up with solutions the first thing you should do is apply the basic definitions.
@daffadaniirfan59284 жыл бұрын
Semangat pak guru!
@pandawagurning3 жыл бұрын
Yea it used Mostly to solve a problem😂
@udayomsrivastava90522 жыл бұрын
That is a really good advice. Would be really helpful for them
@rivsol4 жыл бұрын
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*
@JesusAlbertoPinto5 жыл бұрын
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!
@AnuragKumar-io2sb5 жыл бұрын
U r amazing wow😮
@wec32705 жыл бұрын
somya shree swain noyou cant lol
@yourlordandsaviouryeesusbe29985 жыл бұрын
Great job👏👏
@Ts1nd5 жыл бұрын
you realize you didnt need vectors and could use variables instead right? Calling it vectors make it cooler or what
@arturoortiz59075 жыл бұрын
is more easy use theorem of Pythagoras
@Raxer_th5 жыл бұрын
For students at age 14 and 15. Even I'm 20 I can't solve this. 😭
@ashutoshkaushik73755 жыл бұрын
It is an average question for a high school student in india at least bcz these type of question comes a lot in exams
@unyielding375 жыл бұрын
yeah, what the hell?
@AMa-us8de5 жыл бұрын
"For students at age 14 or 15" So we are the 20 ones dont count haha
@justabeardedguythatisahero98485 жыл бұрын
you need some geometry classes , there are alot of information which you missed , or maybe your educational system focuses on a collage degree preps
@marcoantonioloureiro58835 жыл бұрын
Im am 14 years old and that's the problem that i need to know to solve to go to Brazils air Force (the high school)
@shouvickjoardar45585 жыл бұрын
This question came in JEE MAINS this year 😁!
@shouvickjoardar45585 жыл бұрын
All the very best :)
@usmanakhtar39925 жыл бұрын
I also saw. In jee they solve by equation of tangent to circle.
@shouvickjoardar45585 жыл бұрын
I used basic geometry. Using equation of tangents will make this much more complicated because we will have to use constraints such as that of direct common tangents which i would not definitely study for "mains".
@MindYourDecisions5 жыл бұрын
That's great to hear. For those that don't know, the JEE Main is an important standardized test in India. You have about 2 minutes per problem, so it really helps if you can solve some problems quickly to have time for other problems. If you watched this video you would have solved question 80 of the 2019 JEE Main nearly instantly. I know people request I cover problems from particular tests. But math is universal: the video is from a test in Indonesia, but the same concept appeared on a test in India.
@anuragsingh-gj4vn5 жыл бұрын
@@MindYourDecisions .....you always come with best tricky mathematics questions....I always pause your video first to try solve the problem myself....and it really feel exciting while solving.....thanks for teaching us....
@martint17755 жыл бұрын
Love the videos, hate the comment sections
@c319798395 жыл бұрын
This is pretty typical for a lot of the videos on KZfaq.
@DD-rl7xo5 жыл бұрын
Then why the hell are you making it even more hateful for others by writing this in comments section.
@jimb46775 жыл бұрын
It happens when you divide yourself by zero (Love/Hate relationship...) :-p
@Tehom15 жыл бұрын
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.
@Tehom15 жыл бұрын
The second solution does not correspond to a big white circle around the other two, as I thought, but to a big white circle that's so far to the right that it touches both other circles from above. This would contradict the diagram but satisfy the equations.
@zlotyinternet15 жыл бұрын
Who cares
@HavenAWilliams4 жыл бұрын
The graphic design on this was so satisfying that I almost forgot to pay attention to the math
@prins0075 жыл бұрын
"Did yiu figure that out?"..... Ehhhhh.. NO!!
@birb16865 жыл бұрын
Wow
@andrew741275 жыл бұрын
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.
@liliacfury5 жыл бұрын
Do you think you could make a video on how to approach challenging problems, to figure them out on your own?
@shivanigandroli82493 жыл бұрын
Yes plz make video
@andrewcorrie89365 жыл бұрын
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.
