finding r3 in terms of r1 and r2: circles, C1 and C2, C1 larger than C2; C1 tangent to C2, y-axis, and x-axis; C2 tangent to C1 and x-axis

[mario99]

This problem is very difficult and by just understanding it is a brilliancy!

I have to draw two circles, [imath]C_1[/imath] and [imath]C_2[/imath]. [imath]C_1[/imath] is larger than [imath]C_2[/imath]. [imath]C_1[/imath] is tangent to [imath]C_2[/imath], y-axis, and x-axis. [imath]C_2[/imath] is tangent to [imath]C_1[/imath] and x-axis.

After that I have to draw a third circle, [imath]C_3[/imath]. [imath]C_3[/imath] is tangent to [imath]C_1[/imath], [imath]C_2[/imath], and x-axis.

Let [imath]r_1[/imath], [imath]r_2[/imath], and [imath]r_3[/imath] be the radiuses of the circles [imath]C_1[/imath], [imath]C_2[/imath], and [imath]C_3[/imath], respectively.

Show that:

[imath]\displaystyle \sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath]

 

[Dr.Peterson]

This problem is very difficult and by just understanding it is a brilliancy!

I have to draw two circles, [imath]C_1[/imath] and [imath]C_2[/imath]. [imath]C_1[/imath] is larger than [imath]C_2[/imath]. [imath]C_1[/imath] is tangent to [imath]C_2[/imath], y-axis, and x-axis. [imath]C_2[/imath] is tangent to [imath]C_1[/imath] and x-axis.

After that I have to draw a third circle, [imath]C_3[/imath]. [imath]C_3[/imath] is tangent to [imath]C_1[/imath], [imath]C_2[/imath], and x-axis.

Let [imath]r_1[/imath], [imath]r_2[/imath], and [imath]r_3[/imath] be the radiuses of the circles [imath]C_1[/imath], [imath]C_2[/imath], and [imath]C_3[/imath], respectively.

Show that:

[imath]\displaystyle \sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath]

Start by drawing a picture, including radii to each point of tangency. (You'll probably need a couple tries for the last circle (or at least its center) to fit on the page.)

If you don't understand the problem enough to draw a picture, at least give it a try, and show us where you are struggling.

 

[mario99]

Start by drawing a picture, including radii to each point of tangency. (You'll probably need a couple tries for the last circle (or at least its center) to fit on the page.)

If you don't understand the problem enough to draw a picture, at least give it a try, and show us where you are struggling.

I apologize to you Dr.Peterson for keeping you waiting for long. And I thank you a lot for being so patient. I was studying.

Before I try my attempt and provide some diagrams, I just want to tell you more about this problem. You can ignore what I will say now. It is related to the problem, but we don't need to answer it in this post. I might create a new thread if I needed a continuation to the answer. This problem was given to me by a relative who thinks that I am a math Genius, but she doesn't know that I hate geometry. I have always considered geometry and trigonometry as kids' stuff (I hope that I am not offending anyone). Therefore, I have given them a little attention. I thought that I was born and designed only and only for differential equations. But now I am struggling with simple geometry (may be very difficult geometry).

This problem does not look as easy as you think. It is a big problem and I have divided it into parts. This thread is just a fragment of them. Other parts are, the distance between the centers of C1 and C2, drawing a fourth circle tangent to C2 and C3, drawing a fifth circle tangent to C3 and C4, and so on. The pattern continues forever, and they are asking for a formula to find [imath]r_n[/imath] in terms of [imath]F_n[/imath], the nth Fibonacci sequence by starting with [imath]r_1 = r_2 = 1[/imath].

I tried to sketch the circles in Desmos and this is what I got.



The green circle (the smallest circle) appears to be tangent to both circles, but if you zoom in, you will discover that it isn't. At least for now this diagram appears perfect to the eyes to see.

After drawing the three circles, I came up with the idea to connect their centers.




