Solution 78: Infinitely Many Tangent Circles

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• Published on Feb 7, 2019
• We apply infinite geometric series formula three times to establish connections between circumferences, areas, and length of an important segment.
Congratulations to Theodore Leebrant, Cobalt314, PRAKHAR AGARWAL, Arun Bharadwaj, Veeryan Bhatia, PhigNewton1, Jeremy Weissmann, fmakofmako, Aswini Banerjee, and Mahesh Kumar for successfully solving this math challenge question! Theodore Leebrant was the first person to solve the question.
Your support is a heartfelt source of encouragement that propels the channel forward.
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For more Weekly Math Challenges:

• LetsSolveMathProblems  4 months ago +10

Best of luck to those who are taking the AMC 10/12 this week and next week! =)

• Liam Cannon 3 months ago

Amc was very challenging

• Antonio Banda 4 months ago

I take it (AMC 12) Wednesday. I am sure I'm not good enough still but it is always fun to try and learn from it.

• Tuanicular 4 months ago

Took it today, 10A

• Siddharth Singh 4 months ago +1

Inside a square ABCD pick a random point P, such that AP is of length 10 cm, BP is of length 6 cm and CP is of length 14 cm. Find the area of the square ABCD.
Answer is 232 cm^2 but I can't figure it out how. Please make video for the solution.

• Siddharth Singh 4 months ago +2

Love from India!

• S0RA 24 days ago

I tried the problem on my own beforehand and made it all the way to the part where you square C... and I distributed the exponent inside of my summation notation. OOPS. Thought the answer was 4pi

• Sathvick Satish 4 months ago

Where is your accent from ? I love it cuz it’s such a unique accent that I have never heard before!

• Johannes H 4 months ago

I dont know why, but I tried it exactly like this and the result was wrong🤔. Also I calculated the ratio of the first radius to the length on the bottom which is possble because we know the angle of 22,5° and that turned out to be too much as well... However, cool problem!

• Bulldawg 4 months ago

A lot to digest in this one ... I struggled to keep up and finally fell behind ... I'll have to rewind and view again ...

• Omar 4 months ago

I appreciate the way you explain things clearly. Also I would be so grateful if you do some number theory challenges if
you don't mind. Thanks again!

• LetsSolveMathProblems  4 months ago

I have decided on the "rough draft" version of this week and the next week's challenges, both of which unfortunately do not require number theory. I will try my best to come up with interesting number theory problems in the upcoming weeks. Thank you for your compliment! =)

• salelltd 4 months ago

I don't understand how the radii become 1, r, r², .... I have another proposition here about the radii. How will you disprove this: pasteboard.co/I0eDxZy.png

• salelltd 4 months ago

@LetsSolveMathProblems Because if my proposition is true, then (1-sina)/(1+sina)=c=constant and this means that r2=c.r1 for each circle. This contradicts your statement. If we try this with your solution, we will end up with this problem: r2=c.r1 and r2=r1². Solution to these equations would be r1=c, r2=c². But we wanted to start with r1=1, so this is a contradiction. It doesnt matter if we dont start with r1=1, because we cannot use any other value as r1. There is only one solution and it is r1=c and r2=c². This equation system does not get satisfied with any other values, so we cannot use it for this problem. I mean for example; for the next circle, we would need to use r1=c² and try to find r2=c^4, but this would not satisfy the equation.

• LetsSolveMathProblems  4 months ago

For now, I will write down an intuitive argument and a quick proof using homothety. If you are not familiar homothety (or dilation) yet, I can provide you with a proof that only depends on triangle similarity.

First, here is an intuitive way of thinking about the situation. We know the radii are proportional because the diagram itself, intuitively speaking, is constructed in a proportional manner. That is, no matter how much we magnify the diagram, its overall shape stays the same. Let's say we have found r such that radius of w_1 = r. Then, consider "zooming in" the diagram while keeping the coordinate system fixed, until the radius of w_1 becomes 1. Now, it is easy to see that the radius of w_2 must be r in this magnified version: That is, we are multiplying by r again (since "zooming in" doesn't change proportions).