@ramonchan97325 жыл бұрын
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)
@ramonchan97325 жыл бұрын
@@SuperNovaBass73 Yeah, expanding the compound sine, you will see there is only r left in the equation: (4+r) / sin(θ₁ + θ₂) = (2+r) / sin(θ₃ - θ₁) (4+r) / (sinθ₁cosθ₂ + cosθ₁sinθ₂) = (2+r) / (sinθ₃cosθ₁ - cosθ₃sinθ₁) (4+r) / ((1/3)√(8r) / (2+r) + (√8/3)(2-r) / (2+r)) = (2+r) / (((4-r) / (4+r))(√8/3) - (4√r / (4+r))(1/3)) r = 12 - 8√2
@abdelhakbenkaba10575 жыл бұрын
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
@bag.a.64652 жыл бұрын
imagine an auto-pilot engineering machine that just analyzes the envinronment and builds random things
@sujalarora89473 жыл бұрын
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. 😇😇😇😇😇
@adveshdarvekar77333 жыл бұрын
You have opened the door to genius and beautiful mathematics!
@thewatcherinthecloud5 жыл бұрын
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.
@petrahack5 жыл бұрын
thewatcherinthecloud shouldn’t it be R1-2*R2?
@petrahack5 жыл бұрын
Nope:)
@kostya37125 жыл бұрын
For the first i was confuse, but then i can figure it out
@vsichnirucenahoru3 жыл бұрын
I think it should be R2+(R1-R2)
@stevenvanhulle72424 жыл бұрын
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.
@MikeBTek3 жыл бұрын
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.
@rodrigorocha82125 жыл бұрын
I'm a student of mechatronics engineer and I really like these problems. 👏🏾👏
@shivanshtiwari67944 жыл бұрын
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
@jackash24245 жыл бұрын
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.
@mikesuzio25665 жыл бұрын
same here got a 740 math SAT minored in math in college now cant do much other than simple algebra
@burner8875 жыл бұрын
Take an online course. Any kind. Brilliant is said to hav a great math course, but don't quote me on that.
@the_midnight_blues5 жыл бұрын
Buy some used math books then pursue it at your leisure.
@angeloli77915 жыл бұрын
Maybe khan academy
@nightmare_rex5 жыл бұрын
Khan academy if you want remaster everything abour math
@vjfperez3 жыл бұрын
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
@dannypipewrench533 Жыл бұрын
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.
@fernandomendez695 жыл бұрын
And we can thank Descartes again for inventing analytic geometry... The easiest way to solve this... by far (and for any radii).
@luigidealfaro88313 жыл бұрын
How?
@fernandomendez693 жыл бұрын
@@luigidealfaro8831 You can write the coordinates of the centers of the circles as (0,R1), (a,r), and (a+b,R2), for the centers of the large, small and medium circles. Then, use the Pythagoras theorem to write: a^2+(R1-r)^2=(R1+r)^2 b^2+(R2-r)^2=(R2+r)^2 (a+b)^2+(R1-R2)^2=(R1+R2)^2 solving
@luigidealfaro88313 жыл бұрын
@@fernandomendez69 thx
@vjekokolic90573 жыл бұрын
Presh: "We can solve for big R2" Me: What about D2?
@michaelsorensen75673 жыл бұрын
Diameter is twice the radius, so if you know R2 you also know D2 😝
@Simio_Da_Tundra2 жыл бұрын
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.
@argonwheatbelly6375 жыл бұрын
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.
@shreyashdeogade88695 жыл бұрын
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.
@farendrasingh23345 жыл бұрын
Thank you sir for your creativity and sharing with students
@forestpepper36215 жыл бұрын
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!
@savitajha63755 жыл бұрын
This question was in my book and my teacher solved it.. The question was to prove the equation u gave for sooving..
@sarvesh_soni3 жыл бұрын
same this was teached to me at my coaching in 10th class
@kaziemarieong99535 жыл бұрын
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 😂😂
@Hiyaza25 жыл бұрын
Until the exam paper hits you with that "not to scale"
@kaziemarieong99535 жыл бұрын
Petras Danys HAHAHA then I’m dead 😂 don’t worry, I always get the highest score
@robbierotten22155 жыл бұрын
@@Hiyaza2 or, the radius of R1 is x and the radius of R2 is y, what is r in term of x and y.
