Six circles in a rectangle: show that two lengths are equal

Question

The diagram shows a rectangle, six circles, and a red line segment joining the centres of two circles. Wherever things look tangent, they are tangent. (The tangencies imply that there are three pairs of congruent circles.)

Show that the length of the red line segment equals the height of the rectangle.

I will post my solution, which uses polynomials. I am looking for other solutions, in particular purely geometrical solutions.

Context: I sometimes create diagrams involving circles and discover that there are unexpected equal lengths, for example in a triangle with three circles.

Answer 1

Proof using polynomials

Define:

• height of rectangle $=1$
• $L=$ length of red line segment
• $x=$ square root of radius of upper-left and lower-right circles
• $y=$ square root of radius of two middle circles
• $z=$ square root of radius of lower-left and upper-right circles

Summary of method: We will get two expressions for $y$ in terms of $x$, then show that $x$ is a root of a certain cubic equation. Then we will find an expression for $L^2-1$ in terms of $x$, a factor of which is the cubic expression found earlier, which will imply that $L=1$.

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Use Pythagorus with the right triangle in the left half of the blue rectangle:

$$(1-x^2-z^2)^2+(x^2-z^2)^2=(x^2+z^2)^2$$

$$z^4-(2x^2+2)z^2+(x^2-1)^2=0$$

$$z=1-x$$

Use Pythagorus to equate the bottom and top side lengths of the blue rectangle:

$$\sqrt{(y^2+z^2)^2-(y^2-z^2)^2}=x^2-z^2+\sqrt{(x^2+y^2)^2-(1-x^2-y^2)^2}$$

Rearrange as a quadratic in $y$, solve for $y$, and use the previous result $z=1-x$:

$$y=\frac{\left(\sqrt{2}-2\right)x^2+\left(3-\sqrt{2}\right)x-1}{2x^2-4x+1}$$

Factorise the numerator and denominator and cancel common factors:

$$y=\frac{\left(\frac{1}{\sqrt2}-1\right)(x-1)}{x-1+\frac{1}{\sqrt2}}\tag{1}$$

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Use Pythagorus to equate the bottom and top side lengths of the green rectangle:

$$\sqrt{(x^2+y^2)^2-(1-x^2-y^2)^2}+\sqrt{(2y^2)^2-(1-2y^2)^2}=\sqrt{(x^2+y^2)^2-(x^2-y^2)^2}$$

Rearrange as a quadratic in $y^2$ and solve for $y^2$:

$$y^2=\frac{x^2\left(2x^2-2\sqrt2x-1\right)}{4x^4-12x^2+1}$$

Factorise the numerator and denominator and cancel common factors:

$$y^2=\frac{x^2}{2\left(x+1+\frac{1}{\sqrt2}\right)\left(x-1+\frac{1}{\sqrt2}\right)}\tag{2}$$

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Equate $y^2$ from equations $(1)$ and $(2)$:

$$\left(\frac{\left(\frac{1}{\sqrt2}-1\right)\left(x-1\right)}{x-1+\frac{1}{\sqrt2}}\right)^2=\frac{x^2}{2\left(x+1+\frac{1}{\sqrt2}\right)\left(x-1+\frac{1}{\sqrt2}\right)}$$

Cancel common factors and rearrange:

$$\color{red}{f(x)=\left(2-2\sqrt2\right)x^3+\left(3\sqrt2-4\right)x^2+\left(1-\sqrt2\right)x+1-\frac{1}{\sqrt2}=0}$$

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Use Pythagorus with the right triangle above:

$$\left(\frac{L}{2}\right)^2+y^4=(x^2+y^2)^2$$

Use this and equation $(2)$ to express $L^2-1$:

$$L^2-1=\frac{g(x)}{x^2+\sqrt2x-\frac12}$$

where $g(x)=4x^6+4\sqrt2x^5+2x^4-x^2-\sqrt2x+\frac12$.

Using long division, we find that $g(x)$ is divisible by $\color{red}{f(x)}$.

$$L^2-1=\frac{f(x)\left(-\left(2+2\sqrt{2}\right)x^{3}-\left(4+3\sqrt{2}\right)x^{2}-\left(1+\sqrt{2}\right)x+1+\frac{1}{\sqrt{2}}\right)}{x^2+\sqrt2x-\frac12}$$

Since $f(x)=0$, therefore $L=1$.

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Bonus result

The diagram below shows six circles in a rectangle of height $2$. Wherever things look tangent, they are tangent. The radii of the circles in the top row, from left to right, are $a,b,c$.