Now for the formal proof. It is well known (if you have studied homothety) that the intersection point of external tangents of two circles (let's call it point X) is the homothetic center of the circles. That is, we can think of one of the circles as the image of the other circle resulting from dilation with respect to point X. This proves the desired assertion because dilation preserves length proportions.

Also, I believe your proposition is true. I am not sure why you asked me to disprove it. If I missed something, please let me know.

• mrBorkD 4 months ago

7:00 we want our final solution :^)

• Felipe Lorenzzon 4 months ago

I did it recursively. I expressed the radius of the next circle in terms of the previous one. If you do that, you can easily identify that it forms a geometric series!

• Felipe Lorenzzon 4 months ago

Maybe I post it here this week

• JT 4 months ago

Proof please? Or are you just kidding?

• Mokou Fujiwara 4 months ago

same as part b of my solution?

• Felipe Lorenzzon 4 months ago

I actually found my answer in a different form, but it is an equivalent expression

• Phoenix Fire 4 months ago +4

It was weird for me that you used the variable x for (1+r)/(1-r), because we're already on the Cartesian plane, so we have that x=sqrt(1+x^2), but they're different x's. :S

• LetsSolveMathProblems  4 months ago +2

You are right. I apologize if the choice of letter led to confusion. I will keep it in mind for the future videos! =)

• JT 4 months ago

I have just noticed your answer to a previous post....again, disappointing. The "intuition" argument does not sit well, the "zoom" argument is very unsatisfying and hand wavy.......and you seem dismissive of giving an actual proof. Maybe its just one of thoos e facts, handy to know....you know, without going to the trouble of explaining. And this is what is so weird,\....you are famous for your brilliant multi-tangent explanations, side proofs in long questions that are sometimes as long as the main part itself....but here? You ignore it. Please rectify, thank you.

• RGP Maths 4 months ago +1

Some Newsflashes for you.
The grammatical passive is a voice, not a mood.
Its use was appropriate in those sentences which had no particular agent under consideration.
Gender pronouns are not illegal in Canada, only their application to those who don't want them.
Here in the UK, I am not subject to Canadian law.
Addressing a 60-year-old teacher of Mathematics as "buddy", and advising me to "run along" makes you sound pompous and silly.
Nobody is supporting your vituperative rants: there may be a lesson there.
Finally, I have maintained all along that the geometric ratio of radii was adequately covered in the first place. Your refusal to accept an elementary concept such as dilation is merely classless nitpicking.

• JT 4 months ago

@RGP Maths Seriously? Why would you write such an impassioned post in the passive mood? Did your Grade 9 Public speaking teacher once tell you that it would be an effective tool to impart gravitas? Newsflash buddy, it makes you sound pompous and just a little bit silly. I expect to be forgiven if you think this is rude. And don't forget @Phoenix Fire could be Canadian, and therefore it is illegal to impose a specifically gendered pronoun on their good person.

"It could realistically be expected that if any readers did require such details to be filled in, then the advanced parts of the proof would go over their heads anyway."......woosh!!!!!

that's my whole point, but obviously too subtle for you..................this channel is almost "famous" for filling in the gaps, at every turn, at every point, at every sign of possible confusion. It just wasn't done in this case. And I called him out. That's all.

And there was nothing advanced in this particular problem in any case.

Now run along.

