# Challenge 78: Infinitely Many Tangent Circles

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• Published on Jan 31, 2019
• Congratulations to Rishav Gupta, Kartik Sharma, Xu Chen Tan, Francesco Debortoli, Simon Armstrong, and The vast universe for successfully solving the last week's math challenge question! Rishav Gupta was the first person to solve the question.
Your support is truly a huge encouragement.
Every subscriber and every like are wholeheartedly appreciated.
Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
For more Weekly Math Challenges:

• sarang shri 7 months ago

Infinite sum of radius in gp will be the length of line joining origin to centre.
This will solve in a giffy.

• Fernando Peñaherrera 7 months ago

• Mokou Fujiwara 7 months ago

I did it by dividing this problem into 3 parts.
The steps are very long for me, and I need to draw some diagram. it will be more convenient to post my solution in an image.
URL: i.imgur.com/gnPXkxG.png

• Mokou Fujiwara 7 months ago

One thing to note: the line passing through origin and all centers of circles in the diagram is labelled as Lc, and in the steps, I wrote L2.
L2 and Lc is the exact same line, y=xtan(π/8)

• Rahul Patil 7 months ago

Plz help me to integrate e^sinx

• yowdheya 7 months ago

a=4, b=6, c=2

• Manda Thool 7 months ago

all centre(c0,c1,..) lie on equation y=(√2 - 1)x [since circles are tangent to y=x and y=0] ,distance between Co and C1 equals R1+Ro and is also equal to difference in length of OCo and OC1(where O is origin),
OCo = Ro. √[(√2+1)^2 +1]
OC1 = R1. √[(√2 +1)^2+1]
now, OCo - OC1 = R1+Ro
on solving R1= [√(4 + 2√2) -1]/[√(4 + 2√2) + 1]
therefore radius form an infinite GP with
first term = Ro = 1
and common ratio = R1
C = 2π[Ro+R1+R2+ ...]
on solving infinite GP
C = π[(4+2√2) +1]
now , A = π[(Ro)^2 + (R1)^2 + (R 2)^2 + ....]
on solving the following GP
A = π [(√(4+2√2) + 1)]^2/[4√(4+2√2)]
on solving C^2/A = 4π[√(4+2√2)]
a=4
b=4
c=2

• Anurag Gupta 7 months ago

I think the answer is 4*pi*sqroot(5). As the radius of circle w1 is (3-sqroot(5))/2 .

• Charan Sankar 7 months ago

Step 1:
>To know that the radii form a geometric sequence
>To Calculate the angle formed between the line connecting the origin to the centre of Omega 0 which equals 22.5
Step 2:
The distance from the centre of Omega 1 to Omega 0 is the radius of omega 1 (which we will call r) + the radius of Omega 0 (which is 1). Therefore it equals r+1
Step 3:
The height between the centre if Omega 1 and Omega 2 needs to be found. This is 1-r since you need to take away the radius of Omega 1 and Omega 0.
Step 4:
C^2 = 2*pi*r*(1+r+r^2+...) =4*pi^2/(1-r)^2 and A=pi*(1+r^2+r^4+...) = pi/(1-r^2). Therefore C^2/A = 4pi*((1+r)/(1-r))
Step 5:
From before we can evaluate (1+r)/(1-r) as sine of the angle formed between the line connecting the origin to Omega 0 and the x-axis which is sine of 22.5
Step 6: Putting everything together we get 4*pi*sin(22.5) and using half angle identities we evaluate this to be 4pi(sqrt(4+2sqrt2))
Done!!!

• Serengeti Ghasa 7 months ago

hi
a=4 ;b=4 and c=2
as sin22•5= (1-r₁)/(1+r₁)=(r₂-r₁)/(r₂+r₁) and so on we can found
if r₁=r then
r₂=r² r₃=r³ like this rₙ=rⁿ
then simplify
(2πΣrⁿ)²/πΣ(rⁿ)²
=4π√(4+2√2)
With love from
Prabhat kumar Sahu
odisha india

• Alguien que hable español ? Xd me siento solito

• Arnold Joseph Sanchez Munayco 7 months ago

Christopher Marley no era lo que esperaba ,pero queria ver si un latino tambien ve estos videos xd

• Christopher Marley 7 months ago +1

Arnold Joseph Sanchez Munayco Pues, yo hablo ingles, pero estoy aprendiendo español en mi escuela.