Using methods similar to my answer to the OP, we get the following elegant result:

$a+b=abc+1$

Answer 2

For a geometric approach, let $EF$ intersect the large circles at $G$, $H$. Through $E$, $F$ draw $JEKL$ and $ONFM$ perpendicular to base $AB$ and join $KO$, $LN$.

If it can be shown that $KO$, $LN$ pass through $G$, $H$, then since $EK=EG$, and by symmetry $KO\parallel LN$, then$$KL=GH$$and therefore$$EF=EG+FH+GH=EJ+EK+KL=BC$$

Addendum: Through center $E$ of rectangle $ABCD$ draw $HJ\perp AB$ and complete quadrilateral $HFJG$. Thus $J$, $H$ are midpoints of $AB$, $CD$, respectively.

Since by symmetry $FH\parallel JG$ and $FJ\parallel HG$, then $HFJG$ is a parallelogram. If it is also a rectangle as it appears, then again$$FG=HJ=BC$$

Note: The posted construction seems to require a rectangle with a particular length/width ratio$$\frac{L}{W}\approx\frac{29}{19}$$Further, it appears rectangle $FJGH\sim ABCD$, and hence likewise admits three pairs of tangent circles, as seen below. Thus the construction can be made to spiral around point $E$, inward clockwise and outward counterclockwise, without limit.

Answer 3

I edited the original answer, to correct the mistake OP pointed out in the orignal version. The proof is not "purely geometrical", as probably OP hoped, but it hints and some interesting geometrical properties of the proposed configuration.

Furthremore, we show that once the thesis (i.e. the fact that the distance between the larger circles' centers equals the vertical side of the rectangle) is reached, one can use this result to obtain exact values of all the involved measures, by solving a cubic equation.

Let us consider a more general situation (depicted below), where the middle circles (assume they have radius $1$) are tangent to each other, and each of them is tangent either to the larger circle (with radius $R_1$) or to the smaller one (with radius $R_3$).

We are first going to prove that in all these configurations, the central axis of the middle circles $CD$ is parallel to the common tangent to larger and smaller circle $IP$.

Define $$s = \sqrt R_1 + \sqrt R_3$$ and $$d = \sqrt R_1 -\sqrt R_3.$$

Since $\overline{FN} = 2\sqrt R_1$, and $\overline{LM} = 2\sqrt R_3$, we have $$\overline{DE} = 2\cos \alpha = d(s + 2).\tag{1}\label{1}$$

From $\overline{GH} = 2\sqrt{R_1 R_3}$ we obtain $$\overline{CE} = 2\sin \alpha = s^2-2.\tag{2}\label{2}$$ The fundamental trigonometric identity, together with \eqref{1} and \eqref{2} yields $$d = s\cdot\sqrt\frac{2-s}{2+s}.\tag{3}\label{3}$$

Since $AGI\sim BHI$, we get $$\overline{HI} = \frac{2\sqrt{R_1R_3}}{R_3(R_1-R_3)},$$ whence \begin{eqnarray} \tan \frac{\beta}2 &=& \frac{2sd}{(s-d)(s+d)}\\ &\stackrel{\eqref{3}}{=}&\frac{\sqrt{(2-s)(2+s)}}{s}\tag{4}\label{4}, \end{eqnarray} and, therefore, \begin{eqnarray} \cos\beta &=& \frac{1-\tan^2\frac{\beta}2}{1+\tan^2\frac{\beta}2}\\ &=& \frac{s^2}2-1 \\ &=& \sin\alpha, \end{eqnarray} which proves our statement.

Now let us go back to the original problem and construct two replicas of the original rectangle as depicted below, where $A$ and $D$ are the center of the circles with radius $R_1$, $B$ and $C$ are the points where the rectangles intersect, $E$ is the center of the smaller circle on the bottom-left corner, and $F$ and $G$ are the orthogonal projections of $E$ and $D$ on $PQ$, the rectangle's side. $R$ is the top-left corner of the rectantle, and, finally $S$ and $T$ are the center of one of the radius-$1$ circles, and their tangency point, respectively.