• RGP Maths 4 months ago +2

JT It was I who posted the line you considered rude, not Phoenix Fire. It seems fairly innocuous in comparison to the excessive sarcasm with which you have greeted his inordinately patient attempts to explain the similarity argument to you. As for "most of your audience is silent": I don't think so. You may need to compare the number of up votes conferred on his posts with your own.
LetsSolveMathProblems is not a channel aiming at a general readership. It should not be expected that the solutions will fill in every elementary step, just as a university lecturer would not trouble to establish the basics of the subject which would have been covered in school. It could realistically be expected that if any readers did require such details to be filled in, then the advanced parts of the proof would go over their heads anyway.
@Phoenix Fire. I have made an unsupported assumption in referring to you with a male pronoun. Should this be incorrect, I humbly apologise.

• LetsSolveMathProblems  4 months ago +1

@JT I have posted a response in one of your other comments.

• JT 4 months ago

​@Phoenix Fire "I already gave you the proof, and the fact that you don't understand what a dilation is doesn't mean that it's not a proof. Maybe, you should take the time to try to understand the proof and what it's saying. I guess I'll be nice and explain what a dilation is, and what similarity, congruence, and transformations are while I'm at it."

No you haven't. You have appealed to authority. Look it up.

"We say that two shapes are congruent, when they are almost exactly the same shape."

Really? Almost the same shape? OK.

"don't worry it's not the most precise definition I could have given, and I don't normally teach grade school geometry, so I can't explain it that well"

Well, you are right there. Although deep down you think you do.

"Now, to re-explaining how the proof works."

So, up to this point you have used a lot of time explaining something which hasn't taken us very far, but hey ho, you said the 'proof" was TRIVIAL"

" We'll call the scale factor 1/r because of the video."

WHAT????? Why not call it (1/ square root of kermit the frog squared)....times eleven?

"...because of the video"

My favourite line so far............... Because.

Final points:
Your "explanation" was hopeless. Sorry but it just is. It was longer than the video, but you insist on calling it a trivial part of the whole problem.

You insisted that the explanation was soooooo trivial it did not EVEN require explaining. You then tried to explain because I asked you to, took 800 words to do this trivial explanation, and still failed, due to lack of clarity. If most of your audience is silent, its because they don't understand too.

You said, very rudely I think "The trouble is, nobody knows what kind of proof DOES make sense to you." Her's a clue, gie me one based on geometry, or using similar triangles....that does not involve setting up one circle with radius 1 and one with radius 1/r" Are you up to this "trivial task"

You are clearly not the go-to person for getting answers to questions.

• JT 4 months ago +2

A very poor explanation of the most important point....why is the radius of each circle 1, r, r squared, r cubed....the rest of the problem is simple algebra, yet this most important part is not explained, An unusually poor explanation. Disappointing.

• Omar 4 months ago

@JT I would like to point out that in the problem itself, It was said that circles are inductively constructed which immediately should imply that radii differ by a common factor. Therefore it must be a geometric series. Thus (in my point of view) it's not required to prove that fact not because it is obvious but because the problem just assumes it. I also would like to say that in the video he did not say that the fact is intuitive but rather said " by symmetry" of the diagram shown above which is quite reasonable. I also would like to point that you'll never (ever) find such a perfect proof for any problem that you may encounter, so that any proof would contain gaps but the most important thing is that you get the idea of any problem discussed. I think this is the goal of the channel.

• JT 4 months ago

@LetsSolveMathProblems It was unfortunately evident to me that the only way to elicit a meaningful reply to my initial questions was to adopt this approach. Again, I respectfully submit, that the worst approach to dealing with questions, is to simply appeal to a higher authority.
BTW your link to homethety was both interesting and insightful. So thank you for that.

• LetsSolveMathProblems  4 months ago +3

Before you posted this comment, I took time to provide a geometric proof based only on triangle similarity in one of your many other comments as you desired. In that comment, I also made counterarguments addressing some of your past remarks, many of which you essentially repeated in this another rather censorious reply. Having stated that, I do respectfully ask you to express your opinions without blatantly denouncing other commenters with excessive sarcasm and mockery. Although I am very open to having respectful debates on mathematics or my teaching styles (whether with me as a participant or an audience), I would like such discussions to be professional and mature, if possible.