It seems that I have a triangle. If this was a right triangle, I might be able to relate the sides to each other. Unfortunately, it looks like it is not a right triangle. Is there a secret formula to relate the sides without knowing the angles, like in our case?

 

[khansaheb]

I apologize to you Dr.Peterson for keeping you waiting for long. And I thank you a lot for being so patient. I was studying.

Before I try my attempt and provide some diagrams, I just want to tell you more about this problem. You can ignore what I will say now. It is related to the problem, but we don't need to answer it in this post. I might create a new thread if I needed a continuation to the answer. This problem was given to me by a relative who thinks that I am a math Genius, but she doesn't know that I hate geometry. I have always considered geometry and trigonometry as kids' stuff (I hope that I am not offending anyone). Therefore, I have given them a little attention. I thought that I was born and designed only and only for differential equations. But now I am struggling with simple geometry (may be very difficult geometry).

This problem does not look as easy as you think. It is a big problem and I have divided it into parts. This thread is just a fragment of them. Other parts are, the distance between the centers of C1 and C2, drawing a fourth circle tangent to C2 and C3, drawing a fifth circle tangent to C3 and C4, and so on. The pattern continues forever, and they are asking for a formula to find [imath]r_n[/imath] in terms of [imath]F_n[/imath], the nth Fibonacci sequence by starting with [imath]r_1 = r_2 = 1[/imath].

I tried to sketch the circles in Desmos and this is what I got.

View attachment 36966

The green circle (the smallest circle) appears to be tangent to both circles, but if you zoom in, you will discover that it isn't. At least for now this diagram appears perfect to the eyes to see.

After drawing the three circles, I came up with the idea to connect their centers.

View attachment 36967


It seems that I have a triangle. If this was a right triangle, I might be able to relate the sides to each other. Unfortunately, it looks like it is not a right triangle. Is there a secret formula to relate the sides without knowing the angles, like in our case?

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.....

 

[Dr.Peterson]

After drawing the three circles, I came up with the idea to connect their centers.

Start by drawing a picture, including radii to each point of tangency.

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.

This problem was given to me by a relative who thinks that I am a math Genius, but she doesn't know that I hate geometry. I have always considered geometry and trigonometry as kids' stuff (I hope that I am not offending anyone). Therefore, I have given them a little attention. I thought that I was born and designed only and only for differential equations. But now I am struggling with simple geometry (may be very difficult geometry).

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.

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?

This problem does not look as easy as you think. It is a big problem and I have divided it into parts. This thread is just a fragment of them. Other parts are, the distance between the centers of C1 and C2, drawing a fourth circle tangent to C2 and C3, drawing a fifth circle tangent to C3 and C4, and so on. The pattern continues forever, and they are asking for a formula to find [imath]r_n[/imath] in terms of [imath]F_n[/imath], the nth Fibonacci sequence by starting with [imath]r_1 = r_2 = 1[/imath].

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.

 

[The Highlander]

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.

I will make a few minor adjustments to what I'd already written in light of the further contributions above but, now that I see what you have posted and your own diagrams & comments, the main thing I would 'add' is that you don't need any "secret formula to relate the sides without knowing the angles" because the triangle itself (and its internal angles) aren't where you need to be focusing your attention; only the lengths of its sides, which are given to you (albeit in terms of the variables r_{1, 2 & 3}) are what you need to consider and there's not much "Geometry" involved in arriving at a solution either; this problem simply isn't as complicated as you seem to think it is.

So here's what I prepared earlier; I trust it might set you in the right direction...
(Please remember: I had expected it to be Post #3 in the thread!)

---oooOOOooo---​

If you are having difficulty drawing a diagram of the situation described (or still can't concieve of what it looks like) then you might wish to follow my suggestions, below.

I would avoid naming the circles as "C₁", "C₂" & "C₃" and just call them 1, 2 & 3 (so that the letter "C" may be used to name the points at their centres).

You will need squared paper (¼" squared or similar would be ideal) and a pair of compasses (and a straight edge, of course).

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 that you need to turn your sheet of paper or workbook into "landscape" orientation. (I did.)

