I found in a book of Sangakus the following problem.

Three Circles

Let $R_b$, $R_g$ and $R_r$ the radiuses of the blue, green and red circles $C_b$, $C_g$ and $C_r$.

Prove that $$\frac{1}{\sqrt{R_r}}=\frac{1}{\sqrt{R_b}}+\frac{1}{\sqrt{R_g}}\,.\quad (1)$$ And this I can do. But then

I would like to draw the figure myself with only a ruler and a compass.

I know it is possible as I can construct inverses, sums and square roots with only a ruler and a compass, but when I tried to draw the figure with a "simple" or "natural" construction, I failed.

Does someone have an idea of how to draw it "naturally"?


Answer to a comment of Amzoti: To prove Relation (1) I first prove the relation: $$AB^2=4R_gR_b\quad (2)$$ Two Circles Relation (2) is a consequence of Pythagoras' theorem in the triangle $O_bO_gH$: $$AB^2+(R_g-R_b)^2=(R_b+R_g)^2\,.$$ (It was the previous Sengaku in the book.)

We thus get the relations $$ \begin{align} AB^2 & =4R_gR_b\quad (2) \\ AC^2 & =4R_gR_r\quad (3) \\ BC^2 & =4R_bR_r\quad (4) \\ \end{align} $$ where $A$, $B$, $C$ are the orthogonal projections of the centers of the circles $C_b$, $C_g$ and $C_r$ on the line $d$. The relation $AB^2=AC^2+BC^2+2AC.BC$ then yields, using Relations (2) to (4), $$4R_gR_b=4R_gR_r+4R_bR_r+8\sqrt{R_bR_g}R_r$$ Divided by $4R_rR_gR_b$ this equation is $$\frac{1}{R_r}=\Big(\frac{\sqrt{R_g}+\sqrt{R_b}}{\sqrt{R_bR_g}}\Big)^2$$ which is in fact Relation (1) squared.

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    $\begingroup$ If you have two of the circles and you want to draw the third, I believe you'll have to use one of the methods here. $\endgroup$ – Rahul Jan 3 '13 at 21:48
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    $\begingroup$ It seems the conventional English spelling is Apollonius. $\endgroup$ – GEdgar Jan 4 '13 at 16:06
  • $\begingroup$ @RahulNarain I don't know much about geometry and I didn't know this Apollonius' problem, it is true that the problem I quoted above can be seen as one of the limit cases of Apollonius' problem, the case with two circles and one line. And indeed the general problem of Apollonius is very interesting! Thanks for your comment. $\endgroup$ – Sebastien B Jan 4 '13 at 17:53

enter image description here

I found another straightforward method which can be perfectly show the meaning!

  • $\begingroup$ Are you sure this construction is right? I tried to do it but it did not seem to give the right circle. $\endgroup$ – Sebastien B Feb 4 '13 at 7:53
  • $\begingroup$ yes, I review it again and 100% right. it can be easily proof the radius is satisfy the formula. $\endgroup$ – chenbai Feb 4 '13 at 10:12
  • $\begingroup$ $DC=2\sqrt{r1*r2}. $ $\endgroup$ – chenbai Feb 4 '13 at 10:20
  • $\begingroup$ I checked it again, and it works indeed. Marvelous. How did you come to this construction? $\endgroup$ – Sebastien B Feb 4 '13 at 10:38
  • $\begingroup$ it is my 3rd version. the 2nd version post first which I think it is fine. then I check it again when I review my comments, I realize the DC is what we need. then rest thing is simple. BTW, you have to convert r formula in normal way, ie, r=XXXX, then you will find the a factor $ r1+r2+2\sqrt{r1*r2}$. it take me sometime to find this factor. Anyway, I am very happy to find this solution and thank you for the question. $\endgroup$ – chenbai Feb 4 '13 at 11:54

Perhaps see section 9.3 of Alter Mundus by Alain Matthes

Section 9 also contains a proof of your stated problem and it would be nice to compare to your proof if you would be so kind to add it to your question.

It looks like the web site is still under construction, but also has some items of interest.

I have not tried the construction in 9.3 above, but and am curious if it works.

Update: by the way, you might also want to check out some software and there are MSE recommendations on hyperbolic-geometry-software-programs listed here.