• Pete Berg 7 months ago

Because iam so stupid that after lots of weekly math challenge I've attempt, this is the first time i know what to do. Sorry because i put the answer in the drive because i can't explain things well

• Lawgx 7 months ago

• Czeckie 7 months ago

it's interesting to note, that the answer is actually equal to 4*pi*D, where D is distance from origin to the centre of the largest circle.

• Nicola C 7 months ago

Let's first find the radius of a circle r with respect to the radius R of the previous circle.
With a proportion we can say that
rR/sin(α)=r²/sin(α)+r²+rR
and, with some work,
r=(1-sin(α))/(1+sin(α))*R
with sin(α)=sin(π/8)=√(2-√2)/2.
Let (1-sin(α))/(1+sin(α)) be E.
C²/A=4*π*(1+E)/(1-E)=4*π*1/sin(α)=4*π*2/√(2-√2).
With some simplification we get
4*π*√(4-2√2) and we are done.

• Bisham Dewan 7 months ago

Here the radius of two consecutive circles satisfy a recurrence relation from where we get the radius of n th circle =k.k.k... n times, k=(1-sin(22.5))/(1+sin(22.5)).Now simple use of sum of infinite geometric series gives the answer in the desired form.

• Marben James Baculpo 7 months ago

Solution
Create a line from (0,0) to the center of wo. this line is bisecting the angle formed by the x axis and the line y =x.
We will designate D as the sum of diameters.
Observations about the line (now called AB)
it can be seen that line ab connects the circles and passes through their tangent points, meaning its length is the sum of diameters-except for the other "radius" of wo, so we will have to add one to get the D . It is also bisecting the x=y line, thus having an angle(22.5). As abc is a right triangle, ab must be bc/sin(22.5). Bc is one so AB is 1/sin22.5, still missing the other radius so we add it to get 1/sin 22.5 +1, which is the correct value of D.
Designate r as ratio of the sequence of diameters.
it should also be known that do, di, ... form a geometric sequence. We know D is (1/sin 22.5 +1), and that D = do/1-r as it is a geometric sequence. It follows that r is 1-( 2/D) and thus thar r^2 =(1-2/D))^2.
(this is the formula for g. Series, S = first term divided by (1- ratio))
Letting Ď be the summation of D^2, we can use the fact that do, etc.... form a geometric sequence , so
Ď = 4/(1-(1-2/D)^2)
This is because the squares also form a geometric sequence except with g,ratio and first term squared
So S = do^2/ 1-r^2
Solving Time:
The summation of circumference is pi*D, squared is (pi*D)^2.
The summation of area is Ď*pi/4, dividing one by the other we get
C2/A equals 4*pi * D^2/Ď, which simplifies to (i am skipping the algebra)
4*pi*(D-1),
Seeing D= 1/sin 22.5+1, an seeing that it has been subtracted by 1 above, we get
c^2/a = 4*pi/sin 22.5
Using half angle identity, sin 22.5 is sqrt(1-cos 45/2.) This simplifies to 2/sqrt(2-sqrt2)). Using more algebra that I have chosen to skip, c^2/a simplifies to
4pi* sqrt(4+ 2*sqrt(2))).
Hooray

• LetsSolveMathProblems  7 months ago +1

@Marben James Baculpo I very rarely penalize a solution for minor error(s), especially if you post a quick correction as a follow up. However, I do remark that your answer probably will not be recognized because it is not within the first ten chronologically.

• Marben James Baculpo 7 months ago +2

+LetsSolveMathProblems Oh no! I forgot to specify that BC was the line connecting wo's center to the x axis at their tangent point. Will this response still be accepted?

• Arth Raj 7 months ago +2

Use similarity and trigonometry makes it easier and just solve in 5 min
Construction of a line centre of second circle perpendicular to radius on first circle. And then apply similarity and use trigonometry taking theta π/8

• Joël Ganesh 7 months ago +1

@Arth Raj Ah, now I get your explanation totally. Thanks!

• Arth Raj 7 months ago +1

@Joël Ganesh I guess using trigonometry gives us geometric progression ratio which gives us area and circumference as well ;) 4πcosec(π/8)

• Joël Ganesh 7 months ago +1

I tried this also, the circumferences can be easily calculated this way. However, you need to calculate the sum of areas of the individual circles as well. This is not a linear process so the strategy won't work for the areas.