@leif10755 жыл бұрын
The last triangle he draws is not justified..how do you know there's a line that goes through the smaller circle that will continue to,be the same line when it passes into the larger circle? You don't ...not enough evidence..
@robbierotten22155 жыл бұрын
@@leif1075 Well, I also have noticed not these are not always rigorous proofs.
@TheMusicalTech5 жыл бұрын
Please can you tell which software you used to edit the video?
@JuliusCesar27813 жыл бұрын
The formula for the solution is brilliant ! I don't expected that this is so easy.
@Uni-Coder5 жыл бұрын
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.
@prateekpal94265 жыл бұрын
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
@brotherzaq5 жыл бұрын
Is this using some sort of tangent theorem, pl explain a little, thanks Andrew
@kachunwong81845 жыл бұрын
I draw these 3 circles in autocad and find the radius as 0.6863
@kancumaniseng4 жыл бұрын
@@ic6406 yes you can
@icarokaue73344 жыл бұрын
@@ic6406 yes you can
@alienpioneer4 жыл бұрын
Yes, the answer is 0.6862915
@eliesaber39484 жыл бұрын
@@ic6406 YES YOU CAN
@MewOneIn4 жыл бұрын
Wait, this cheating
@krishnakalita36344 жыл бұрын
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
@MrProy33 Жыл бұрын
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.
@jimtwisted19845 жыл бұрын
1/r=1/a+1/b reminds me of the resistance formula for parallel electric circuits
@vunguyenkhanh96155 жыл бұрын
Kissing circles theorem
@aryanvnn5 жыл бұрын
Looks like a threesome.
@blizzbee5 жыл бұрын
Threesome circles theorem, then.
@pi174 жыл бұрын
Is it internationally recognised?
@mrminer0711664 жыл бұрын
"Four circles to the kissing come...." except one was flat, had an infinite circumference.
@artsmith1035 жыл бұрын
For 14 year olds!? Amazing. I think this was one of the hardest doable puzzles I've seen Pesh do.
@patricepouget12525 жыл бұрын
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.
@gamingpuzzled75322 жыл бұрын
ur wrong because this question assumes that the 2nd largest circle is half the radius of the largest.
@heliocentric17565 жыл бұрын
(*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.
@user-us5cw3eq8y5 жыл бұрын
That's in my mind, the simplest solution! Only Pythagorean theorem!
@mogu-mogu23354 жыл бұрын
This is in our FIITJEE package....in chapter circle Came in ntse stage 2
@thomasraahauge52315 жыл бұрын
It's half past midnight. I just sat by, baffled, bewildered, and babbling. No inflated ego here. Not even by a long shot.
@callum81474 жыл бұрын
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).
@treyquattro4 жыл бұрын
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!
@hbushnell8 ай бұрын
By multiplying the result, 4/(3+2sqrt(2)), by (3-2sqrt(2))/(3-2sqrt(2)), the denominator simplifies to 9-8=1, creating a much simpler solution!
@kapil_ag__5 жыл бұрын
JEE Mains 2019 ka question h
@vineetmankani84984 жыл бұрын
Which paper? Slot?
@godlysting51834 жыл бұрын
Yes please tell
@crazy4cgl7665 жыл бұрын
Using direct common tangent 2√(4*2)=2√(4r) + 2√(2r) Square both side
@jamessanchez30324 жыл бұрын
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.
@gedlangosz11275 жыл бұрын
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.
@kancumaniseng4 жыл бұрын
I am a simple Indonesian. I see "Indonesia" in title, I upvote..
@Gruntled20014 жыл бұрын
I am a simple Russian. I see "Indonesia" in title, I upvote, too ;)
@keshavxb39033 жыл бұрын
Hello there, I am 14 and I completed it half (you taught me the other one) thank you!
@arthurg.machado68035 жыл бұрын
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
@kimkim-mh5wb5 жыл бұрын
nice formula at the end
@mauriciomon6005 жыл бұрын
Why is it so similar to the formula of the projected image of an object in convex/concave mirror?
