Pappus chain - triangle of tangency points

Question

$A,B,C$ are collinear points. Starting with 3 circles with diameter $AB,AC,BC$, there is a chain of tangent circles, all tangent to one of the two small interior circles and to the large exterior one.

The question is to prove a relation of the internal angles $α,β,γ$ of $\triangle B_nC_nD_n\ (n=1,2,\dots)$ $$\tag0\frac{\cot(α) + \cot(β) - 2}{\cot(γ)}=\frac1n-2$$ This can be rewritten in the side lengths and area $$\tag1 \frac{2n-1}n(|D_nB_n|^2+|D_nC_n|^2)+\frac1n|B_nC_n|^2=8\text{Area}(\triangle B_nC_nD_n)$$

My proof for $n=1$:

Denote $b:=|D_1C_1|,c:=|D_1B_1|,d:=|B_1C_1|$.
Using Heron's formula $16\text{Area}(\triangle B_1C_1D_1)^2=4(b^2c^2+b^2d^2+c^2d^2)-(b^2+c^2+d^2)^2$, equation (1) becomes $$b^2+c^2+d^2=2\sqrt{4(b^2c^2+b^2d^2+c^2d^2)-(b^2+c^2+d^2)^2}$$ squaring and rearranging $$\tag2 16 (b^2c^2+b^2d^2+c^2d^2) = 5 (b^2+c^2+d^2)^2$$ Let $A(0,0),B(r,0),C(1,0),r=AB/AC$, using this post we get \begin{array}{lll} B_1=\left(\frac{r}{{1+(1-r)^2}},\frac{r(1-r)}{{1+(1-r)^2}}\right)& C_1=\left(\frac{r^2}{{1-2r+2r^2}},\frac{r(1-r)}{{1-2r+2r^2}}\right)& D_1=\left(\frac{r(1+r)}{{1+r^2}},\frac{r(1-r)}{{1+r^2}}\right)\\ b^2= \frac{{\left(r - 1\right)}^{2} r^{2}}{{\left(2r^{2} - 2r + 1\right)} {\left(r^{2} + 1\right)}}& c^2= \frac{{\left(r - 1\right)}^{2} r^{2}}{{\left(r^{2} - 2r + 2\right)} {\left(r^{2} + 1\right)}}& d^2= \frac{{\left(r - 1\right)}^{2} r^{2}}{{\left(2r^{2} - 2r + 1\right)} {\left(r^{2} - 2r + 2\right)}} \end{array} plug in (2) and simplify,

I think, for $n>1$, it can be proved using same method.

Is there a method to avoid heavy computation?

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An interesting fact:
For $n=1$ the equation $(0)$ becomes $$\frac{\cot(α) + \cot(β) - 2}{\cot(γ)}=-1\\⇔\cot(α) + \cot(β)+\cot(γ)=2$$ By the formula the Brocard angle of $\triangle B_1C_1D_1$ is $\cot^{-1}(2)$.

Answer 1

Inversive geometry.

We use an inversion centered in $A$, the point of intersection of the two circles of tangency of all circles in the chain. The power of the inversion is the product: $$ K:=AB\cdot AC\ , $$ so that $B$ and $C$ transform in each other. We denote this inversion by a star, so so far $B^*=C$, $C^*=B$.

The red circle with diameter $AB$ is tangent to the ray $ABC$. It transforms into a line, since the center of the inversion $A$ is on it, which passes through $B^*=C$, this line is again orthogonal to the ray $ABC$ in $B^*=C$. (The inversion preserves angles.) In the picture we also use the red color for the transform.

Similarly, the indigo circle with diameter $AC$ is transformed in the indigo line through $C^*=B$ which is orthogonal in this point to $ABC$.

The chain of circles tangent to the two circles transforms into a chain o circles tangent to the two (red and indigo) parallel lines.