1. Because of the simmetry, $AB$ and $CD$ lie on the central axes of the large and small circles, therefore $A$, $B$ and $E$ are aligned, and $$AB\parallel CD.$$
2. We also have $$\measuredangle BCD = 90^\circ+\frac{\beta}2$$ and $$\measuredangle BAD = \frac{\beta}2 + \alpha,$$ so that $$\measuredangle BCD + \measuredangle BAD = 90^\circ + \beta + \alpha = 180^\circ,$$implying that $ABCD$ is cyclic. Since it is a trapezoid, because of 1., it must be an isoceles trapezoid. Hence $$AD \cong BC.$$
3. Now we have \begin{eqnarray} \overline{BC} &=& \overline{PQ}-\overline{PF} + \overline{BF}\\ & & - \overline{CG}-\overline{GQ}, \end{eqnarray} where \begin{eqnarray}\overline{PQ} &=& R_1 + R_3 + 2\sqrt R_1 + 2\sqrt R_3\\ \overline{PF} &=& R_3\\ \overline{BF} &=& R_3 \tan\frac{\beta}2\\ \overline{CG} &=& R_1 \tan\frac{\beta}2\\ \overline{GC} &=& R_1. \end{eqnarray} Hence we can write, using \eqref{3} and \eqref{4} \begin{eqnarray} \overline{BC} &=& 2\sqrt R_1 + 2\sqrt R_3 - (R_1-R_3)\tan\frac{\beta}2\ &=& 2s - sd \tan\frac{\beta}2\\ &=& 2s - s^2\cdot \sqrt{\frac{2-s}{2+s}} \cdot\frac{\sqrt{(2-s)(2+s)}}s\\ &=& s^2. \end{eqnarray}
4. Since the vertical side of the rectangle is $$\overline{PR} = R_1 + R_3 + 2\sqrt{R_1 R_3} = s^2$$ we have our thesis, i.e. $$\boxed{AD \cong PR}.$$

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The numerical values of $s$ and $p$ which correspond to OP's configuration can be determined by solving the simultaneous conditions $$ \begin{cases} d = s\cdot \sqrt{ \frac{2-s}{2+s}}\\ \frac{s^4}4 + 1 = \left[1+\left(\frac{s+d}2\right)^2\right]^2, \end{cases} $$ where we added to \eqref{3} Pythagorean Theorem applied to $ATS$. The resulting sextic equation is $$s^6+4s^5+4s^4−8s^3−24s^2−32s−16=0,$$ which we can factor (thanks to mezzoctane comment) as \begin{eqnarray} &&(s^3+2s^2-4)^2 -8(s+2)^2=0\\ &&(s^3+2s^2-2\sqrt 2 s -4-4\sqrt 2)\\ &&\cdot (s^3+2s^2+2\sqrt 2 s -4+4\sqrt 2)=0 \end{eqnarray} The only admissible solution comes from the first factor above. Using Cardano's formula yields \begin{eqnarray} s &=& -\frac23 + \sqrt[3]{\frac{46+36\sqrt 2+6\sqrt{72\sqrt 2 + 105}} {27}}\\ & & + \sqrt[3]{\frac{46+36\sqrt 2-6\sqrt{72\sqrt 2 + 105}} {27}}\\ &\approx& 1.9592641159631635. \end{eqnarray} which corresponds to $$d \approx 0.19873498225185185.$$ This results in $$R_1 = \frac{(s+d)^2}4 \approx 1.16424002697420486,$$ and $$R_3 = \frac{(s-d)^2}4 \approx 0.77486570766157536.$$ Note that the dimensions of the rectangle circumscribing the six circles are $$\overline{PR} = s^2,$$ as you already meantione, and, using \eqref{3}, \begin{eqnarray} \overline{PQ} &=& R_1 + R_3+ 2(\sqrt R_1 +\sqrt R_3)\\ &=& \frac{4s(1+s)}{2+s}, \end{eqnarray} so that \begin{eqnarray} \frac{\overline{PQ}}{\overline{PR}} &=& \frac{4(1+s)}{s(2+s)}\\ &\approx& 1.525977382531713. \end{eqnarray}

Answer 4

This answer was started with the intention to show the claimed property by using a less involved calculus done by a human hand with a human eye. However, the complexity cannot be avoided. The only solution that is a fair complete solution is the computer path. It is simple, based on Groebner bases algorithms, and does the job immediately. It also shows why a naive approach cannot lead to simple quick solution. (Except for finding a trick, that i am maybe missing.) So i will also insert a human computational start, and a geometric point of view, also incomplete.

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Computer view:

Psychologically it is a benefit to see first the computer solution, mainly because it makes clear, there is only one way to realize the picture. Here the human hand is imposing algebraic conditions, letting the computer solve a resulted system of algebraic equations. Its complexity equals the complexity of the stated problem.

How do we translate the picture into equations? We are placing the solution box with the center in the origin of axes, and let its width and height be $2w$ and $2h$.