Once you have drawn your axes you can then draw the circles by assuming that the equation provided is true and selecting perfect squares for the radii of circles 1 & 2. To fit the axes above I chose r₁ = 9 (units, ie: boxes) and r₂ = 4 units.

Now, assuming that [imath]\sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath] then r₁ = 9 and r₂ = 4 gives us r₃ = 1.44.

[I seem to have run into a problem on this forum that there are only so many words allowed in a single post so I will continue in the next post; please read on there.]

 

[The Highlander]

[Continued from previous post immediately above...]

So you can now plot the centre of circle 1 at the point C₁ (9, 9) and, using your pair of compasses, draw circle 1 with radius r₁ = 9, so that it is tangential to both the x- & y- axes.

You can then plot the centre of circle 2 at the point C₂ (21, 4) and, again using your pair of compasses, draw circle 2 with radius r₂ = 4, so that it is tangential to both the x-axis and circle 1.

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.)

Now that you have drawn circles 1 & 2 you can use your pair of compasses to locate (approximately) the centre of circle 3 on your sketch because you know it is 10.44 (r₁ + r₃) units away from C₁ and 5.4 (r₂ + r₃) units away from C₂ and will lie on the line y = 1.44. Plotting that point (C₃) will allow you to draw in circle 3 and you are now ready, sketch completed, to add a few more straight lines to your diagram and show that [imath]\sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath] is, indeed, true.

​

NB: The method above does not prove or even show the outcome required by the problem, it just provides a way for a (reasonably) accurate diagram to be constructed on paper.

One last hint: I was able to determine a value for the the horizontal distance between the centres of the two larger circles (purely for the purpose of drawing a sketch) because I assigned arbitrary values to their radii (9 & 4, which also happened to be perfect squares). I then got that distance using the (Pythagorean) formula:-

\(\displaystyle \sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\)​

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? Also: how else could that distance of 12 units be described on my diagram?)
And, finally, there is no need for the use of any Trigonometric (Ratios or Identities) to solve this problem; Pythagoras gave you everything that is needed!

Hope that helps.

 

[The Highlander]

I tried to sketch the circles in Desmos and this is what I got.

​

The green circle (the smallest circle) appears to be tangent to both circles, but if you zoom in, you will discover that it isn't. At least for now this diagram appears perfect to the eyes to see.

If you want Desmos to show the circles "correctly" then you will need to use appropriate values for the centres & radii of each circle, for example, see here.

(The coordinates of each circle's centre are indicated in its equation, eg: C₁ lies at the point (9, 9) hence the equation of that circle is: \(\displaystyle (x-\pmb{\color{red}9}\color{black})^2+(y-\pmb{\color{red}9}\color{black})^2=81\))

 

[Dr.Peterson]

If you want Desmos to show the circles "correctly" then you will need to use appropriate values for the centres & radii of each circle, for example, see here.

(The coordinates of each circle's centre are indicated in its equation, eg: C₁ lies at the point (9, 9) hence the equation of that circle is: \(\displaystyle (x-\pmb{\color{red}9}\color{black})^2+(y-\pmb{\color{red}9}\color{black})^2=81\))

I'm confused? Who said C1 has to be at (9,9)? His is clearly at (3,3).

The big issue, if one wants to make an accurate picture, is how to choose the center C2, and then C3, to make them tangent. (The problem tells us how we can find that third radius, but not how to start it off, or how to find the centers.) Do you have any advice on that?

In any case, a picture intended to help in talking about the problem doesn't have to be exact, so that wasn't important.

I tried to sketch the circles in Desmos and this is what I got.

View attachment 36966

The green circle (the smallest circle) appears to be tangent to both circles, but if you zoom in, you will discover that it isn't. At least for now this diagram appears perfect to the eyes to see.

After drawing the three circles, I came up with the idea to connect their centers.