There is a simple way to draw the transformed chain inside this strip, see the figure. The transformed tangency points $B_n^*,C_n^*,D_n^*$ can be easily figured out. Let us denote by $2R,2r$ the distances $$ \begin{aligned} 2R &= AB\ ,\\ 2r &= BC\ . \end{aligned} $$ So $AC=2(R+r)$, and the power of the inversion is $K=AB\cdot AC=2R\cdot2(R+r)$ Let us use complex numbers $z$ to describe the locations (affixes) of points $Z$ in the plane, here $Z$ is the one or the other point shown in the figure. For the affix, we use the lower case notatin. The origin of the complex plane is $A=0$ and we arrange so that $B=2R$, $C=2(R+r)$. Then the affixes of $B_n^*,C_n^*,D_n^*$ are respectively: $$ \begin{aligned} b_n^* &= 2R + 2nri\ ,\\ c_n^* &= 2(R + r) + 2nri\ ,\\ d_n^* &= 2R + r + (2n-1)ri\ ,\\ \end{aligned} $$ Our inversion is given algebraically in this complex $z$-plane by $$ z\to z^*=\frac K{\bar z}\ . $$ So we have formulas for the (affixes of the) points $B_n,C_n,D_n$, $n\ge 1$.

Time to switch to algebra to show the OP equality $(1)$.

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We start a computation of all involved terms in the relation to be shown, then explicitly find formulas for the LHS (weighted sum of squared distances) and the RHS (eight times an area). For the RHS we use a formula for the signed area in terms of a determinant, this is the reason for inserting a $\pm $ sign when starting the computation, of course, finally we adjust the sign to have a positive number. $$ \begin{aligned} |D_nB_n|^2 &= |d_n-b_n|^2=\left|\frac K{b_n^*}-\frac K{d_n^*}\right|^2 =\frac {K^2}{|b_n^*|^2\;|d_n^*|^2}\cdot\underbrace{|d_n^*-b_n^*|^2}_{2r^2} \ , \\[2mm] |D_nC_n|^2 &=\frac {K^2}{|c_n^*|^2\;|d_n^*|^2}\cdot\underbrace{|d_n^*-c_n^*|^2}_{2r^2} \ , \\[2mm] |B_nC_n|^2 &=\frac {K^2}{|b_n^*|^2\;|c_n^*|^2}\cdot\underbrace{|c_n^*-b_n^*|^2}_{4r^2} \ , \\[2mm] LHS &= \frac{2n-1}n\Big(|D_nB_n|^2+|D_nC_n|^2\Big)+\frac1n|B_nC_n|^2 \\ &=\frac {2r^2\; K^2}{n\;|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \Big( (2n-1)|b_n^*|^2+(2n-1)|c_n^*|^2+2|d_n^*|^2 \Big) \\ &=\frac {2r^2\; K^2}{{\color{gray}n}\;|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 16{\color{gray}n}\;(R^2+Rr+n^2r^2)\ , \\[2mm] \pm RHS &=\pm 8\text{Area}(\triangle B_nC_nD_n) \\ &=8\cdot\frac i4 \begin{vmatrix} 1 & b_n &\bar b_n\\ 1 & c_n &\bar c_n\\ 1 & d_n &\bar d_n \end{vmatrix} =2i \begin{vmatrix} 1 & \displaystyle \frac K{\bar b_n^*} & \displaystyle \frac K{b_n^*}\\ 1 & \displaystyle \frac K{\bar c_n^*} & \displaystyle \frac K{c_n^*}\\ 1 & \displaystyle \frac K{\bar d_n^*} & \displaystyle \frac K{d_n^*} \end{vmatrix} = 2i \begin{vmatrix} 0 & \displaystyle \frac K{\bar b_n^*} - \frac K{\bar d_n^*} & \displaystyle \frac K{b_n^*} - \frac K{d_n^*}\\ 0 & \displaystyle \frac K{\bar c_n^*} - \frac K{\bar d_n^*} & \displaystyle \frac K{c_n^*} - \frac K{d_n^*}\\ 1 & * & * \end{vmatrix} \\ & = 2i\; K^2 \begin{vmatrix} \displaystyle \frac 1{\bar b_n^*} - \frac 1{\bar d_n^*} & \displaystyle \frac 1{b_n^*} - \frac 1{d_n^*}\\ \displaystyle \frac 1{\bar c_n^*} - \frac 1{\bar d_n^*} & \displaystyle \frac 1{c_n^*} - \frac 1{d_n^*} \end{vmatrix} = 2i\; K^2 \begin{vmatrix} \displaystyle \frac 1{\bar b_n^*\bar d_n^*}(\bar d_n^* - \bar b_n^*) & \displaystyle \frac 1{b_n^*d_n^*}(d_n^* - b_n^*) \\ \displaystyle \frac 1{\bar d_n^*\bar d_n^*}(\bar d_n^* - \bar c_n^*) & \displaystyle \frac 1{c_n^*d_n^*}(d_n^* - c_n^*) \end{vmatrix} \\ & = \frac{2i\; K^2}{|d_n^*|^2} \begin{vmatrix} \displaystyle \frac 1{\bar b_n^*}(r+ir) & \displaystyle \frac 1{b_n^*}(r-ir) \\ \displaystyle \frac 1{\bar c_n^*}(-r+ir) & \displaystyle \frac 1{c_n^*}(-r-ir) \end{vmatrix} = \frac{2ir^2\; K^2}{|d_n^*|^2} \begin{vmatrix} \displaystyle \frac {b_n^*}{|b_n^*|^2}(1+i) & \displaystyle \frac {\bar b_n^*}{|b_n^*|^2}(1-i) \\ \displaystyle \frac {c_n^*}{|c_n^*|^2}(-1+i) & \displaystyle \frac {\bar c_n^*}{|c_n^*|^2}(-1-i) \end{vmatrix} \\ &= \frac{2ir^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \begin{vmatrix} b_n^*(1+i) & \bar b_n^*(1-i) \\ c_n^*(-1+i) & \bar c_n^*(-1-i) \end{vmatrix} \\ &= \frac{2ir^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 2i\operatorname{Im}\Big( b_n^*(1+i) \cdot \bar c_n^*(-1-i) \Big) \\ &= \frac{2ir^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 2i\operatorname{Im}\Big( b_n^*\bar c_n^*(-2i) \Big) \\ &= \frac{2r^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 4\operatorname{Re}\Big( b_n^*\bar c_n^*\Big) \\ &= \frac{2r^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 4\operatorname{Re}\Big( (2R +2nri)(2(R+r)-2nri)\Big) \\ &= \frac{2r^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 4\Big( (2R)\cdot2(R+r) + (2nr)^2\Big) \\ &= \frac{2r^2\; K^2}{|b_n^*|^2\;|c_n^*|^2\;|d_n^*|^2} \cdot 16( R^2 +Rr + n^2r^2) \ . \end{aligned} $$ So the expressions for $LHS$ and $RHS$ coincide.