  • $\begingroup$ It seems to be exactly what I want. I will try it. Many thanks. $\endgroup$ – Sebastien B Jan 3 '13 at 23:02
  • $\begingroup$ @SebastienB: excellent, please let us know and I'd love to also see your proof of the result in your problem. Have fun and regards! $\endgroup$ – Amzoti Jan 3 '13 at 23:03
  • $\begingroup$ I wrote the proof after the question above as it was to long to fit in a comment. Regards. $\endgroup$ – Sebastien B Jan 4 '13 at 0:21
  • $\begingroup$ @SebastienB: thank you kindly. Looks like Ricardo also provided a very nice response. Regards $\endgroup$ – Amzoti Jan 4 '13 at 16:01
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    $\begingroup$ I tried the construction, it works, but I feel the one in the link of GEdgar is more elegant algthough this is very subjectiv, and I don't fully understand the proof of why the construction in the link of GEdgar is valid. Anyway many thanks, I also have some other Sangakus to play with now. $\endgroup$ – Sebastien B Jan 7 '13 at 22:17

We'll solve the following problem:

Given, a straight line $r$, a point $A$ such that $A \in r$ and the radii $R_b$ and $R_g$, construct three circles $\lambda_b$,$\lambda_g$ and $\lambda_r$, such that they are tangent to each other and to $r$, and $R_r \leq min(R_b, R_g)$.($R_b$, $R_g$, and $R_r$ are the radii of $\lambda_b$,$\lambda_g$ and $\lambda_r$ respectively).

See the figure below: SangakuConst

  1. Draw a straight line $t$ such that $A \in t$ and $ t \perp r$.
  2. Mark $O_b$ such that $O_b \in t$ and $O_bA = R_b$.
  3. Draw $\lambda_b$.
  4. Draw $s$ such that $s \parallel r$ and $d(s,r)=R_g$.
  5. Draw an arc $\mu$ centered at $O_b$ and radius $R_b + R_g$ such that $\mu$ intesects $s$. The intersection point is $O_g$.
  6. Draw $\lambda_g$.
  7. Find out $R_r$. (Auxiliary construction).
  8. Find out $O_r$. ($O_r$ is the intersection point of two arcs: one of them has $O_b$ as center and $R_b + R_r$ as radius and the other one has $O_g$ as center and $R_g + R_r$ as radius).
  9. Draw $\lambda_r$.

Auxiliary construction

Let $d=d(A, B)$ , where $\{B\}= r \cap \lambda_g$.

  1. From a point $Q$ draw two arbitrary rays $w$ and $h$.
  2. Mark a point $N$ such that $d(N,Q) = R_b + d + R_g$ and $N \in h$.
  3. Mark a point $M$ such that $d(M,Q) = R_b$ and $M \in w$.
  4. Draw a straight line $u$ such that $u$ passes through $M$ and $N$.
  5. Mark a point $P$ such that $d(P,Q) = R_b + d + R_g + R_g $ and $P \in h$.
  6. Draw a straight line $v$ such that $v \parallel u$ and $v$ passes through $P$.
  7. $R_r = d(M,T)$ where $\{T \}= v \cap u$.
  • $\begingroup$ (1) The diagram is beautiful. How did you make it? (2) It would be helpful to mention that the rays $w$ and $h$ are arbitrary and their angle doesn't matter. But what is $d$ in $R_b+d+R_g$? I don't think you have defined it. $\endgroup$ – Rahul Jan 4 '13 at 17:57
  • $\begingroup$ @RahulNarain. Thanks for the comments, Rahul. I made the changes you pointed out. As regards the diagram, I used Compass and Ruler (C.a.R.) and after that I made some changes with Inkscape. $\endgroup$ – RicardoCruz Jan 4 '13 at 18:30
  • $\begingroup$ Thanks for the info! You could add it to the question about software for drawing geometry diagrams on this site. (Also I see that you had already labeled $d$ in the diagram, but I missed it among all the other letters.) $\endgroup$ – Rahul Jan 4 '13 at 18:53

Perhaps look in the Geometry Junkyard http://www.ics.uci.edu/~eppstein/junkyard/tangencies/apollonian.html

  • $\begingroup$ I tried the construction and it works (although I needed a little bit of time to adapt it to my "limit" case). But when trying to understand the proof under the construction in your link I got stuck at some point: "The blue perpendicular we constructed is a third parallel line halfway between a and b". My problem is that so that the blue line gives again a line by inversion it shall go through the center of D, that is the tangency point of A and B, which is not the case. Well I'm not very used to think with inversion, so perhaps I misunderstood something... $\endgroup$ – Sebastien B Jan 7 '13 at 18:58

enter image description here]![enter image description here

I think my method is the one you may want. it is nature.

it is not difficult to proof. just care $KE=\sqrt{r1*r2}$ and KM=MB


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