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The circle of Apollonius

How to Construct an Apollonian Circle?

Proving Equivalency

Let $B$ and $C$ be two different fixed points, and let $A$ be any point such that \begin{equation} \tag{1} \label{eq1} \frac{\overline{AB}}{\overline{AC}}=k, \end{equation} where $k$ is any positive constant, different from one. We start with the following problem

Problem 1

Let $M$ be any point lying on the segment $BC$.

Prove:     Segment $MA$ bisects angle $\angle BAC$     if     $\frac{\overline{MB}}{\overline{MC}} = k$

Proof:

Using the parallel axiom we can pass through $B$ a line that is parallel to $\overleftrightarrow{AC}$.

Since $\angle BMD \cong \angle CMA$ (vertical opposite angles) and $\angle MDB \cong \angle MAC$ (alternating angles between $\overleftrightarrow{DB} \parallel \overleftrightarrow{AC}$), we get \begin{equation} \tag{2} \triangle MDB \sim \triangle MAC \end{equation}

$\Rightarrow$

Since $\frac{\overline{MB}}{\overline{MC}} =k $ (given), we can use similar property (2) to get $\frac{\overline{BD}}{\overline{CA}}=k$, using equality (1) we can replace $\overline{CA}$ with $\frac{1}{k}\overline{BA}$ to get $\overline{BD}=\overline{BA}$. Thus $\angle BAM \cong \angle MDB$ (base angles in isosceles triangle), but $\angle MDB \cong \angle MAC$ (alternating angles between $\overleftrightarrow{DB} \parallel \overleftrightarrow{AC}$), therefor $\angle BAM \cong \angle MAC$ and as a result of this $AM$ bisects $\angle BAC$.

$\Leftarrow$

Since $MA$ bisects $\angle BAC$ (given), we get $\angle BAM \cong \angle MAC$. In addition $\angle MDB \cong \angle MAC$ (alternating between parallels), thus $\angle BAM \cong \angle MDB$ and this implies that $\overline{BD}=\overline{AB}$. We can thus replace $\overline{AB}$ in equation (1) with $\overline{BD}$ and get $\frac{\overline{BD}}{\overline{AC}}=k$. From the last equality and similarity property (2) we get $\frac{\overline{MB}}{\overline{MC}}=k$. $■$

The result of problem (1) suggesting that the bisector of angle $\angle BAC$ is itersecting with segment $BC$ at a fixed point $M$. It doesn't matter whether we move point $A$ or not, as long as equality (1) preserved (for the point $A$), the intersection point $M$ (of $BC$ and the bisector of $\angle BAC$) is not going to move. We can now bisect the complementary angle of $\angle BAC$ to get the diameter of our Apollonian circle. However we are not done yet, we still don't know whether the locus of all points $A$, satisfying equation (1), form more than one circle.

Problem 2

Let $L$ be any point lying on the line $\overleftrightarrow{BC}$ and outside of the segment $BC$.

Prove:     Segment $LA$ bisects the supplementary angle of $\angle BAC$     if     \begin{equation} \tag{3} \frac{\overline{LB}}{\overline{LC}} = k \end{equation}

Proof:

Using the parallel axiom we can pass through $L$ a line that is parallel to $\overleftrightarrow{AB}$ and intersect with the ray $\overleftrightarrow{CA}$ at a point $H$.

Since $\angle LHC \cong \angle BAC$ and $\angle CLH \cong \angle CBA$ (alternating angles between $\overleftrightarrow{LH} \parallel \overleftrightarrow{AB}$), we get \begin{equation} \tag{4} \triangle ACB \sim \triangle HCL \end{equation}