Fig. $1$:

The three circles have the radii $a,b,c$, and the upper $b$-radius circle has its $x$-coordinate $x_u$. The code below considers a polynomial ring $R$ in the six free variables $a,b,c;h,w;x_u$, immediately sets the half-height to one, $h=1$, and let $J$ be the ideal generated by the equations which reflect:

• $h=1$,
• upper $a$-circle is tangent to the upper $b$-circle,
• upper $b$-circle is tangent to the upper $c$-circle,
• upper $b$-circle is tangent to the lower $a$-circle,
• upper $c$-circle is tangent to the lower $a$-circle,
• upper $b$-circle passes through the origin
• the line between the two centers of $a$-circles is tangent in the origin to the line between the two centers of $a$-circles.

There are seven equations, six variables, we check the dimension of the ideal, it is zero, and ask for its "variety" with coordinates in the real algebraic numbers, which is a set of points.

Sage Code: (Please skip if it feels annoying.)

R.<a,b,c,h,w,xu> = QQ[]

O = vector([0, 0,])
Oa = vector([-w + a, h - a])
Ob = vector([    xu, h - b])
Oc = vector([ w - c, h - c])
Oaa, Obb, Occ = -Oa, -Ob, -Oc
J = R.ideal([h - 1
, d2(Oa, Ob) - (a+b)^2
, d2(Ob, Oc) - (b+c)^2
, d2(Oc, Oaa) - (a+c)^2
, d2(Ob, Oaa) - (a+b)^2
, d2(Ob, O) - b^2, (Oa-Oaa)*Ob])
for sol in J.variety(AA):
if sol[w] <= 0:     continue
if sol[a] <= 0:     continue
if sol[b] <= 0:     continue
if sol[c] <= 0:     continue
print(f'SOLUTION:')
print(f'    a  = {sol[a]} with minimal polynomial {sol[a].minpoly()}')
print(f'    b  = {sol[b]} with minimal polynomial {sol[b].minpoly()}')
print(f'    c  = {sol[c]} with minimal polynomial {sol[c].minpoly()}')
print(f'    w  = {sol[w]} with minimal polynomial {sol[w].minpoly()}')
print(f'    xu = {sol[xu]} with minimal polynomial {sol[xu].minpoly()}')

And here are the solutions:

SOLUTION:
    a  = 1 with minimal polynomial x - 1
    b  = 0.50000000000000000? with minimal polynomial x - 1/2
    c  = 5.828427124746190? with minimal polynomial x^2 - 6*x + 1
    w  = 2.414213562373095? with minimal polynomial x^2 - 2*x - 1
    xu = 0 with minimal polynomial x
SOLUTION:
    a  = 0.606577858091834? with minimal polynomial x^6 - 2*x^5 - x^4 + 8*x^3 + 3*x^2 - 6*x + 1
    b  = 0.5210075620474032? with minimal polynomial x^6 - 2*x^5 - x^4 - x^3 + 2*x^2 - x + 1/4
    c  = 0.4037108932628932? with minimal polynomial x^6 - 6*x^5 + 23*x^4 - 48*x^3 - 5*x^2 + 6*x + 1
    w  = 1.525935796142292? with minimal polynomial x^6 + 8*x^4 - 8*x^3 + 16*x^2 - 32*x - 16
    xu = 0.2049759110110414? with minimal polynomial x^6 - 2*x^5 + 3*x^4 - 8*x^3 + 7*x^2 - 6*x + 1

The first "solution" found is not a solution to our problem, but a solution of the problem of drawing six circles tangent to each other as specified and (somehow) to the axes of a box. Here is a plot of it, since it can be easily plotted with exactly specified coordinates.

Fig. $2$:

The missing circle is the reflection in the origin of the big brown $c$-circle.

The second solution was already plotted in the Figure $1$, and the values of $a,b,c;w,x_u$ are algebraic of degree $6$. This complexity cannot be eliminated.

The moral of the story is: We have exactly one configuration realizing the posted OP picture, and when normed to a height $H=2h=2$ it involves for the radii and the width complicated algebraic numbers (living in algebraic fields of degree six).

Note: Using similar eliminations, the computer shows that for $w\ne 0$ and $h=1$ from the given equations we have "only" $(e^2-1)(e^2-2)=0$. This relation is involved for the human eye, and also confirms the complexity of reaching the wanted $e=1$, after discarding $e=\sqrt 2$.

