If I was late to reply, bear with me. I was studying. Am I lying? If you don't believe me, ask the Epic professor Dan (he's the most awesome person and always all in) about me. He knows exactly how hard working student I am.
You know the lengths of 3 sides - hence the triangle is unique, Use the laws of sines and cosines in triangles (and other trigonometric identities) to proceed.....
Thank you professor Khan for passing by. Aren't these laws and identities involving angles? Then, how would we be able to use them? Let us just hope that what Highlander claimed was true that we don't need any of these. Just a
Pythagorean!
You haven't done all that I suggested. Draw in segments to the points where the circles touch the axes, and then label all these segments in terms of the three radii. Then you can use various methods to write equations. I used right triangles formed by adding in segments parallel to the axes (more or less equivalent to the distance formula), and wrote an equation for each of the three lines you drew.
I have drawn the segments. I just did not labeled them. I didn't do that because I thought it was not necessary as I already know these sides as a combination of [imath]r_1[/imath], [imath]r_2[/imath], and [imath]r_3[/imath]. I don't see how I will use right triangles from a non-right triangle. Also the distance formula will relate only two points together which will involve only one side that I already know its length without the formula.
I'm glad you are realizing that your impressions were wrong. Geometry tends to be downplayed today, but there is a lot there, far deeper than what we learn in high school, and it can be very challenging. Challenge is good. So is humility.
If I had a lot of skills in geometry, and I faced this problem, It would probably be not challenging. I lack the skills of this topic, but I am the kind of person who loves to get stuck in a problem for weeks to figure its answer by my own. After failing, I would refer to this forum or other facilities. In simple words, I love the Challenge, but not the Humility.
What makes geometry and trigonometry challenging is that everything can be looked at from multiple perspectives (e.g. identities in trig), and nothing is routine as in algebra (and its relatives in the line of calculus). I consider probability to be similar; have you tried doing much with that?
I have done some probability, and I might even use geometry there without noticing. It is not so difficult, it is just a confusing topic!
The problem does not look easy to me (though, from my perspective, it is more or less straightforward). We'll probably be interested in seeing the full statement of the original problem; it's quite possible that the part you've asked for is the hard part, and the result just has to be applied repeatedly.
I am glad that you find it not easy, but of course with your giant skills in this geometry, once you start attacking the problem, everything gets solved by itself.
I don't have the exact statement of the problem, but if I was ambiguous somewhere, just let me know, I will dig deep to explain again with full strength.
Since the main goal of the main problem is to find [imath]r_n[/imath], I am sure that there is a pattern, we still don't see it, but it will appear eventually.
I am sorry Dr. Peterson for not showing any mathematical attempt so far as I still don't understand the relation between the circles and this rectangle which one of its legs passes through center of the smallest circle. I think that it is the main key to form right triangles or similar triangles.
Note: I did not provide a diagram with labeling as mine would be similar to the diagram you have given but without the rectangle. (Thank you for the shortcut.)
I started writing my 'answer' (below the ---oooOOOooo---) when there was only one response to your original post in the thread (by
@Dr.Peterson at Post #2) but, now I've come back to post
my work, I see there have been several new posts in the thread (I had expected
mine to be Post #3
). However, since it took me a while to compose it and make it all "look" right (and I was prevented from posting for several hours because the site stopped 'working' for me!), I am now going to post it anyway as I hope at least some of it may help you towards a solution.
Thank you a lot Highlander for this one thousand word explanation. Your attack to the problem depends on compasses. Unfortunately, I don't have them and I have never used them, except as a kid. I am a digital student, so I do all my work in the computer. If there were a way to use them online, I would be so glad to learn that. First of all, I am so impressed when you said that this problem is not difficult and it can be solved without any identities at all.
No geometry,
No trigonometry: this blows my mind and your idea to solve the problem without them proves that you are a Genius. It was a brilliancy to draw the three circles so accurately. But sometimes precision is not required. It has always been the fight between engineers and mathematicians. Engineers accept pi as 3.14 and mathematicians accept it only and only [imath]\pi[/imath]. Therefore, I am 100% sure that you are a mathematician.
I am so sorry Highlander if I disappointed you as I will not be able to sketch the diagram by hand. However, what you have explained has given me a lot of ideas. I loved your so precise sketch in Desmos (I would never be able to sketch one like it) and now I am plugging numbers in your formulas trying to figure out a pattern that will lead me to the required formula. Maybe this is not the purpose of the sketch and maybe this is not the right approach, but at least trying to do something is better than nothing.
To follow my suggestions your y-axis will need to be at least 18 boxes high and the x-axis at least 25 boxes in length, so 20 x 30 for the axes would be ideal but that will likely mean
What do you mean by boxes here? Does it mean 18 units, for example?
I am a noob, bear with me.
The centre of circle 2, C2 (21, 4), is obtained from the fact that the distance between C1 & C2 is 13 (r1 + r2) and that gives rise to a Pythagorean (5, 12, 13) triple because C1 is 5 units "higher" than C2 and, thus, the horizontal distance between them works out (by Pythagoras) to be 12 units.
(This use of Pythagoras' Theorem is a BIG hint as to how the whole problem may be solved and I have also added some further points, A, B, D & E (on my diagram, qv), that provide a further hint as to how to proceed to a solution.)
This is so difficult for me to understand. I will try my best to squeeze my brain to come up with something from it.
But note also that \(\displaystyle \sqrt{144}=\sqrt{4\times 36}= \sqrt{4}\sqrt{36}=2\sqrt{9\times 4}\). (Does that look in any way familiar?
No. It does not. Should I be familiar with these numbers?
Thank you a lot Highlander for this great work. I know that you spent a lot of time collecting and organizing your ideas, latex, and diagrams. This itself deserves a brilliancy. If this problem got solved only by Pythagoras, I would break Gauss's picture on my wall, and I would put in its place Pythagoras' picture. Until now, I consider Gauss as the legend of all.
I am sorry that your post was not
#3. It is not only you who gets problems in accessing this website. Me too, I had a lot of them, but thanks to professor Khan. Once I mentioned that I had difficulties to access the site as it was so slow, he fixed the problem immediately and since then, I got no issues.