• JT 4 months ago

@LetsSolveMathProblems. So still too trivial to actually show some type of geometric proof eh? Again, I will note, you spend oodles and oodles of time talking and deriving almost every single side issue or interesting occurrence in your other videos, which are almost incidental or tangential to the original problem....... yet, in this instance, you still fail to address my question. Very disappointing that you have maintained this belligerent stance. Or have you now started believing in all the hype and praise that abound your comment section, and that you are above reproach?

" I believed that, if I wished to present more foundational materials, it would be better to prove that the segment from the origin to the center of w_0 passes through all the centers and points of tangency"

You assumed wrongly. This part of the video truly WAS THE TRIVIAL part of the whole problem that nobody cares about.
Wow, a, b, c and d are in a straight line. Literally, everyone is rolling their eyes at this 2 minute episode.

But, hey everyone...it's OBVIOUS that circles formed in this manner have their radii in a geometric ratio, given by this totally made up formula which I am now plucking from the book of knowledge and righteous authority..........and, by the way, I'm not even sure if this is interesting enough to pause for a moment and repeat that. Isn't that kind of cool. Let's explore. No, no, no no ........its too trivial. OK everyone. No questions? Let''s move on

Still no proof apart from the appeal to authority....and the trick that very poor teachers employ when asked a question form one student...." well all the others understand (nudge nudge wink wink) so we'll just move on shall we?"

• LetsSolveMathProblems  4 months ago +4

During the planning stage for this video, I presumed including a formal proof is unnecessary since it is not difficult (intuitively or otherwise) to see that the radii must form a geometric sequence: the circles are being dilated in a symmetric manner. I believed that, if I wished to present more foundational materials, it would be better to prove that the segment from the origin to the center of w_0 passes through all the centers and points of tangency.

As other commenters have pointed out, a proof using dilation is perfectly acceptable. In fact, dilation and other transformations (inversion, reciprocation, etc.) are in the heart of modern geometry.

• el tapa 4 months ago +2

My abilities in geometry are==NULL. AND I'M ON 2ND YEAR OF ENGINEERING. So sad. Chilean education it's not very good but What i like Is that public universities are the best of the best in this beautifull country

• gu4t4f4c 4 months ago

buena qlo dale pa delante nomas

• matyourin 4 months ago +6

Nice :)
Maybe note that the geometric sum only works because r

• Kh Bye 4 months ago +5

1.How do you prove that the circles are proportional by a ratio of r?
2. Or is that a conclusion you get from the qn itself stating the circles are 'constructed inductively'?
3. If 'constructed inductively' wasn't stated, can we still prove it from the diagram?
4. Or is there not enough evidence/facts to prove it?
Sorry for the long rant hahahaha

• JT 4 months ago

@LetsSolveMathProblems There you go, that wasn't too hard now was it?

Avoid using the term trivial....it is used in a pejorative way, a dismissive way....designed to flatter those in the know and subjugate those who are not yet as enlightened.

• LetsSolveMathProblems  4 months ago +1

​@JT Please allow me to make a few important counterarguments before I proceed to show you the proof that you desire (which, as you will see, is indeed very elementary).

As you likely know if you have explored classic college level math textbooks, numerous renowned authors frequently use the terms "trivial," "easy to see," and "evidently" to encourage the readers to actively fill in the easier portions of the proof or the explanation. Walter Rudin's "Principles of Mathematical Analysis" is a quintessential example. Many of his proofs have intentional gaps (signified by aforementioned keywords) that some of them, for me, sometimes took hours to fill in. Nevertheless, he is considered one of the best instructors in introductory real analysis, and many argue his extreme conciseness in his proofs is what makes him so outstanding. In contrast to hours of work often required for a student to understand one assertion in "Principles," what you call the missing portion in my explanation should not take an astute viewer (without an experience with dilation) more than 20 minutes to prove. (If you have an experience with dilation, it is indeed trivial.)

