# Tangent circles in a rectangle $$ABCD$$ is a rectangle with 4 semicircles tangent to each other. $$AB$$ is the diameter of a circle, $$FC$$, $$GD$$ are radii. Points $$E$$, $$F$$, $$G$$, $$H$$ are on the sides of the rectangle. The circle with radius $$HI$$ is tangent to the other circles. What is the ratio of $$AB:BC$$?

• It looks like $AB$ is a diameter of one circle, and $FC$ and $GD$ are radii of two of the others. It's this the case? (In other words, it looks like three of the circles go through the corners of the rectangle. Is this the case?) Oct 1, 2020 at 9:35
• Yes you're right and $HI$ is one of the radius. I've tried Pythagoras Theorem, but it came with loads of equations Oct 1, 2020 at 9:37
• Yeah, but $I$ doesn't look like a special point in any way. So it's not that relevant to clarify. The fact that the corners of the rectangle lies on the circumferences is very relevant. That's why I asked about those and not $I$. Oct 1, 2020 at 9:40
• Just an observation: $EFGH$ is a tangential quadrilateral. Oct 3, 2020 at 10:12

Let $$\bigcirc E$$, $$\bigcirc F$$, $$\bigcirc G$$, $$\bigcirc H$$ have respective radii $$e$$, $$f$$, $$g$$, $$h$$, and define $$u := |BF|$$ and $$v := |AH|$$. Then each tangency gives a Pythagorean relation:

\begin{align} e^2+u^2 &= (e+f)^2\tag{1} \\ e^2+v^2 &= (e+h)^2 \tag{2} \\ f^2+(2e-g)^2 &= (f+g)^2 \tag{3} \\ g^2+(u+f-v)^2 &= (g+h)^2 \tag{4} \\ (u-v)^2 + (2e)^2 &= (f+h)^2 \tag{5} \end{align}

It's possible to chug through the system eliminating variables one by one, but the symbol-crunching seems to be a mess best left to a computer. If we're going to resort to that anyway, we might as well just let the CAS tackle the whole system in one go. Mathematica readily generates two solutions: the extraneous $$(f,g,h,u,v) = (0,e,-2e,0,0)$$ and also $$(f,g,h,u,v) = \left(\frac98 e, \frac{16}{25} e, \frac{25}{28} e, \frac{15}8 e, \frac{45}{28} e\right) \tag{6}$$

from which we find

$$|AB|:|BC| = 2e : u+f = 2 : 3 \tag{\star}$$

The fact that all of the values in $$(6)$$ are rational multiples of $$e$$ suggests that there might be a clever way to get to the solution, but I'm not seeing it.

• Thank you very much for this solution. I have added a graphical remark to my answer ; I am curious to know if it could be a mean to find out a solution... Oct 1, 2020 at 16:53

Let us take a coordinate system with $$A$$ the origin, $$AB$$ the $$x$$-axis, $$AD$$ the $$y$$ axis. We can assume WLOG that the abscissa of $$B$$ is $$2$$.

Let use notations: $$r$$ and $$s$$ for the radii of circles centered in $$H$$ and $$F$$ resp. and

$$D=(0,d), \ H=(0,h), \ G=(g,d)$$

As a consequence $$F=(2,d-s)$$.

The 5 circles' contacts give the 5 conditions:

$$\begin{cases}(1)&h^2+1&=&(1+r)^2\\ (2)&1+(d-s)^2&=&(1+s)^2\\ (3)&g^2+(d-h)^2&=&(g+r)^2\\ (4)&4+(d-s-h)^2&=&(r+s)^2\\ (5)&(2-g)^2+s^2&=&(g+s)^2\end{cases}$$

(a big thank to Jan-Magnus Økland who has spotted the errors in my initial system. Thank you as well to Blue who has alerted me about the existence of a solution, whereas I thought at first there was none).