$\Rightarrow$

From equality (3) and similarity property (4) we get \[ k=\frac{\overline{LB}}{\overline{LC}}=\frac{\overline{LC}-\overline{BC}}{\overline{LC}}=1-\frac{\overline{BC}}{\overline{LC}}, \] thus $\frac{\overline{BC}}{\overline{LC}}=1-k$. We can use this equality and similarity property (4) to get \[ \frac{\overline{AB}}{\overline{HL}}=\frac{\overline{BC}}{\overline{LC}}=\frac{\overline{AC}}{\overline{HC}} =1-k. \] Using these chain of equalities and equalities (1) and (3) we get \[ \begin{array}{lll} \frac{\overline{HA}}{\overline{HL}} &= \frac{\overline{HC}}{\overline{HL}}-\frac{\overline{AC}}{\overline{HL}} \\ &= \frac{(1-k)^{-1}\overline{AC}}{\overline{HL}}-\frac{k^{-1}\overline{AB}}{\overline{HL}} \\ &= (1-k)^{-1}k^{-1}\cdot \frac{\overline{AB}}{\overline{HL}}-k^{-1}\cdot \frac{\overline{AB}}{\overline{HL}} \\ &= (1-k)^{-1}k^{-1}\cdot (1-k)-k^{-1}(1-k) \\ &= k^{-1}\big(1-(1-k)\big) \\ &= 1. \end{array} \] Thus, $\overline{HA}=\overline{HL}$ and this implies that $\angle CLH \cong \angle HAL$ (the base angles in isosceles triangle are congruent). But, $\angle LAB \cong \angle CLH$ (alternating angles between $\overleftrightarrow{LH} \parallel \overleftrightarrow{AB}$), therefor $\angle LAB\cong HAL$, and indeed, segment $AL$ bisects $\angle HAB$.

$\Leftarrow$

Since $\angle LAB\cong \angle ALH$ (alternate angles between $\overleftrightarrow{HL}\parallel\overleftrightarrow{AB}$), we get $\overline{HL}=\overline{HA}$. Let $u$ be the proportionality constant of property (4), then \[ \begin{array}{lll} u^{-1} &=\frac{\overline{HL}}{\overline{AB}} =\frac{\overline{HA}}{\overline{AB}} \\ &=\frac{\overline{HC}}{k\cdot\overline{AC}}- \frac{\overline{AC}}{k\cdot\overline{AC}} \\ &= \frac{1}{k}\left(\frac{\overline{HC}}{\overline{AC}}-1\right) \\ &= \frac{1}{k}\left(u^{-1}-1\right). \end{array} \] Thus $u^{-1}=\frac{1}{k}\left(u^{-1}-1\right)$ which give us $u^{-1}=\frac{1}{1-k}$ or $u=1-k$. We may now use similarity property (4) to get \[ \begin{array}{ll} \frac{\overline{LB}}{\overline{LC}} = \frac{\overline{LC}}{\overline{LC}} - \frac{\overline{BC}}{\overline{LC}} &= 1-\frac{\overline{BC}}{\overline{LC}} \\ &= 1-u =1-(1-k)= k. \quad ■ \end{array} \] From the last equality we get $1-\frac{\overline{BC}}{\overline{LC}}=k$, thus \begin{equation} \tag{5} \overline{LC}=\frac{\overline{BC}}{1-k}. \end{equation} Using similarity (2) we get $\frac{\overline{BM}}{\overline{MC}}=k$ or $\overline{BM}=\overline{MC}\cdot k$, in addition, $\overline{BC}=\overline{BM}+\overline{MC}$, thus $\overline{BC}=\overline{MC}\cdot k+\overline{MC}$ or \begin{equation} \tag{6} \overline{MC}=\frac{\overline{BC}}{1+k} \end{equation} From equalities (5) and (6) we get \[ \begin{array}{ll} \overline{ML} &=\overline{LC}-\overline{MC} \\\\ &=\frac{\overline{BC}}{1-k}-\frac{\overline{BC}}{1+k} \\\\ &=\left(\frac{2k}{1-k^2}\right)\cdot \overline{BC}. \end{array}\] The length $\overline{ML}$ is the diameter of our Apollonian circle, thus the radius of our Apollonian circle is equal to \[ \left(\frac{k}{1-k^2}\right)\cdot \overline{BC}. \] Notice that $k$ was chosen as a positive real number that is less than one. In the case where $k$ is greater than one the radius is equal to $ \left(\frac{k}{k^2-1}\right)\cdot \overline{BC}$. In any case the length of the radius of an Apollonian circle with positive proportion $k\not = 1$ is equal to \[ \displaystyle \left(\frac{k}{\left|1-k^2\right|}\right)\cdot \overline{BC}. \]