@Shivam-kz2dg5 жыл бұрын
@@mauriciomon600 it is also similar to the formula of resistors connected in parallel or capacitor connected in series
@rantlord83735 жыл бұрын
@@Shivam-kz2dg Effective radius of two bubbles, resistance, spring constant of 2 springs in series and list goes on.
@Shivam-kz2dg5 жыл бұрын
@@rantlord8373 self inductance, coefficient of thermal conductivity etc etc etc. Edit:- making the list will be fun
@swingardium7065 жыл бұрын
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.
@chinareds545 жыл бұрын
I found the triangles, too, but I messed up in setting up one of the equations and ended up with a ridiculous radical equation [ 12 * sqrt (r+1) + 2*sqrt (2r) = sqrt (32) ] and I gave up.
@gavin_n5 жыл бұрын
Finally some honest comments!
@madhavgaur54125 жыл бұрын
You know what this happens only to geniuses
@Hexanitrobenzene5 жыл бұрын
SbAsAlSe HONRe I squared (r+2) and got r^2 +2r+4, then I squared (r+4) and got r^2 +16r+16, luckily after that I understood I need to recheck this... :D However, when calculating the square of length of line which is divided in two lines, I still managed to add squares of lengths, instead of adding lengths and then squaring. I got that r=4/3, nice and simple :) I also remember a test during high school, where I completely forgot the sine theorem, used cosine theorem and got lost in radicals... It happens. By the way, judging by username, you like chemistry ?
@Hexanitrobenzene5 жыл бұрын
chinareds54 This equation is actually solvable. If we square both sides, we get three linear terms and one root. Then we arrange only one root on one side of equation, and square again. We get quadratic equation. Roots must be checked in the original equation, because squared equation has more solutions than the original.
@ashleyr3896 Жыл бұрын
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. 🙂
@Sh4dowbanned Жыл бұрын
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
@wanyinleung9125 жыл бұрын
I like how he says keep watching the video for "a" solution
@nirmalajagdish89013 жыл бұрын
Wonderful wayof proceeding thanks. Dr.JPP
@kyleyoung9965 жыл бұрын
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.
@jo_nm94844 жыл бұрын
Nobody: Random dude with a huge ego: *_lOl tHaT waAs EeeEEEeZY 😂😂😂_*
@aakashvashistha59425 жыл бұрын
This question came in ssc cgl 2017
@hdrevolution1235 жыл бұрын
I cannot express my gratitude for such a wonderful problem and clear proofs.
@MehdiHuseynzad5 ай бұрын
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
@shreyansjain48424 жыл бұрын
Mind your decision: Challenging problem from indonesia Jee aspirants:Hold my beer
@eleethtahgra71825 жыл бұрын
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....
@shubhamtiwari81195 жыл бұрын
this question has been asked in so many competitive exam in india..2 or 3 years ago.. right now I'm obsessed
@surfer8555 жыл бұрын
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).
@carlanderson085 жыл бұрын
at 4:50: i don´t understand how to get to the term on the bottom right?
@swingardium7065 жыл бұрын
It took me a while but I got there (I think, there may be some errors but since I got the right answer I'm fairly sure I did everything correctly). I started by writing out the Pythagorean Theorem for the purple triangle. From there, it's a lot of expanding brackets and substituting in for the values of (x1)^2=4*r*R1 and (x2)^2=4*r*R2, but you eventually get to the equation: R1*r + R2*r + 2*r*√(R1*R2) = R1*R2. Dividing this equation through by R1*R2*r gives an equation which is of the form 'a^2 + 2*a*b + b^2 = c^2' (though this is not immediately obvious). Simplifying from there gives the desired equation.
@walexandre94525 жыл бұрын
From the first equation: x1+x2=2√R1√R2 From the second: x1=2√R1√r From the third: x2=2√R2√r Putting x1 and x2 into the first one, and factoring √r: √r(√R1+√R2) = √R1√R2. Then: √r = √R1√R2/(√R1+√R2) Inverse: 1/√r = √R2/(√R1√R2)+√R1/(√R1√R2) So: 1/√r = 1/√R1+1/√R2
@vijayasekhar20225 жыл бұрын
W Alexandre stop nonscense
@walexandre94525 жыл бұрын
@@vijayasekhar2022 ????