• el tapa 7 months ago

Looks interesting

• Kartik Sharma 7 months ago

a=4. b=4 c=2
My first recognition was in this video ..😊😊😊😊

• Mahesh Kumar 7 months ago +2

Considering the first two tangent circles let the radius of the smaller one be r.The distance between the centres of the circles=r+1 and the difference between their heights=1-r.The line y=x makes a 45° angle with the x-axis and since the circles are tangent to both x axis and y=x the angle bisector of the 45° angle will clearly pass through the centres of all the circles.Therefore it can be easily seen that sin(45/2)=(1-r)/(1+r). Similarly the radius of the third circle will be r^2.Now c=2π(1+r+r^2+....UpTo infinity) and A=π(1+r^2+r^4+....)
We can evaluate A and c using sum of an infinite G.P. and we get c^2/A=4π×sqrt(4+2√2)
Hence a=4,b=4 and c=2

• Mahesh Kumar 7 months ago

Just because they're tangent to each other

• Mahesh Kumar 7 months ago

The length of the line joining the centres is equal to the sum of their respective radii

• Jeet Sharma 7 months ago

Can u give a lil bit more explanation abt how distance between two centres is r+1 ?

• Aswini Banerjee 7 months ago

Let centre of w_0 is A and of w_1 is B , perpendicular from A to x axis is C and perpendicular from B to AC is D.
Now if redius of w_1 is r then from ABD right angled tringle (1-r)/(1+r)=sin

• fmakofmako 7 months ago

for a given circle of radius r the position along the x axis of tangency to the y=x line is at r+r/sqrt(2) and the position along the x axis of tangency to the x axis is at r(1+sqrt(2)).
the difference in x positions of adjacent circles w0 and w1 is then (1-r)(1+sqrt(2)), the difference in y positions is (1-r) and the total distance between the centers is (1+r).
You can do the same type of analysis for any adjacent circles and come up with the same equation where r is the ratio of the smaller circle's radius to the larger circle's radius.
Therefore r is not only the radius of w1, but also the geometric ratio for all the circles. You can then see that A = pi + pir^2+pi r^4+.... = pi/(1-r^2) and C = 2 pi + 2 pi r + 2 pi r^2 +... = 2 pi/(1-r).
The sums are convergent because r (1+r)^2/(1-r)^2 = 4+2sqrt(2)
so C^2/A = 4 pi sqrt(4+2sqrt(2))
a = 4, b = 4, c =2

• Jeremy Weissmann 7 months ago

I claim C²/A = 4π · √( 4 + 2√2 ).
In my solution, I assume that the radii of the circles form a geometric sequence {rₙ}. We are given r₀ = 1, and the ratio of the sequence is therefore r₁/r₀ = r₁.
First, let us look at the expressions we are being asked to compute. Assuming the sequence of radii is geometric with ratio r₁, we calculate:

=
( Σ 2π · rₙ )²
=
4π² / (1 - r₁)²
and
A
=
Σ π · rₙ²
=
π / (1 - r₁²)
where in the second calculation we used that {rₙ²} is geometric if {rₙ} is. Now some simple algebra gives:
C²/A
=
4π · (1 - r₁²) / (1 - r₁)²
=
4π · (1 + r₁) / (1 - r₁)
which is a fine start to our calculation!
In order to calculate further, we must take into account the geometry of the figure. Consider circle ω₀: Extending the radius drawn in the given diagram up to the line y = x, we form a 45º-45º-90º triangle. Further, connecting the center of circle ω₀ to the point of tangency on line y = x, we form another 45º-45º-90º triangle. The proportions of this triangle being known (namely, 1:1:√2), we can easily fill in the missing segment lengths, and in particular we conclude that circle ω₀ is tangent to the x-axis at a distance of 1 + √2 from the origin.
Next, draw in radius r₁ from the center of circle ω₁ down to the x-axis. Because circles ω₀ and ω₁ are tangent to each other, their centers may be connected with a line segment of length 1 + r₁. Following the argument in the previous paragraph, circle ω₁ is tangent to the x-axis at a distance of (1 + √2) · r₁ from the origin, and hence the distance between the points of tangency of circles ω₀ and ω₁ on the x-axis is (1 + √2) · (1 - r₁).
The line segments under discussion form a trapezoid, and, by drawing a line segment parallel to the x-axis, we can form a right triangle with legs (1 + √2) · (1 - r₁) and 1 - r₁, and hypotenuse 1 + r₁. Dividing the side lengths by the shared factor 1 - r₁, and applying the Pythagorean theorem to solve for the length of the hypotenuse ( c = √(a² + b²) ), we conclude:
(1 + r₁) / (1 - r₁) = √[ (1 + √2)² + 1² ] = √( 4 + 2√2 ) .
With this relation in hand, we can easily complete the above calculation of C²/A as follows:
C²/A
=
4π · √( 4 + 2√2 )
which is what we wished to show.