We have a system of 5 non linear equations in 5 real unknows $$d,g,h,r,s$$ with positivity constraint (in fact there are other restrictions if we want to comply with the positions given in the figure).

In fact, using (1) and (2), one can obtain:

$$r=\sqrt{1+h^2}-1 \ \ \text{and} \ \ s=\dfrac{d^2}{2(d+1)}.$$

Plugging these expressions in the remaining equations (3),(4),(5), we get a system of 3 equations in 3 unknowns.

We could study the triple points of intersections of the implicit surfaces with these equations.

I have done differently by submitting system (1) to a Computer Algebra System which has given two $$5$$-tuples of solutions. Only one of them was physically meaningful...

$$d=3, \ \ g=16/25 \approx 0.64, \ \ h=45/28 \approx 1.6071, \ \ r=25/28 \approx 0.8929, \ \ s=9/8 \approx 1.1250$$

Therefore: AB/BC = 2/3.

Edit 1: Matlab program with symbolic variables:

syms d g h r s
[D,G,H,R,S]=solve(...
h^2+1==(1+r)^2,...
1+(d-s)^2==(1+s)^2,...
g^2+(d-h)^2==(g+r)^2,...
4+(d-s-h)^2==(r+s)^2,...
(2-g)^2+s^2==(g+s)^2,...
d,g,h,r,s)


Edit 2: Using 3 symmetries on the initial figure (situated on the bottom left) gives rise to a new figure that can be used (by suppressing the external half-circles) as a base for a non-classical periodic arrangement of mutually tangent disks of 3 different sizes on the plane (see for example "Unequal circles" in https://en.wikipedia.org/wiki/Circle_packing • FYI: I get that the final ratio is $|AB|:|BC|=2:3$.
– Blue
Oct 1, 2020 at 14:16
• @Blue Thanks. Therefore, I must have an error in my computations. Oct 1, 2020 at 14:18
• R=QQ[d,h,g,r,s] I=ideal(h^2+1-(r+1)^2,1+(d-s)^2-(1+s)^2,g^2+(d-h)^2-(r+g)^2,4+(d-s-h)^2-(r+s)^2,(2-g)^2+s^2-(s+g)^2) primaryDecomposition I -- {ideal(8*s-9,28*r-25,25*g-16,28*h-45,d-3), ideal(r+3*s+2,g*s+2*g-2,d*s-h*s-3*s^2-4*g-2*s+4,h^2-9*s^2-6*s,d*h-h*s-3*s^2,d^2-2*h*s-6*s^2-8*g-6*s+8)} So $d=3.$ You just wrote $d$ for $g$ in two places. Oct 1, 2020 at 16:40
• @Jan-Magnus Økland Thank you very very much for spotting my error (coming from a late change of names). May I ask you which software do you use ? Please transform this remark into a solution ; I would be happy to upvote it. Besides, I just added in my nswer a remark with an accompaining figure about a possible use of this configuration. Oct 1, 2020 at 16:46

Just to add the number crunching in M2. Jean Marie had two misprints in his equations (3) and (5). The solution point $$(d,g,h,r,s)=(3,\frac{16}{25},\frac{45}{28},\frac{25}{28},\frac98)$$ is found by the following few lines:

R=QQ[d,g,h,r,s]
I=ideal(h^2+1-(r+1)^2,1+(d-s)^2-(1+s)^2,g^2+(d-h)^2-(r+g)^2,4+(d-s-h)^2-(r+s)^2,(2-g)^2+s^2-(s+g)^2)
primaryDecomposition I -- {ideal(8*s-9,28*r-25,28*h-45,25*g-16,d-3), ideal(r+3*s+2,g*s+2*g-2,d*s-h*s-3*s^2-4*g-2*s+4,h^2-9*s^2-6*s,d*h-h*s-3*s^2,d^2-2*h*s-6*s^2-8*g-6*s+8)}


So to answer the question, the ratio is $$2:3$$