View attachment 36967


It seems that I have a triangle. If this was a right triangle, I might be able to relate the sides to each other. Unfortunately, it looks like it is not a right triangle. Is there a secret formula to relate the sides without knowing the angles, like in our case?

Here is a version of your drawing that I made with GeoGebra; I can set the first two radii arbitrarily and it will generate the third. This is developed using the result (and also intermediate results) of the exercise, so I had to sketch it first on paper, inaccurately, in order to do the work.

​

Note the three right triangles I drew in (right angles at A, B, C), which are the basis of the work.

 

[mario99]

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.

 

[Dr.Peterson]

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 r1, r2, and r3. 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.

No, you didn't draw all the segments I recommended: every radius from a center to any point of tangency. Those are all present in my new drawing, as well as the right triangles you need. Start from there. Write an equation for each right triangle.

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 π\piπ. Therefore, I am 100% sure that you are a mathematician.

No precision is needed at all! You're exactly right. But therefore you don't need a compass to sketch the important relationships.

But of course geometry is needed; that's what Pythagoras did.

 

[The Highlander]

I'm confused? Who said C1 has to be at (9,9)? His is clearly at (3,3).

I'm a bit confused myself now. Are you saying that it was I who confused you?
I didn't say anywhere that C₁ had to be at (9, 9).

I thought I had made it perfectly clear that I had only (arbitrarily) chosen the radii of the two larger circles to be 9 & 4 respectively to facilitate the production of a sketch on paper (using compasses et al) and those numbers being perfect squares also simplified the arithmetic.

In the first two posts in the the thread there appeared to be a suggestion that making such a drawing would prove a bit problematic which was what prompted me to produce my ("thousand word"! ) explanation of how one might go about it.

My use of 9 for the radius of the first circle, thus placing its centre at (9, 9), was only an attempt at just that.
I hope that clears up any confusion?

@mario99: You seem to be doggedly avoiding taking the hints that have been supplied above. Perhaps I could have one last go at sending you in the right direction?

No. It does not. Should I be familiar with these numbers?

When I said: "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? Also: how else could that distance of 12 units be described on my diagram?)" I had hoped that you would notice that 9 was r₁ and 4 was r₂! (And that AB equalled 12.)
Maybe the significance of that will dawn on you if you follow the rest of what I say here...

You only need to concentrate on the horizontal distances between the circles' centres. Look at this diagram...

​

Again, I have added points A - F (for clarity and ease of reference). I have also added (greyed out) the numbers that I (arbitrarily) chose for ease of drawing the situation; they prove nothing other than showing that, for one example, the result you need to prove 'works'.

However, you can see that, by setting the first two circles' radii to 9 & 4, I could easily get the horizontal distance between their centres as 12 (by Pythagoras) but can you provide a generalized expression for that distance using only r₁ & r₂?

That's how you need to start towards a solution. Once you've done that then repeat it for the horizontal distances between the centres of the other circle pairs. (Circles 1 & 3 and 2 & 3.)

You will then have algebraic expressions for: AB (=FC₂), DC₃ & C₃E. (I do hope you're not as averse to Algebra as you are to Geometry! )

Now think about how you could use the algebraic expressions of those three lengths. (What is the relationship between them?)

Hope that does help towards a final solution here.

 

[Dr.Peterson]

I'm a bit confused myself now. Are you saying that it was I who confused you?
I didn't say anywhere that C1 had to be at (9, 9).

I thought I had made it perfectly clear that I had only (arbitrarily) chosen the radii of the two larger circles to be 9 & 4 respectively to facilitate the production of a sketch on paper (using compasses et al) and those numbers being perfect squares also simplified the arithmetic.

In the first two posts in the the thread there appeared to be a suggestion that making such a drawing would prove a bit problematic which was what prompted me to produce my ("thousand word"! ) explanation of how one might go about it.

My use of 9 for the radius of the first circle, thus placing its centre at (9, 9), was only an attempt at just that.
I hope that clears up any confusion?

