Thanks

@DaFlashPro

Жыл бұрын

welcome

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."

@mehujmehuj2229

5 жыл бұрын

Triangles are the basics of all geometry, just continue drawing them until you draw the right ones :D

@sanjeevgoel8865

4 жыл бұрын

😂

@pablomalaga4676

4 жыл бұрын

@@mehujmehuj2229 I think also of sailing and navigation

@vivekkedia2140

4 жыл бұрын

Thanks to let me know i will try to use it in future

@DarthAlphaTheGreat

3 жыл бұрын

Triangles make everything easier.

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

@user-us5cw3eq8y

5 жыл бұрын

This is a wonderful idea to use the geometry of circles to analyze electrical circuits🤔

@chaos.corner

4 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

@matteobruneau7757

4 жыл бұрын

I thought about it too

@arielfuxman8868

4 жыл бұрын

Made me think of optical lenses.

@Walczyk

3 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.

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".

@victorgorelik7383

8 ай бұрын

“Solve without pen. 419”

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

@ashutoshkaushik7375

5 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

@unyielding37

5 жыл бұрын

yeah, what the hell?

@AMa-us8de

5 жыл бұрын

"For students at age 14 or 15" So we are the 20 ones dont count haha

@justabeardedguythatisahero9848

5 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

@marcoantonioloureiro5883

5 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)

This question came in JEE MAINS this year 😁!

@shouvickjoardar4558

5 жыл бұрын

All the very best :)

@usmanakhtar3992

5 жыл бұрын

I also saw. In jee they solve by equation of tangent to circle.

@shouvickjoardar4558

5 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".

@MindYourDecisions

4 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-gj4vn

4 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....

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 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?"

@hassanakhtar7874

4 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.

@daffadaniirfan5928

4 жыл бұрын

Semangat pak guru!

@pandawagurning

3 жыл бұрын

Yea it used Mostly to solve a problem😂

@udayomsrivastava9052

Жыл бұрын

That is a really good advice. Would be really helpful for them

Love the videos, hate the comment sections

@c31979839

5 жыл бұрын

This is pretty typical for a lot of the videos on KZread.

@DD-rl7xo

5 жыл бұрын

Then why the hell are you making it even more hateful for others by writing this in comments section.

@jimb4677

5 жыл бұрын

It happens when you divide yourself by zero (Love/Hate relationship...) :-p

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-io2sb

5 жыл бұрын

U r amazing wow😮

@wec3270

5 жыл бұрын

somya shree swain noyou cant lol

@yourlordandsaviouryeesusbe2998

5 жыл бұрын

Great job👏👏

@Ts1nd

5 жыл бұрын

you realize you didnt need vectors and could use variables instead right? Calling it vectors make it cooler or what

@arturoortiz5907

5 жыл бұрын

is more easy use theorem of Pythagoras

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

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

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

@shivanigandroli8249

3 жыл бұрын

Yes plz make video

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.

You have opened the door to genius and beautiful mathematics!

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).

Thank you sir for your creativity and sharing with students

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.

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. 😇😇😇😇😇

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

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.

@petrahack

5 жыл бұрын

thewatcherinthecloud shouldn’t it be R1-2*R2?

@petrahack

5 жыл бұрын

Nope:)

@kostya3712

5 жыл бұрын

For the first i was confuse, but then i can figure it out

@vsichnirucenahoru

3 жыл бұрын

I think it should be R2+(R1-R2)

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

@birb1686

5 жыл бұрын

Wow

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

thanks. you always help to expand our knowledge

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

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.6465

2 жыл бұрын

imagine an auto-pilot engineering machine that just analyzes the envinronment and builds random things

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

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

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.

@mikesuzio2566

5 жыл бұрын

same here got a 740 math SAT minored in math in college now cant do much other than simple algebra

@burner887

5 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_blues

5 жыл бұрын

Buy some used math books then pursue it at your leisure.

@angeloli7791

5 жыл бұрын

Maybe khan academy

@nightmare_rex

5 жыл бұрын

Khan academy if you want remaster everything abour math

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

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!

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.

@gamingpuzzled7532

2 жыл бұрын

ur wrong because this question assumes that the 2nd largest circle is half the radius of the largest.

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.

@Tehom1

5 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.

@zlotyinternet1

5 жыл бұрын

Who cares

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

@luigidealfaro8831

3 жыл бұрын

How?

@fernandomendez69

3 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

@luigidealfaro8831

3 жыл бұрын

@@fernandomendez69 thx

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

Great problem, great solution. Thanks for the video.

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)

@ramonchan9732

5 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

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

I have this question as a homework assignment, thanks so much for the help!

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!

Me gustó la generalización del problema

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

@michaelsorensen7567

3 жыл бұрын

Diameter is twice the radius, so if you know R2 you also know D2 😝

I am amaze the approach for solving this problem

love the art of your explaining and animation

(*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-us5cw3eq8y

5 жыл бұрын

That's in my mind, the simplest solution! Only Pythagorean theorem!

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

@sarvesh_soni

3 жыл бұрын

same this was teached to me at my coaching in 10th class

Since it was drawn to scale so nicely, by visualization I would have guessed 0.7. Great generalized solution (proof?) away from the digital crutch ! I guess it might be useful for packaging - na, they just draw a circle in CAD and just get a rough idea.

Great video! I'd like a more detailed explination for the solution, it took me quite a while to find out why I was wrong (forgot the ² in 32=r(4+2√2)²...)

Kissing circles theorem

@aryanvnn

5 жыл бұрын

Looks like a threesome.

@blizzbee

5 жыл бұрын

Threesome circles theorem, then.

@pi17

4 жыл бұрын

Is it internationally recognised?