@fuseteam5 жыл бұрын
hmm (1) (R1+R2)²-(R1-R2)²=(x1+x2)² (2) (R1+r)²-(R1-r)²=x1² (3) (R2+r)²-(R2-r)²=x2² We need 1/√r=1/√R1+1/√R2 if we expand the right term of (1) we get (R1+R2)²-(R1-R2)²=x1²+2*x1*x2+x2² now we can substitude (2) and (3) for x1² and x2² so we get (R1+R2)²-(R1-R2)²=(R1+r)²-(R1-r)²+2√(((R1+r)²-(R1-r)²)((R2+r)²-(R2-r)²))+(R2+r)²-(R2-r)² If a=R1+r, b=R1-r and c=R2+r then b+c=R1+R2, b+c-a=R2-r and a-c=R1-R2 thus (b+c)²-(a-c)²=a²-b²+2√((a²-b²)(c²-d²))+c²-(b+c-a)² b²+2bc+c²-a²+2ac-c²=a²-b²+2√((a²-b²)(c²-(b+c-a)²))+c²-(b+c-a)² -2(a²-b²)+2ac+2bc-1(c²-(b+c-a)²)=2√((a²-b²)(c²-(b+c-a)²)) ......... i miss my compose key now and I may have overcomplicated it .-. hmm from (R₁+r)²-(R₁-r)²=x₁² we can get R₁²+2R₁r+r²-(R₁²-2R₁r+r²)=x₁² R₁²+2R₁r+r² - R₁²+2R₁r-r²=x₁² x₁²=4R₁r x₁=2√R₁r i really did overcomplicate it o.O considering we can just substitude R₂ for R₁ to get (R₂+r)²-(R₂-r)²=x₂² we can conclude that x₂=2√R₁r thus (R₁+R₂)²-(R₁-R₂)²=(2√R₁r+2√R₂r)² R₁²+2R₁R₂+R₂²-(R₁²-2R₁R₂+R₂²)=4R₁r+4R₁R₂r²+4R₂r 2R₁R₂+2R₁R₂=4R₁r+4R₁R₂r²+4R₂r 4R₁r+4R₁R₂r²+4R₂r=4R₁R₂ R₁R₂r²+R₁r+R₂r=R₁R₂ R₁R₂r²+R₁r+R₂r-R₁R₂=0 R₁R₂r²+(R₁+R₂)r-R₁R₂=0 r=(-(R₁+R₂)±√((R₁+R₂)²-4R₁R₂R₁R₂))/2R₁R₂ r=(-(R₁+R₂)±√(R₁²+2R₁R₂+R₂²-4R₁²R₂²))/2R₁R₂ uhhh hmmmm
@jlsmatejuanluisramirez5 жыл бұрын
I work out this problem with my students in Mexico
@paominlienguite23975 жыл бұрын
Creative ideas...thanks to you, I Love Mathematics as much as I love my Life.
@andiailmann19835 жыл бұрын
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!
@kamallochandash80865 жыл бұрын
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
@jnvc72705 жыл бұрын
That's the same as his answer
@sarbanibanerjee29335 жыл бұрын
But how?
@jnvc72705 жыл бұрын
Kamallochan Dash 5:00
@saikoraito5 жыл бұрын
this video explain how to get that formula
@umutkosar89835 жыл бұрын
@name2 it gives 0.686. Redo it..! I found the correct result with the formula too. Btw, systems push us to MEMORIZE the formulas, not understand where they come from and how! That is why, instead of memorizing them, it is better to find out where and how they are found. It is more useful and catchy for the future..!
@jimweldizon67045 жыл бұрын
can you teach me how r = 32 -(4 + 2√2)^2 became r = 12 - 8√2?
@blothhunder93055 жыл бұрын
I'd also like to know!