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An algebraic path (incomplete solution): (Done humanly.) We are trying to solve the same system in a human manner: $$ \left\{ \begin{aligned} x_u^2 +(h-b)^2 &= b^2 & & \text{i.e.} &d(O_b,O) &= b\\ (x_u + w-a)^2 +(a-b)^2 &= (a+b)^2 & & \text{i.e.} &d(O_a,O_b) &= a+b\\ (-x_u + w-c)^2 +(b-c)^2 &= (b+c)^2 & & \text{i.e.} &d(O_b,O_c) &= b+c\\ (a-c)^2 +(2h-a-c)^2 &= (a+c)^2 & & \text{i.e.} &d(O_c,O'_a) &= a+c\\ x_u(w-a) & = (h-a)(h-b)& & \text{i.e.} & Oa_O&\perp O_bO\\ e^2 &= (a+b)^2 - b^2 \end{aligned} \right. $$ Below we are setting $h=1$, and trying to eliminate everything but $e$. The first variable we eliminate below is $x_u$. We have $x_u=(1-a)(1-b)/(w-a)$, and $x_u^2=2b-1$. So the above system implies: $$ \left\{ \begin{aligned} (1-a)^2(1-b)^2 &= (2b-1)(w-a)^2\\ (2b-1) + 2(1-a)(1-b) + (w-a)^2 &= 4ab\\ (2b-1) - 2(1-a)(1-b)\frac{w-c}{w-a} + (w-c)^2 &= 4bc\\ (2-a-c)^2 &= 4ac\\ e^2 &= (a+b)^2 - b^2 \end{aligned} \right. $$ We eliminate $w$ by using the expression for $(w-a)^2=(1-a)^2(1-b)^2/(2b-1)$ from the first equation in the second one. Also, we can subtract above the second and third equations from each other to cancel the term in $w^2$.

This gives $\displaystyle 2(1-a)(1-b)\left(2+\frac{a-c}{w-a}\right)-2w(a-c) +(a^2-c^2) =4b(a-c)$. We divide by $(a-c)$ and use the radical formula for $(w-a)$. $$ \left\{ \begin{aligned} 0 &= a^2b^2 - 2a^2b - 6ab^2 + a^2 + 2ab + b^2 \\ 4b &= 2(1-a)(1-b)\left(\frac2{a-c}+\frac{\sqrt{2b-1}}{(1-a)(1-b)}\right) -2\frac{(1-a)(1-b)}{\sqrt{2b-1}}-a-c\\ (2-a-c)^2 &= 4ac\\ e^2 &= (a+b)^2 - b^2 \end{aligned} \right. $$ The equation in $a,c$ only lets us write $c$ as $a+2\pm\sqrt{8a}$. Because of $c<2$ we have $c=a+2-\sqrt{8a}$. This leads to: $$ \left\{ \begin{aligned} 0 &= a^2b^2 - 2a^2b - 6ab^2 + a^2 + 2ab + b^2 \\ 0 &= (1-a)(1-b)\cdot\frac1{\sqrt{2a}-1} +\sqrt{2b-1} -\frac{(1-a)(1-b)}{\sqrt{2b-1}} +1 -\sqrt{2a} -2b\\ e^2 &= (a+b)^2 - b^2 \end{aligned} \right. $$ Now the first equation can be written as $0=a^2(1-b)^2-2ab(3b-1)+b^2$, and we consider it as an equation of second degree in $a$. This gives $$ a_\pm =\frac b{(1-b)^2}\Big[3b-1\pm2\sqrt{b(2b-1)}\Big]\ . $$ The right sign is minus, so we have a formula for $a$ in terms of $b$: $$ a =\frac b{(1-b)^2}\left(\sqrt b -\sqrt{2b-1}\right)^2\ . $$ In order of making computations without radicals, we are searching for a parametrization of $b$ so that $\sqrt b$ and $\sqrt{2b-1}$ are rational expressions. One such parametrization is $$ \small \begin{aligned} b &= \left(\frac{1+u^2}{1+2u-u^2}\right)^2 & a &=\frac 14\left(\frac{1+u^2}{1-u}\right)^2 \\ \sqrt b &= \frac{1+u^2}{1+2u-u^2} & \sqrt a &= \frac{1+u^2}{2(1-u)} & \sqrt{2b-1} &= \frac{1-2u-u^2}{1+2u-u^2} \\[2mm] 1-a &= \frac {(1-2u-u^2)(3-2u-u^2)}{4(1-u)^2} & 1-b &= \frac {4u(1-u)(1+u)}{{(1+2u-u^2)^2}}\ ,\\[2mm] e &=\frac 14\cdot\frac{(3-2u+u^2)(1+u^2)^2}{(1+2u-u^2)(1-u)^2}\ , \end{aligned} $$ where $u$ is a suitable real value. Using $u$ in the remained equation (involving radicals, now made to no longer involve them), we obtain: $$ \begin{aligned} 0 &= \frac{(1-a)(1-b)}{\sqrt{2a}-1} -\frac{(1-a)(1-b)}{\sqrt{2b-1}} -\sqrt{2a} +\sqrt{2b-1} -(2b-1)\\ &=\frac{(1-a)(1-b)}{\sqrt{2a}-1} - \frac {u(1+u)(3-2u-u^2)}{(1-u)(1+2u-u^2)} -\sqrt{2a} + \frac{4u(1-2u-u^2)}{(1+2u-u^2)^2} \end{aligned} $$ This can continue, but i have to stop here, the effort does not compensate for the result.