I do additionally remark it would not be pragmatic for me to *explicitly* justify every part of my explanation in my videos. Certainly, a video on Vieta's relations need not review the definition of polynomials, nor a video on cyclic quadrilaterals need an in-depth exposition of angle chasing. Since I have such a diverse audience in terms of mathematical maturity, I strive to construct my videos in a manner that most of my target audience (which varies from video to video) should be familiar or competent enough in prerequisites to intuitively understand or prove the more foundational pieces.

If it is not clear from the context, I stated "diagram is proportional" to signify that the "diagram is constructed proportionally." I apologize if the ambiguity in that statement made it difficult to derive the intended meaning.

Having addressed your arguments, let me finally contend that, *to viewers familiar with dilation, the fact that radii are proportional is indeed obvious.* My assertion may be hard to believe until you study dilation yourself. As an introduction, I direct you to an excellent article on the blog of Richard Rusczyk (a former US Olympiad winner and the founder of "Art of Problem Solving"): artofproblemsolving.com/community/c864h980383. For a more thorough, but still elementary, treatment of transformations, Chapter 4 of "Geometry Revisited" by Coxeter and Greitzer is unsurpassed.

However, if you have not studied dilation before, and thus have not taken time to formally prove the properties of dilation, it is necessary for us to establish a bridge of formality using triangle similarity or other methods. We will do so right now.

Proposition: Consider any three consecutively constructed circles X, Y, Z in the diagram in this video, with radii x, y, z, respectively, such that 0 < z < y < x. Then, x/y = y/z.
Proof: Let A be the origin; B, E, D be the centers of X, Y, Z, respectively; and C, G, F be the feet of altitudes to the x-axis from points B, E, D, respectively. Then, DE = y+z and BE = x+y (as proven in the video). Furthermore, by triangle similarity, AG = AF * y/z and AC = AF* x/z. Since DE/EB = FG/GC (prove this yourself if your high school geometry class did not discuss it), we get (z+y)/(AF*(y/z-1)) = (x+y)/(AF*(x/z-y/z)). Quick simplification yields (y+z)/(y-z) = (x+y)/(x-y), which after cross-multiplication and cancellations results in 2xz - 2y^2 = 0. This is equivalent to x/y = y/z, completing the proof.

• Mokou Fujiwara 4 months ago

But my solution is not perfect. I actually did something wrong in my solution. I should correct my symbol L2 to Lc.
I will try to do every challenge if I have time, even if I cannot become a winner. I will try to explain my steps fully.

• JT 4 months ago

@Mokou Fujiwara Someone who actually did the work the that author of the video should have done.Thank you.

• JT 4 months ago

@LetsSolveMathProblems "We know the radii are proportional because the diagram itself is proportional"

How can a single object (the diagram) be proportional???? ....to itself?????? Separate objects can be proportional to each other, but an object cannot be proportional to itself. That is like saying "Monday is proportional to Monday"

" Let's say we have found r such that radius of w_1 = r. Then, "zoom in" the diagram while keeping the coordinate system fixed, until the radius of w_1 becomes 1. Now, it is easy to see that the radius of w_2 must be r in this magnified version:"

That is about as clear as mud. Beware the teacher who says" ..easy to see" and "obvious"

"A formal proof using triangle similarity (I came up with one just now after reading your comment) requires a little bit of work, although it does not require anything advanced. "

Please provide such a proof........ or it so trivial that it can be written in the margin of your digital textbook?

"For this particular problem, however, the intuitive argument above is probably more than sufficient."

No. It isn't. Honest.

" Every step I make in the video can be proved and justified..."

Except this particular part obviously.

You just didn't prepare yourself adequately for this video solution I'm afraid.

• ZeT 47 4 months ago +13

That was nice. every single idea needed o solve it is given by grade 11. Sadly, I don’t think any of my grade 11 students will have the stamina to tackle this problem.