• rhythm mandal 7 months ago

Soln: a=4, b=4, c=2.

• PhigNewton1 7 months ago +2

Part 1: One must realize that the radii of circle ω_n form a geometric series. By scaling down the graph so that ω_0 is over where ω_1 was originally produces the same set of circles, except with the first one missing. Realizing this, we can find the sum of the areas and the sum of the circumferences.
Part 2: Assuming the r_0=1, with r_n being the radius of ω_n, we derive the following formulas for area and circumference: C_n=2πrⁿ, A_n=πr²ⁿ, with r being the scaling factor between consecutive circles. Finding the total sum gives us C=2πΣrⁿ=2π/(1-r) and A=πΣr²ⁿ=π/(1-r²). C²/A=4π(1+r)/(1-r).
Part 3: Let's look at a right triangle. This triangle is made by a line connecting two neighboring centers and a horizontal and vertical line coming from each of them. The hypotenuse of this triangle has length (1+r), and the vertical side has length (1-r). We know that the angle opposite the vertical side is 22.5°, since the hypotenuse runs along a line equidistant from the x-axis and y=x. From this, we can conclude that (1+r)/(1-r)=1/sin(22.5°).
Part 4: We can find csc(22.5°) using the double angle formula - sin(45°)=2sin(22.5°)cos(22.5°) → sin(45°)=2sin(22.5°)√(̅1̅-̅s̅i̅n̅²̅(̅2̅2̅.̅5̅°̅)̅) → ½=4sin²(22.5°)(1-sin²(22.5°)) → 0=(sin²(22.5°))²-sin²(22.5°)+⅛ → (sin²(22.5°))=½-½√½ → 1/sin(22.5°)=√(̅4̅+̅2̅√̅2̅).
Part 5: Plugging the results from the last two parts into our C²/A equation gives us C²/A=4π/sin(22.5°)=4π√(̅4̅+̅2̅√̅2̅).

• Veeryan Bhatia 7 months ago +1

Solution: write the equation of the line y=x in standard form. Now we can use the distance from a point to a line formula. If we let the x coordinate of the circle with radius 1 be x then we use the distance from a point to a line to see that x = sqrt (2) +1. Now since all the centers of circles lie on the same line, we can set up a right triangle between 2 centers, set some variables, and solving tells us that the ratio between the radius of w_i and w_(i-1) = (sqrt (4+2sqrt (2))-1)÷(sqrt (4+2sqrt (2))+1). We let that quantity be n (common ratio). Now we have 2 infinite geometric sequences. The sum of all areas is easily seen to be = pi÷(1-n^2). The sum of all circumference = 2pi÷(1-n).
We seek: (4pi^2÷(1-n)^2)÷(pi÷(1-n^2)). This simplifies to: 4pi*(1+n)/(1-n). Plugging in the value for n we get the above answer.

The solution is a=4, b=6, c=4.
First, by letting the circle be centred at (a,r) then the point tangent to y=x is (a-r/sqrt(2),r+r/Sqrt(2)) so since y=x, a = r (1+sqrt(2)). Then the line between the centres of any circle and the smaller one to its left has length r_n + r_(n-1) which is the hypotenuse of a triangle of angle 45 and adjacent side (r_n - r_(n-1) ) * (1+sqrt(2)). That lets us find r_(n-1) = lambda r_n, where lambda = (1+sqrt(2))/(3+sqrt(2)). Then C = 2*pi*(1+lambda+lambda^2+... ) = 2*pi/(1-lambda) and A similarly = pi/(1-lambda^2). Thus C^2/A = 4*pi*(2+sqrt(2)). To make Prof LSMP happy we can write 2+sqrt(2) = sqrt(6+4*sqrt(2) ) and read off the answer.
Edit: looking at the other answers, I see I made a boneheaded error. Of course the triangle at the end of my calc does not have angle 45 (as a casual glance at the diagram confirms).
In fact it is a triangle with hypot = r_n+r_n-1, opposite r_n - r_(n-1) and adjacent (r_n - r_(n-1) ) *(1+sqrt(2)) and then Pythagoras gives the ratio of r_(n-1) to r_n. Rats :(