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Geometric approach (incomplete): (But geometric properties are collected.)

We start with the following situation, which is a subconstellation in the given constellation of circles:

Here, the "same" notation $a,b$ is used for the radii of the two pairs of equal (here in the sense of congruent) circles. The "red segment" $AA'$ matches the OP red segment. Explicitly, we introduce the involved points in the following order, so that everything is well defined. (We use cartesian axes only as orientation, no involved analytic geometry is needed.)

Let $O$ be a point, here taken as the origin of the axes. Denote by a prime the reflection w.r.t. $O$, so it is a map of points $X\to X'$.

Let $B$ be a point on the $Oy$-axis at distance $b$ from $O$. Let $A$ be a point on the $Ox$-axis at distance $a+b$ from $B$. In particular, its distance $e$ from $O$ satisfies $e^2=(a+b)^2-b^2=a^2$. (This $e$ has nothing to do with the later point $E$, nor with $\exp(1)$, i am just needing a new letter...)

(The picture was drawn with $OB=b=36$, $a=49$, $AB=a+b=85$, $e=77$. The triangle $\Delta AOB$ is a $36-77-85$ pythagorean triangle. So $a,b$ are perfect squares, just in case...)

Let $D\in AB$ be the tangency point of the circles $\mathcal A=\operatorname{Circle}(A,e)$ and $\mathcal B=\operatorname{Circle}(B,b)$.

Construct all reflected objects, the points $A',B',D'$, and the circles $$\mathcal A',$\mathcal B'$.

Reflect $D$ w.r.t. $OB$ to obtain $E$, and build $E'$.

So $D'\in A'B'$ is the tangency point of the circles $\mathcal A'=\operatorname{Circle}(A',\alpha)$ and $\mathcal B'=\operatorname{Circle}(B',b)$. And $E,E'$ are tangency points for the pairs of circles $(\mathcal A',\mathcal B)$ and $(\mathcal A,\mathcal B')$.

The exterior tangents of $(\mathcal A',\mathcal B)$ intersect in a point $S$, and these tangents are $SJK$, $SLM$, with $J,L\in \mathcal B$, and $K,M\in \mathcal A'$. Build also their reflections.

Arrange in notation of the two tangents that $SJK$ and its reflection $S'J'K'$ delimit a region of the plan (intersection of two half-planes) where all the picture lives in. (And the needed box will be the minimal rectangular box in this strip between the parallels $SJK\|S'J'K'$. We will build perpendiculars on them which are tangent to $\mathcal A,\mathcal A'$ on the far side w.r.t. $O$).

(This was a long exact description of what could be omitted, given the picture and a wink with the hands.)

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We would want to get the sizes of the box.

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As a warm up, let us compute the coordinates of some easy involved points. $$ \begin{aligned} B &=(0,b) %& B'&=-B=(0,-b) \\ A &= (e,0) %& A' &=-A=(-e,0)\ , \qquad e^2=(a+b)^2 - b^2 = a(a+2b)\\[2mm] D &= \frac b{a+b}(e,a) %& D' &= -D = -\frac b{a+b}(e,a) \\ E &= \frac b{a+b}(-e,a) %& E' &= -E = -\frac b{a+b}(-e,a)= \frac b{a+b}(e,-a) \\[2mm] S &= \left(\frac{be}{a-b}\ ,\ \frac{ae}{a-b}\right) %&S' &= -S =-\left(\frac{be}{a-b}\ ,\ \frac{ae}{a-b}\right) \end{aligned} $$ For $D=(x_D,y_D)$ we have used $y_D:b=y_D:OB=a/(a+b)$, and $x_D:b=x_D:DB=AO:AB=e:(a+b)$. For $S=(x_S,y_S)$ we have built similar to $\Delta A'BO$ a similar triangle $\Delta A'SP$, $P$ being the projection of $S$ on $AA'$. Then $y_S:(x_S+e)=SP:A'P=BO:A'O=b:e$, and a second relation can be obtained from the trapezium $A'BJK$. Project $B$ on $A'K$, we obtain a right triangle with sides $A'B=(a+b)$, and catheti $(a-b)=A'K-BJ$, and $2\sqrt {ab}$ (same as $JK$). This triangle is similar to $\Delta SJB$, so $SB:b=SB:JB=(a+b):(a-b)$, leading to a formula for $SB$, which can be used to obtain $x_B$ from the similarity $x_B:SB=e:(a+b)$.