@ddc99995 жыл бұрын
32-(16+8V2’+8)=8-8V2’ He’s wrong I guess... OH BOI It’s a division not a subtraction
@kostya37125 жыл бұрын
It is multiplication, so you have to divide it both side to get the r
@rahmahz5 жыл бұрын
It's a division r = 32/(4+2√2)^2 = 12-8√2 That's correct
@eleinaedelweiss62155 жыл бұрын
So r=32/(4+2 square-root-of 2)^2 =32/24+16 square-root-of 2 =4/3+2 square-root-of 2 Than to eliminate the square root = 4 (3 - 2 square-root-of2) -------------------------------------------- (3+2 square-root-of2) (3-2 square-root-of2) = 12 - 8 square-root-of2 / 9-8 = 12 - 8 square-root-of2 (cant find the square-root-of symbol on my damn phone keyboard)
@maalikserebryakov Жыл бұрын
Quite straightforward with the apparatus of cartesian geometry
@IG00Channel3 жыл бұрын
I love the way you say Devan's school! XD Love from Indonesia. 🇮🇩
@TheGrinningSkull5 жыл бұрын
Just when I solved it, you one up me with the general solution. Ahhhhhh.
@Samuel-kc1pg5 жыл бұрын
I had to put the purple triangle up side down to understand why its hight is R1-R2 😅
@ajdajd4 жыл бұрын
Thank you! I was staring for quite a while and couldn't see it!
@21nod4 жыл бұрын
Or you can see that it is R2
@MohamedGaber1114 жыл бұрын
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.
@themathaces83704 жыл бұрын
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.*
@amitbadal525 жыл бұрын
1/✓c=1✓a+ 1✓b Where c is the radius of smaller circle and a,b the radius of other 2 circles
@lemonadeloverz5 жыл бұрын
amit Badal I’m guessing u just copied the answer from the end of the video...
@amitbadal525 жыл бұрын
This is written in my maths note book
@akshiyadav68135 жыл бұрын
@@lemonadeloverz every indian student knows it
@isolayou5 жыл бұрын
I think this was studied by me when i was in 9th or 10th...but had totally forgotten it... basically, when it comes to mathematics, we indians u see....... ........….........…………….… ……………………………. i hope u got it 😉
@niku11935 жыл бұрын
I think you haven't seen the questions of an Indian extrance exam CAT 😂
@wombat4191 Жыл бұрын
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.
@fawadmirza.5 жыл бұрын
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.
@rpa92py155 жыл бұрын
Me gustó la generalización del problema
@Shadowwind44 жыл бұрын
I don’t understand how you knew the hypotenuse of the first triangle was 4 + r
@inktel4 жыл бұрын
have a closer look, the line that connects the centers of the circles, which is our hypothenuse, consists of two radii. as a result, its value is 4+r
@TheLegend-vo7fs4 жыл бұрын
Shadowwind4 what they did was they added both the radius’s together to get the hypotenuse, the larger radius was 4 and the smaller one was r. As a result we got 4+r
@Shadowwind44 жыл бұрын
Ahh okay I see now. Is that always the case for tangent circles?
@jaimecarter39883 жыл бұрын
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!
@PotatoAimMC5 жыл бұрын
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
@uttkarshk79165 жыл бұрын
Well, bro, I got this in my test and surprisingly i cracked it :)
@plaustrarius5 жыл бұрын
Descartes Theorem for tangent circles! a line is a circle of infinite radius!
@parasiticangel83305 жыл бұрын
you could just say the curvature is 0.
@plaustrarius5 жыл бұрын
@@parasiticangel8330 yes, those are equivalent statements. curvature zero sort of loses the essence of it being a circle, at least to me. and the theorem is about tangent circles, that's why I phrased it as such.
@NotYourAverageNothing5 жыл бұрын
how does that help?
@coleozaeta63445 жыл бұрын
Pretty good for a squidbillie. I don’t think I’ve thought of lines like that before
@plutarchheavensbee34835 жыл бұрын
Dont tell this theory to flat earthers. I think their heads would explode if they realized a flat line and an infinite circle have the same curvature.
@CaliforniaBushman3 жыл бұрын
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.
@potterteksmith75484 жыл бұрын
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