@mrminer071166

3 жыл бұрын

"Four circles to the kissing come...." except one was flat, had an infinite circumference.

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!

@hbushnell

6 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!

Really nice one sir, your videos are always good for those who really want to learn

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.

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

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 😂😂

@Hiyaza2

5 жыл бұрын

Until the exam paper hits you with that "not to scale"

@kaziemarieong9953

5 жыл бұрын

Petras Danys HAHAHA then I’m dead 😂 don’t worry, I always get the highest score

@robbierotten2215

5 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.

@leif1075

4 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..

@robbierotten2215

4 жыл бұрын

@@leif1075 Well, I also have noticed not these are not always rigorous proofs.

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!

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

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

@brotherzaq

5 жыл бұрын

Is this using some sort of tangent theorem, pl explain a little, thanks Andrew

nice formula at the end

@mauriciomon600

5 жыл бұрын

Why is it so similar to the formula of the projected image of an object in convex/concave mirror?

@Shivam-kz2dg

5 жыл бұрын

@@mauriciomon600 it is also similar to the formula of resistors connected in parallel or capacitor connected in series

@rantlord8373

5 жыл бұрын

@@Shivam-kz2dg Effective radius of two bubbles, resistance, spring constant of 2 springs in series and list goes on.

@Shivam-kz2dg

5 жыл бұрын

@@rantlord8373 self inductance, coefficient of thermal conductivity etc etc etc. Edit:- making the list will be fun

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.

Nice vídeo !! I wonder which program do you use for the animations along the problems

I work out this problem with my students in Mexico

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

@kancumaniseng

4 жыл бұрын

@@ic6406 yes you can

@icarokaue7334

4 жыл бұрын

@@ic6406 yes you can

@alienpioneer

4 жыл бұрын

Yes, the answer is 0.6862915

@eliesaber3948

4 жыл бұрын

@@ic6406 YES YOU CAN

@fazaazka2887

4 жыл бұрын

Wait, this cheating

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.

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

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

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

@swingardium706

5 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.

@walexandre9452

5 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

@vijayasekhar2022

5 жыл бұрын

W Alexandre stop nonscense

@walexandre9452

5 жыл бұрын

@@vijayasekhar2022 ????

@fuseteam

5 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

thank u sir for these amazing videos...

I opt biology but I love to solve such problems. I love your channel. ❤❤

This question came in ssc cgl 2017

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.

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

Good explanation.Thank you, God bless you.

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

@ajdajd

4 жыл бұрын

Thank you! I was staring for quite a while and couldn't see it!

@21nod

4 жыл бұрын

Or you can see that it is R2

Wonderful wayof proceeding thanks. Dr.JPP

Cool! I had the idea in my head but couldn't quite rationalize an appropriate solution. Thanks for refreshing my mind!

@treyquattro

3 жыл бұрын

you mean ... re- _Presh_ -ing your mind?!?

JEE Mains 2019 ka question h

@vineetmankani8498

4 жыл бұрын

Which paper? Slot?

@godlysting5183

4 жыл бұрын

Yes please tell

I used Pamela's law where both circles intercepted each other

@diegoalvarez5289

4 жыл бұрын

Where can i find more info of that law?

Thank you for this easy explanation sir..

Love all these problems.....💓💓🤓

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

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

@Gruntled2001

4 жыл бұрын

I am a simple Russian. I see "Indonesia" in title, I upvote, too ;)

Very elegant solution!. The general solution is also insightful.

S in India, we hav these types of maths problems in our 8th 9th nd 10th standards... But i had forgotten... thx presh for making me remember those gud old days..!

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....

btw the answer is 12-8*sqrt(2).

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

This is the first problem in your channel I could solve it myself.... This one is 6th or 7th problem I am encountering in your channel before that I gave up easily and saw the solution from next time I will give a try to every problem I will see....

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

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

@parasiticangel8330

5 жыл бұрын

you could just say the curvature is 0.

@plaustrarius

5 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.

@NotYourAverageNothing

5 жыл бұрын

how does that help?

@coleozaeta6344

5 жыл бұрын

Pretty good for a squidbillie. I don’t think I’ve thought of lines like that before

@plutarchheavensbee3483

5 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.

I've been watching your videos for a while now, when I first started I couldn't hope to solve one of these questions. Your channel actually helped me improve in math to the point where I was actually able to solve this question. I feel good man (btw im 15 so me being able to solve this problem actually makes me feel good)

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.

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.

@chinareds54

5 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_n

5 жыл бұрын

Finally some honest comments!

@madhavgaur5412

5 жыл бұрын

You know what this happens only to geniuses

@Hexanitrobenzene

5 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 ?

@Hexanitrobenzene

5 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.

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

@blothhunder9305

5 жыл бұрын

I'd also like to know!

@ddc9999

5 жыл бұрын

32-(16+8V2’+8)=8-8V2’ He’s wrong I guess... OH BOI It’s a division not a subtraction

@kostya3712

5 жыл бұрын

It is multiplication, so you have to divide it both side to get the r

@rahmahz

5 жыл бұрын

It's a division r = 32/(4+2√2)^2 = 12-8√2 That's correct

@eleinaedelweiss6215

5 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)

I m very excited to solve this prblm by another method

You can also solve for the top length of the rectangle as 4√2 (using pythagorean) and then set that equal to √r + (2√2)√r. But the algebra on that is kind of messy.

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

I knew that question,but never know the answer(i'm indonesian,though)

@kanui3618

5 жыл бұрын

Pahamin aja dulu bro👍

@justaman7124

3 жыл бұрын

Perbaiki dulu grammar lu msh ada yg salah

Thought about this one fir a while. Solved for the rough answer of 32/(4+2sqrt(2))^2 in my head. Then used paper to simplify to the 12-8sqrt2 answer.

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