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Lemma A: The (light yellow) circle with diameter $A'B$ is tangent to $SJK$ in the mid point of $JK$, this mid point has same distance $\sqrt {ab}$ to $J,K,E$.

Note: This circle is thus after a reflection w.r.t. $SBA'$ also tangent to $SLM$ in the mid point of $LM$, this mid point has same distance $\sqrt {ab}$ to $L,M,E$.

The two mid points and $E$ are on the perpendicular in $E$ on $SBA'$.

Proof: Consider the trapezium $A'BJK$, and the mid points Let $\Omega,T$ of $A'B$, $KJ$.

It may help to take a copy of this trapezium, and to paste it in reflection w.r.t. $\Omega$, this reflection is denoted by a star, so we have in the picture a rectangle $JKJ^*K^*$ centered in $\Omega$. We start the proof. Project $B$ on $A'K$ in $P$.

Consider the triangles $\Delta A'BP$ and $\Delta \Omega TE$ with $\hat \Omega =\hat{A'}$ and the proportionality of adjacent sides, $$ \frac{\Omega E}{\Omega T} = \frac{(a+b)/2}{(a-b)/2} = \frac{(a+b)}{(a-b)} = \frac{A'B}{A'P}\ . $$ They are similar, so $\hat E=\hat P=90^\circ$, and $TE=BP/2=KJ/2=\sqrt{ab}=TK=TJ$.

$\square$

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Lemma B: The points $J,L,E'$ from Figure $1$ are collinear.

Proof: Let $R$ be the intersection of $JL$ with $SBEA'$. Let $O_1$ be the projection of $O$ on this line. By symmetry it is enough to show $O_1E=O_1R$. So we compute: $$ \begin{aligned} BR &=\frac{BJ^2}{BS}=\frac{b^2}{b(a+b)/(a-b)}=b\cdot\frac{a-b}{a+b} \\ O_1B &= \frac{BO^2}{BA'}=\frac{b^2}{a+b}=b\cdot\frac{b}{a+b} \\ O_1R &= O_1B+BR=b\cdot\frac{(a-b)+b}{a+b}=b\cdot\frac{a}{a+b} \\ O_1E &= BE-BO_1 =b-b\cdot\frac{b}{a+b} =b\cdot\frac{a}{a+b} \end{aligned} $$ so $O_1R=O_1E$.

$\square$

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The height of the box:

Which is the distance $h$ from $O$ to the tangent $SJK$? Its double $H:=2h$ is the height of our box.

Recall that $O$ is one intersection point of the green circle $\odot(B)$ with the yellow circle $\odot(\Omega)$. We denote by $\gamma,\eta$ the angles $\gamma=\widehat{A'SK}=\widehat{PBA'}$, and $\eta=\widehat{OA'B}$. Then $\widehat{OBK^*}=\eta-\gamma$.

The distance $h$ from $O$ to the tangent $SJK$ is then $JB=b$ plus the projection of $OB=b$ on the side $BK^*$, which is $b\cos(\eta-\gamma)$.

If we really want, an algebraic formula for $h$ can be written, $$ \begin{aligned} h &=b+ b\cos(\eta-\gamma)=b+b\cos\eta\cos \gamma+b\sin\eta\sin\gamma \\ &= b +b\cdot \frac e{a+b}\cdot\frac{2\sqrt{ab}}{a+b} +b\cdot \frac b{a+b}\cdot\frac{a-b}{a+b} \ , \end{aligned} $$ which can be simplified, but a geometric view may want to work with $\eta,\gamma$.

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The width of the box:

Which is the width $W=2w$ of the box? For this we draw a parallel to $SJK$ through $A'$. Let $Q$ be its far most intersection with the circle $\mathcal A'$, seen from $O$.