That line was one I thought I'd overwritten; I was only confused at first, when I saw that you were responding to what he wrote before you used (9,9) in your example, and took that to be a correction to how he made his circle, centered at (3,3), and before I looked at what you'd written (which, yes, was rather long).

In any case, the important point about that is that you don't need an accurate picture in the first place. A pencil sketch is more appropriate.

 

[khansaheb]

This problem is very difficult and by just understanding it is a brilliancy!

I have to draw two circles, [imath]C_1[/imath] and [imath]C_2[/imath]. [imath]C_1[/imath] is larger than [imath]C_2[/imath]. [imath]C_1[/imath] is tangent to [imath]C_2[/imath], y-axis, and x-axis. [imath]C_2[/imath] is tangent to [imath]C_1[/imath] and x-axis.

After that I have to draw a third circle, [imath]C_3[/imath]. [imath]C_3[/imath] is tangent to [imath]C_1[/imath], [imath]C_2[/imath], and x-axis.

Let [imath]r_1[/imath], [imath]r_2[/imath], and [imath]r_3[/imath] be the radiuses of the circles [imath]C_1[/imath], [imath]C_2[/imath], and [imath]C_3[/imath], respectively.

Show that:

[imath]\displaystyle \sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath]

Then Law of cosines for triangles will state :

(r₁+ r₂)² = (r₃+ r₁)²+ (r₂+ r₃)² - 2 * cos (C) * (r₃+ r₁) * (r₂+ r₃)
......and other similar equations that can be derived from Pythagoras's theorem, but that will be really long way.

In addition, you'll need to apply Laws of Sines and elementary trigonometric identities.

 

[mario99]

View attachment 36999
Then Law of cosines for triangles will state :

(r₁+ r₂)² = (r₃+ r₁)²+ (r₂+ r₃)² - 2 * cos (C) * (r₃+ r₁) * (r₂+ r₃)
......and other similar equations that can be derived from Pythagoras's theorem, but that will be really long way.

In addition, you'll need to apply Laws of Sines and elementary trigonometric identities.

See what I did below. You would be surprised!

No, you didn't draw all the segments I recommended: every radius from a center to any point of tangency. Those are all present in my new drawing, as well as the right triangles you need. Start from there. Write an equation for each right triangle.

No precision is needed at all! You're exactly right. But therefore you don't need a compass to sketch the important relationships.

But of course geometry is needed; that's what Pythagoras did.

My bad Dr.Peterson.

View attachment 36981​

I will start with Dr.Peterson's circles to write equations.

[imath]r_1 + r_2[/imath] is the length of longest leg.

Another way to write this length is to use the distance formula.

[imath]r_1 + r_2 = \sqrt{(r_2 - r_1)^2 + (c_2 - c_1)^2} = \sqrt{(2 - 3)^2 + (7.9 - 3)^2} = \sqrt{25.01}[/imath]

By following the same idea for other legs, I get:

[imath]r_2 + r_3 = \sqrt{(r_3 - r_2)^2 + (c_3 - c_2)^2} = \sqrt{(0.6 - 2)^2 + (5.7 - 7.9)^2} = \sqrt{6.80}[/imath]

[imath]r_1 + r_3 = \sqrt{(r_3 - r_1)^2 + (c_3 - c_1)^2} = \sqrt{(0.6 - 3)^2 + (5.7 - 3)^2} = \sqrt{13.05}[/imath]

View attachment 36985​

Hope that does help towards a final solution here.

Thanks a lot Highlander for this extra hint. It is so much appreciated.

For your diagram.

[imath]r_1 + r_2 = 13[/imath]

[imath]r_2 + r_3 = \sqrt{(1.44-4)^2 + (16.2 - 21)^2} = \sqrt{29.5936}[/imath]

[imath]r_1 + r_3 = \sqrt{(1.44-9)^2 + (16.2-9)^2} = \sqrt{108.9936}[/imath]

It does not matter if I work with Highlander or Dr.Peterson diagram.

Back to Dr.Peterson equations.

