The kissing circles theorem - challenging problem from Indonesia!
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- čas přidán 11. 07. 2024
- Thanks to Devan from Lilin Bangsa Intercultural School for suggesting this problem! What is the radius of the small circle in between the blue circle (radius 4) and green circle (radius 2)? This was a challenge problem for students aged 14 to 15.
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Thanks
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."
Triangles are the basics of all geometry, just continue drawing them until you draw the right ones :D
😂
@@mehujmehuj2229 I think also of sailing and navigation
Thanks to let me know i will try to use it in future
Triangles make everything easier.
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.
Solution made me think of formula for parallel resistor (caculated the replace resistor Rv):
1/Rv = 1/R1 + 1/R2
😁
This is a wonderful idea to use the geometry of circles to analyze electrical circuits🤔
@@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
I thought about it too
Made me think of optical lenses.
@@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.
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*
The equation 1/a + 1/b = 1/c is the "optic equation" (see Wikipedia) hence is connected to the heptagon and the "heptagonal triangle".
“Solve without pen. 419”
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!
U r amazing wow😮
somya shree swain noyou cant lol
Great job👏👏
you realize you didnt need vectors and could use variables instead right? Calling it vectors make it cooler or what
is more easy use theorem of Pythagoras
No doubt that this is the best CZcams channel ever, very very thank you sir for making such a great work. please go ahead like this
For students at age 14 and 15. Even I'm 20 I can't solve this. 😭
It is an average question for a high school student in india at least bcz these type of question comes a lot in exams
yeah, what the hell?
"For students at age 14 or 15"
So we are the 20 ones dont count haha
you need some geometry classes , there are alot of information which you missed , or maybe your educational system focuses on a collage degree preps
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 😁!
All the very best :)
I also saw.
In jee they solve by equation of tangent to circle.
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".
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.
@@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....
Love the videos, hate the comment sections
This is pretty typical for a lot of the videos on CZcams.
Then why the hell are you making it even more hateful for others by writing this in comments section.
It happens when you divide yourself by zero
(Love/Hate relationship...) :-p
You have opened the door to genius and beautiful mathematics!
I cannot express my gratitude for such a wonderful problem and clear proofs.
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 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?"
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.
Semangat pak guru!
Yea it used Mostly to solve a problem😂
That is a really good advice. Would be really helpful for them
The formula for the solution is brilliant ! I don't expected that this is so easy.
Wonderful wayof proceeding thanks. Dr.JPP
Do you think you could make a video on how to approach challenging problems, to figure them out on your own?
Yes plz make video
Please can you tell which software you used to edit the video?
I love the way you say Devan's school! XD
Love from Indonesia. 🇮🇩
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.
😇😇😇😇😇
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 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.
Creative ideas...thanks to you, I Love Mathematics as much as I love my Life.
The graphic design on this was so satisfying that I almost forgot to pay attention to the math
Thank you sir for your creativity and sharing with students
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.
same here got a 740 math SAT minored in math in college now cant do much other than simple algebra
Take an online course. Any kind. Brilliant is said to hav a great math course, but don't quote me on that.
Buy some used math books then pursue it at your leisure.
Maybe khan academy
Khan academy if you want remaster everything abour math
I'm a student of mechatronics engineer and I really like these problems. 👏🏾👏
Nice vídeo !! I wonder which program do you use for the animations along the problems
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 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.
Who cares
"Did yiu figure that out?"..... Ehhhhh.. NO!!
Wow
thanks. you always help to expand our knowledge
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).
And we can thank Descartes again for inventing analytic geometry...
The easiest way to solve this... by far (and for any radii).
How?
@@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
@@fernandomendez69 thx
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
imagine an auto-pilot engineering machine that just analyzes the envinronment and builds random things
love the art of your explaining and animation
It's half past midnight. I just sat by, baffled, bewildered, and babbling. No inflated ego here. Not even by a long shot.
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.
thewatcherinthecloud shouldn’t it be R1-2*R2?
Nope:)
For the first i was confuse, but then i can figure it out
I think it should be R2+(R1-R2)
1/r=1/a+1/b reminds me of the resistance formula for parallel electric circuits
Great problem, great solution.
Thanks for the video.
Cool! I had the idea in my head but couldn't quite rationalize an appropriate solution. Thanks for refreshing my mind!
you mean ... re- _Presh_ -ing your mind?!?