Then the half-width $w$ is the projection of $OQ$ on the direction $SJK$, $$ \begin{aligned} w &= QA'+PB-OB\sin(\eta-\gamma) = a + 2\sqrt{ab} - b\sin(\eta-\gamma)\ ,\text{ so}\\ w-a &= 2\sqrt{ab}\cdot\frac{e^2}{(a+b)^2} + e\cdot\frac{b(a-b)}{(a+b)^2}\ . \end{aligned} $$ and again, an algebraic formula in $a,b$ can be given.

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The corner tangencies:

We come closer to the fully filled box, the question is when a further circle of radius $c$ in OP notation may fit in the corner of the box, so that it is tangent to the two box sides at the corner, and to both a green circle and a red circle.

We are in the following situation:

The equations extracted from the picture are: $$ \left\{ \begin{aligned} (a-c)^2 +(t-c)^2 &= (a+c)^2\\ (b-c)^2 +(s-c)^2 &= (b+c)^2\\ (a-s)^2 + (b-t)^2 &= (a+b)^2 \end{aligned} \right. $$ and in our case the lengths of $s,t$ are: $$ \left\{ \begin{aligned} s &=w-b\sin(\eta-\gamma)\\ t &=h-a\ . \end{aligned} \right. $$ One can simplify to some extent, but again, algebraic computations of a higher degree are involved.

Answer 5

I keep the same notations as in OP solution and I will propose another one which doesn't require polynomial factorization .
First , we can calculate the width of the rectangle in two ways by orthogonal projection of the red line , and we therefore obtain :

Answer 6

I don't have a geometrical solution.

I think your solution is nice, and using your idea, I'll show another (maybe simpler) way to get $f(x)=0$.

Also, I'll show some results which are similar to the bonus result you wrote.

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As you wrote, we have $z=1-x$ and

$$\sqrt{(y^2+z^2)^2-(y^2-z^2)^2}=x^2-z^2+\sqrt{(x^2+y^2)^2-(1-x^2-y^2)^2}$$ i.e. $$\sqrt{2x^2+2y^2-1}=2y-2xy+1-2x\tag3$$

from which you correctly got $$y=\frac{\left(\frac{1}{\sqrt2}-1\right)(x-1)}{x-1+\frac{1}{\sqrt2}}\tag{1}$$

Next, as you wrote, we have

$$\sqrt{(x^2+y^2)^2-(1-x^2-y^2)^2}+\sqrt{(2y^2)^2-(1-2y^2)^2}=\sqrt{(x^2+y^2)^2-(x^2-y^2)^2}$$

i.e.

$$\sqrt{2x^2+2y^2-1}=2xy-\sqrt{4y^2-1}\tag4$$

Now, we can use $(3)(4)$ to have $$2y-2xy+1-2x=2xy-\sqrt{4y^2-1}$$ i.e. $$\sqrt{(2y+1)(2y-1)}=2x(2y+1)-(2y+1)$$ Dividing the both sides by $\sqrt{2y+1}$ gives $$\sqrt{2y-1}=(2x-1)\sqrt{2y+1}$$ Squaring the both sides, we get $$y=\frac{2x^2-2x+1}{4x(1-x)}\tag5$$

From $(1)(5)$, we obtain $f(x)=0$.

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You wrote an elegant result when the height of the rectangle is $2$.

Letting $h$ be square root of height of the rectangle, we have $$\begin{cases}z=h-x \\\\y=\dfrac{(\frac{1}{\sqrt 2}-1)(x-h)h}{x-(1-\frac{1}{\sqrt 2})h} \\\\(2-2\sqrt 2) x^3+ (3 \sqrt 2 - 4) h x^2+ (1 - \sqrt 2 )h^2 x+(1 - \frac{1}{\sqrt 2}) h^3 =0\end{cases}$$ from which I got $$\color{red}{x^2+y^2=\frac{4}{h^4}x^2y^2z^2+\frac{h^2}{2}}$$ and $$\color{red}{z^2x^2+\frac{ h^2}{2}z^2+\frac{h^4}{4}=\frac{ h^2}{2}x^2+h^2y^2}$$

Let $a=x^2,b=y^2$ and $c=z^2$ as you did.

• For $h=\sqrt 2$, we get your result $$a+b=abc+1$$ and $$ca+c+1=a+2b$$

• For $h=2$, we get $$4a+4b=abc+8$$ and $$ca+2c+4=2a+4b$$