When I looked at this equation,

[imath]r_1 + r_2 = \sqrt{(r_2 - r_1)^2 + (c_2 - c_1)^2} = \sqrt{(2 - 3)^2 + (7.9 - 3)^2} = \sqrt{25.01}[/imath]

I came up with the idea to solve for [imath](7.9 - 3)[/imath], or [imath]4.9[/imath], the horizontal distance between the centers of the circles [imath]C_1[/imath] and [imath]C_2[/imath].

[imath](r_1 + r_2)^2 = (r_2 - r_1)^2 + (c_2 - c_1)^2 = (2 - 3)^2 + (7.9 - 3)^2 = 25.01[/imath]

[imath]4.9^2 = (r_1 + r_2)^2 - (2 - 3)^2 = r_1^2 + 2r_1r_2 + r_2^2 - 1[/imath]

But I already know that [imath]1 = (-1)^2 = (2 - 3)^2 = (r_2 - r_1)^2 = r_2^2 - 2r_1r_2 + r_1^2[/imath]

Then

[imath]4.9^2 = (r_1 + r_2)^2 - (2 - 3)^2 = r_1^2 + 2r_1r_2 + r_2^2 - (r_2^2 - 2r_1r_2 + r_1^2) = 4r_1r_2[/imath]

Or

[imath]4.9 = 2\sqrt{r_1r_2}[/imath]

If I do, the same idea for other legs, I get:

[imath]-2.2 = 2\sqrt{r_2r_3}[/imath]

[imath]2.7 = 2\sqrt{r_1r_3}[/imath]

Since these are meant to be the horizontal distances between the circles, negative signs can be ignored.

[imath]2.2 = 2\sqrt{r_2r_3}[/imath]

Since [imath]4.9[/imath] is the horizontal distance between [imath]C_1[/imath] and [imath]C_2[/imath], it must be equal to the horizontal distances between [imath]C_1[/imath] and [imath]C_3[/imath] AND [imath]C_2[/imath] and [imath]C_3.[/imath]

Or

[imath]4.9 = 2.7 + 2.2[/imath]

Or

[imath]2\sqrt{r_1r_2}[/imath] = [imath]2\sqrt{r_1r_3}[/imath] + [imath]2\sqrt{r_2r_3}[/imath]

[imath]\sqrt{r_1r_2}[/imath] = [imath]\sqrt{r_1r_3}[/imath] + [imath]\sqrt{r_2r_3}[/imath]

[imath]\sqrt{r_1r_2}[/imath] = [imath]\sqrt{r_1}\sqrt{r_3}[/imath] + [imath]\sqrt{r_2}\sqrt{r_3}[/imath]

[imath]\sqrt{r_1r_2}[/imath] = [imath]\sqrt{r_3}(\sqrt{r_1} + \sqrt{r_2})[/imath]

[imath]\displaystyle \sqrt{r_3} = \frac{\sqrt{r_1r_2}}{\sqrt{r_1} + \sqrt{r_2}}[/imath]


Thanks a lot Dr.Peterson, Highlander, and professor Khan for the contribution in this geometry thread.

 

[Dr.Peterson]

I haven't tried to follow all the details, but it looks like you've probably done more or less the right things.

But to really answer the question, you need to do all this without any specific numbers. Just use the variables.

 

[The Highlander]

That line was one I thought I'd overwritten; I was only confused at first, when I saw that you were responding to what he wrote before you used (9,9) in your example, and took that to be a correction to how he made his circle, centered at (3,3), and before I looked at what you'd written (which, yes, was rather long).

In any case, the important point about that is that you don't need an accurate picture in the first place. A pencil sketch is more appropriate.

Good. I'm very glad that any confusion was only temporary.

When I came back to post my first response (which ended up as Post #6) I read the intervening ones (Posts 3 to 5) before I had logged in so I didn't even notice that the OP had placed Circle 1 at (3, 3); I certainly wasn't suggesting that was an 'error' or that placing it at (9, 9) was in some way 'better' except in that doing so it might make it easier to construct a good representation on paper.