• Arun Bharadwaj 7 months ago +6

You can find the equation of the line joining the origin and w_0, by noticing that the angle it makes with the x axis is 45/2. Using half angle identities for tan, we arrive at the x value where w_0 is tangent to x-axis (x=sqrt 2 +1). Constructing the equation is fairly simple now. We arrive at y=(sqrt(2)-1)x. Due to symmetry, it is easy to show that all the centers of the circles lie along this line. This makes it easy to find the distance between origin and w_0 which is sqrt(4+2sqrt(2)). Now, we know that 1+sum_(i=1)^(inf)2r_i = sqrt(4+2sqrt(2)) where r_i is the radius of the i^th circle excluding w_0. We know that the radii forms a geometric series. So it is easy to find the common ratio as a = (sqrt(4+2sqrt(2))-1)/(sqrt(4+2sqrt(2))+1). Therefore r_n = a^n.
From here, we can calculate (sum _(i=0)^inf (2pi*a^i))/(sum_(i=0)^(inf) (pi a^2i). This gives us the result of 4pi(sqrt(4+2sqrt2)

• PRAKHAR AGARWAL 7 months ago

4pie(root2)(root(2+root(2)))
Let the equation of circle be (x-a) ^2+(y-b)^2=c^2
As it is tangent to x axis therfore b=c
As it is tangent to y=x therfore a=b(1+root(2))
Now all consecutive circles are tangent to each other
Therfore the distance between their centre is equal to sum of radius
Which gives us a recursive relation for Bn=B0(root(4+2root2)-1)/(root(4+2root2)+1)
Now for C it is a GP with common difference (root(4+2root2)-1)/(root(4+2root2)+1) and for A the common difference is
((root(4+2root2)-1)/(root(4+2root2)+1)) ^2

• Cobalt314 7 months ago

Let us draw a line through the centers of the circles. This line forms an angle of pi/8 with the x-axis, and looking to each pair of adjacent circles we see that sin(pi/8) = (r_{n-1}-r_n)/(r_{n-1}+r_n). Using half-angle formulas and solving for r_n, we get r_n = r_{n-1}*(2-sqrt(2-sqrt(2)))/(2+sqrt(2-sqrt(2))), the radical fraction which we will call a. Thus, we have r_n = r_{n-1}*a = a^n, as r_0 = 1. Letting C_n = circumference of omega_n and A_n = area of omega_n, we see that C_n = 2*pi*a^n and A_n = pi*a^2n, and summing over these is a geometric series: C = sum(C_n) = 2*pi/(1-a) and A = sum(A_n) = pi/(1-a^2). Thus, we end up with C^2/A = 4*pi^2*(1-a)(1+a)/(pi*(1-a)^2) = 4*pi*(1+a)/(1-a) = 4*pi*sqrt(4+2*sqrt(2)).

• Theodore Leebrant 7 months ago

I found the line joining all the centres, which is y=(√2 -1)x. Considering the length from origin to omega_0 (which is √(2+(2+√2)), and the fact that the radii forms a geometric series, we know 1+2r_1+2r_2+2r_3+...=√(2+(2+√2)).
Therefore, we find r_n=((√(4+2√2)-1)/(√(4+2√2)+1))^n for all positive integer n.
From there just calculate C and A using infinite series.

• Jeremy Weissmann 7 months ago +1

Silly me, I also computed that value for r_n but assumed I wouldn’t be able to solve for C and A unless I simplified that expression!

• Theodore Leebrant 7 months ago

@LetsSolveMathProblems yep I won't edit :)

• LetsSolveMathProblems  7 months ago

Please do *not* edit your comment. I almost always overlook minor errors. =)

• Theodore Leebrant 7 months ago +1

Oops it should be √(2(2+√2)) not √(2+(2+√2))