$\square$

Answer 2

Relabeling $B_n,C_n,D_n$ as $F',E',D'$, they can be obtained by inversion (as explained in @dan_fulea 's answer):

$A,B,C$ are collinear points.
$DEF$ is an isosceles right triangle [i.e. $DE=DF,DE\perp DF$.]
$BCFE$ is a rectangle [i.e. $BE\perp BC,CF\perp BC,|EF|=|BC|$.]
$D',E',F'$ are inversion of $D,E,F$ about a circle centered at $A$.

Changing the radius of inversion circle, $\triangle D'E'F'$ will be scaled by a factor, so equation $(0)$ is unchanged.

Choosing the radius of inversion circle so that it is orthogonal to the circle with diameter $EF$, then $D',E',F'$ are the intersection of $AD,AE,AF$ with the circle with diameter $EF$:

Plugging in $n=\frac{BE}{BC}$, equation $(0)$ becomes $$\tag4\frac{\cot(\angle D'E'F') + \cot(\angle D'F'E') - 2}{\cot(\angle E'D'F')}=\frac{BC}{BE}-2$$ Denote $\overparen{DD'}=\angle DED',\overparen{DE'}=\angle DEE',\overparen{DF'}=\angle DEF'$.

The strategy is to rewrite (4) in only $\overparen{DE'}$ and $\overparen{DF'}$.