I fully agree that any sketch doesn't need to be 'accurately' drawn to aid coming up with a solution, however, choosing perfect squares (9 & 4) for the radii of Circles 1 & 2 certainly helped me to identify a way to an answer.

Just like the OP, once I had constructed the diagram (my original is below) I too was focussed on the triangle formed by the centres which led me to several dead ends but, noticing that the horizontal distance between r₁ & r₂ on my sketch was 12 (because it was a Pythagorean triple), led me then to consider the horizontal distances between all the centres and, thence, a solution to the problem became obvious.

I do tend to write rather lengthy posts but that is because I try to offer as much detail as I think will enable even the youngest (or less mathematically skilled) readers to understand the what I'm saying.

​

I would have redacted or rubbed out most of the 'working' shown on my page but, since this solution has already been posted, I just didn't bother.

 

[The Highlander]

Hope that does help towards a final solution here.
Thanks a lot Highlander for this extra hint. It is so much appreciated.

Yes, you certainly do appear to have arrived at the terminus albeit via a somewhat tortuous route.

As @Dr.Peterson points out, you don't have to use any numeric calculations to get there (or, indeed, an 'accurate' sketch!).

I had already prepared (& written up) my own solution in anticipation of posting it if you didn't come back with a solution in the next day or so and, therefore, since it was already written out (and for the sake of clarity?), I thought I would just add it to the thread anyway....

Referring to my (final ) diagram below...

​

The length \(\displaystyle FC_2\) is the same as \(\displaystyle AB\), therefore (by Pythagoras)...

[imath]     AB^2=(r_1+r_2)^2-(r_1-r_2)^2\\ \\ \implies AB=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}\\ \\ \implies AB=\sqrt{({r_1}^2+2r_1r_2+{r_2}^2) - ({r_1}^2-2r_1r_2+{r_2}^2)}\\ \\ \implies AB=\sqrt{4r_1r_2}=\sqrt{4}\sqrt{r_1r_2}\\ \\ \implies AB=2\sqrt{r_1r_2}[/imath]

Similarly, it can be shown that...

[imath]DC_3=2\sqrt{r_1r_3}[/imath]    and    [imath]C_3E=2\sqrt{r_2r_3}[/imath]

But [imath]AB=DC_3+C_3E[/imath]

Therefore...

[imath]     2\sqrt{r_1r_2}=2\sqrt{r_1r_3}+2\sqrt{r_2r_3}\\ \\ \implies \sqrt{r_1r_2}=\sqrt{r_1r_3}+\sqrt{r_2r_3}\\ \\ \implies \sqrt{r_1r_2}=\sqrt{r_3}\sqrt{r_1}+\sqrt{r_3}\sqrt{r_2}\\ \\ \implies \sqrt{r_1r_2}=\sqrt{r_3}(\sqrt{r_1}+\sqrt{r_2})\\ \\ \implies \sqrt{r_3}=\frac{\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}[/imath]        QED.

Hope that helps.

 

[khansaheb]

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

I would love to take the credit - but
it was NOT me responsible for ameliorating your troubles.

I suspect it was @stapel who FIXED it.

 

[mario99]

I haven't tried to follow all the details, but it looks like you've probably done more or less the right things.

But to really answer the question, you need to do all this without any specific numbers. Just use the variables.

I was about to say how could that be possible?! But then, I saw what Highlander did.

Yes, you certainly do appear to have arrived at the terminus albeit via a somewhat tortuous route.

Hope that helps.

I don't know what was that supposed to mean, but I consider your solution without numeric calculations as an Art of beauty.

I came up with the idea of the horizontal distances because of Dr.Peterson's and your diagrams. Solving with variables only, did not come to my mind, but now after seeing it, I can say it is possible and easier.

Thanks Dr.Peterson and Highlander for the hard work

I would love to take the credit - but
it was NOT me responsible for ameliorating your troubles.

I suspect it was @stapel who FIXED it.

Then, thank you a lot stapel