This is in our FIITJEE package....in chapter circle
Came in ntse stage 2
Presh: "We can solve for big R2"
Me: What about D2?
Diameter is twice the radius, so if you know R2 you also know D2 😝
I have this question as a homework assignment, thanks so much for the help!
I opt biology but I love to solve such problems. I love your channel. ❤❤
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!
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!
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)
@@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
Very elegant solution!. The general solution is also insightful.
Really nice one sir, your videos are always good for those who really want to learn
I like how he says keep watching the video for "a" solution
Me gustó la generalización del problema
I am amaze the approach for solving this problem
For 14 year olds!? Amazing. I think this was one of the hardest doable puzzles I've seen Pesh do.
(*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.
That's in my mind, the simplest solution! Only Pythagorean theorem!
I work out this problem with my students in Mexico
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!
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.
Kissing circles theorem
Looks like a threesome.
Threesome circles theorem, then.
Is it internationally recognised?
"Four circles to the kissing come...." except one was flat, had an infinite circumference.
Well, bro, I got this in my test and surprisingly i cracked it :)
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!
thank u sir for these amazing videos...
Mind your decision:
Challenging problem from indonesia
Jee aspirants:Hold my beer
at 4:50: i don´t understand how to get to the term on the bottom right?
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.
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
W Alexandre stop nonscense
@@vijayasekhar2022 ????
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 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.
ur wrong because this question assumes that the 2nd largest circle is half the radius of the largest.
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!
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 😂😂
Until the exam paper hits you with that "not to scale"
Petras Danys HAHAHA then I’m dead 😂 don’t worry, I always get the highest score
@@Hiyaza2 or, the radius of R1 is x and the radius of R2 is y, what is r in term of x and y.
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..
@@leif1075 Well, I also have noticed not these are not always rigorous proofs.
This question was in my book and my teacher solved it.. The question was to prove the equation u gave for sooving..
same this was teached to me at my coaching in 10th class
this question has been asked in so many competitive exam in india..2 or 3 years ago.. right now I'm obsessed
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
I draw these 3 circles in autocad and find the radius as 0.6863
@@ic6406 yes you can
@@ic6406 yes you can
Yes, the answer is 0.6862915
@@ic6406 YES YOU CAN
Wait, this cheating
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
Is this using some sort of tangent theorem, pl explain a little,
thanks
Andrew
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.
Thankyou for inspiring us with your maths
nice formula at the end
Why is it so similar to the formula of the projected image of an object in convex/concave mirror?
@@mauriciomon600 it is also similar to the formula of resistors connected in parallel or capacitor connected in series
@@Shivam-kz2dg Effective radius of two bubbles, resistance, spring constant of 2 springs in series and list goes on.
@@rantlord8373 self inductance, coefficient of thermal conductivity etc etc etc.
Edit:- making the list will be fun
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.
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.
Just when I solved it, you one up me with the general solution. Ahhhhhh.
JEE Mains 2019 ka question h
Which paper? Slot?
Yes please tell
Hello there, I am 14 and I completed it half (you taught me the other one) thank you!
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 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.
All your videos are awesome
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.
I had to put the purple triangle up side down to understand why its hight is R1-R2 😅
Thank you! I was staring for quite a while and couldn't see it!
Or you can see that it is R2
Nobody:
Random dude with a huge ego: *_lOl tHaT waAs EeeEEEeZY 😂😂😂_*
Thanks ! Couldn’t think of this method
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
I am a simple Indonesian. I see "Indonesia" in title, I upvote..
I am a simple Russian. I see "Indonesia" in title, I upvote, too ;)
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.
Hi
In case:2 for formulae, how did you arrive the side of purple triangle ( R1-R2). Does it works?
Thank you for this easy explanation sir..
I think you haven't seen the questions of an Indian extrance exam CAT 😂
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....
Love all these problems.....💓💓🤓
At time 2:24 in the video how do you get the 2 times the square root of 2 times the square root of r from that Pythagorean theorem above? I’m confused? Thanks
I used Pamela's law where both circles intercepted each other
Where can i find more info of that law?
Descartes Theorem for tangent circles!
a line is a circle of infinite radius!
you could just say the curvature is 0.
@@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.
how does that help?
Pretty good for a squidbillie. I don’t think I’ve thought of lines like that before
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.
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.
hello presh..May i ask?Have you ever took part in International mathematics olympiad?