To rewrite the right-hand side in $\overparen{DE'}$ and $\overparen{DF'}$: \begin{align} \frac{BC}{BE}&=\frac{AC-AB}{BE}\\&=\frac{AC}{CF}-\frac{AB}{BE}\\&=\cot(\angle CAF)-\cot(\angle BAE)\\&=\cot(\angle EFE')-\cot(\angle FEE') \\&=\cot(\overparen{EF'})-\cot(\overparen{FE'}) \\&=\cot(\frac\pi4-\overparen{DF'})-\cot(\overparen{DE'}-\frac\pi4) \\&=\frac{\cot(\overparen{DF'})+1}{\cot(\overparen{DF'})-1}-\frac{\cot(\overparen{DE'})+1}{1-\cot(\overparen{DE'})} \end{align} so the right-hand side of (4) is $$\tag5 \frac{BC}{BE}-2=\frac2{\cot(\overparen{DF'})-1}+\frac2{\cot(\overparen{DE'})-1} $$ For the left-hand side of (4): By adding arcs: \begin{align} \angle D'E'F'&=\overparen{DF'}+\overparen{DD'}\\ \angle D'F'E'&=\overparen{DE'}-\overparen{DD'}\\ \angle E'D'F'&=\pi-\overparen{DE'}-\overparen{DF'} \end{align} so \begin{align} &\cot(\angle D'E'F') + \cot(\angle D'F'E')\\&=\cot(\overparen{DF'}+\overparen{DD'})+\cot(\overparen{DE'}-\overparen{DD'}) \\&=\frac{\cot(\overparen{DF'}) \cot(\overparen{DD'})-1}{\cot(\overparen{DF'})+\cot(\overparen{DD'})}+\frac{\cot(\overparen{DE'}) \cot(\overparen{DD'})+1}{\cot(\overparen{DD'})-\cot(\overparen{DE'})}\\&=\frac{\left(\cot^2(\overparen{DD'})+1\right)\left(\cot(\overparen{DE'})+\cot(\overparen{DF'})\right)}{\left(\cot(\overparen{DD'})-\cot(\overparen{DE'})\right)\left(\cot(\overparen{DD'})+\cot(\overparen{DF'})\right)}\tag6 \end{align} To rewrite $\overparen{DD'}$ in $\overparen{EF'}$ and $\overparen{FE'}$:
Note that the tangent at $D$ is parallel to $AB$, so $\overparen{DD'}=\angle DAB$, so \begin{align} \cot(\overparen{DD'})&=\cot(\angle DAB)\\ &=\frac{\frac{AB+AC}2}{BE-\frac{BC}{2}}\\ &=\frac{\frac{AB+AC}{BE}}{2-\frac{BC}{BE}}\\ &=\frac{\frac{AB}{BE}+\frac{AC}{CF}}{2-\frac{BC}{BE}}\\ &=\frac{\cot(\overparen{EF'})+\cot(\overparen{FE'})}{2-\cot(\overparen{EF'})+\cot(\overparen{FE'})}\\ &=\frac{\cot(\frac\pi4-\overparen{DF'})+\cot(\overparen{DE'}-\frac\pi4)}{2-\cot(\frac\pi4-\overparen{DF'})+\cot(\overparen{DE'}-\frac\pi4)}\\ &=\frac{\frac{\cot (DF')+1}{\cot (DF')-1}+\frac{\cot (DE')+1}{1-\cot (DE')}}{2-\frac{\cot (DF')+1}{\cot (DF')-1}+\frac{\cot (DE')+1}{1-\cot (DE')}}\\ &=\frac{\cot(\overparen{DE'})-\cot(\overparen{DF'})}{2-\cot(\overparen{DF'})-\cot(\overparen{DE'})}\tag7 \end{align} Plugging (6),(7) to the right-hand side of (4) \begin{align} &\frac{\cot(\angle D'E'F') + \cot(\angle D'F'E')-2}{\cot(\angle E'D'F')} \\&=\frac{\cot(\angle D'E'F') + \cot(\angle D'F'E')-2}{-\frac{\cot(\overparen{DE'}) \cot(\overparen{DF'})-1}{\cot(\overparen{DE'})+\cot(\overparen{DF'})}}\\ &=\frac2{\cot(\overparen{DF'})-1}+\frac2{\cot(\overparen{DE'})-1} \end{align} is same as (5), so we proved (4).

$\square$