Submitted by yonatan zilpa on
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 $BC$.
Proof:
Using the parallel axiom we can pass through $B$ a line that is parallel to $\overleftrightarrow{AC}$.
$\Rightarrow$
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 $BC$.
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 $H$.
$\Rightarrow$
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 \[ \displaystyle \left(\frac{k}{\left|1-k^2\right|}\right)\cdot \overline{BC}. \]- Log in to post comments
Comments
spiderTom replied on Permalink
DIABLO IMMORTAL Eternals Generator 2022
vq60 replied onSat, 01/15/2022 - 02:50 Permalink
Hot galleries, thousands new daily.
spiderTom replied onSat, 01/15/2022 - 02:50 Permalink
DIABLO IMMORTAL Hack No Offer 2022
spiderTom replied onSat, 01/15/2022 - 03:28 Permalink
DIABLO IMMORTAL free game download
spiderTom replied onSat, 01/15/2022 - 04:07 Permalink
DIABLO IMMORTAL Eternal Orbs 2022
spiderTom replied onSat, 01/15/2022 - 04:44 Permalink
DIABLO IMMORTAL gold free
spiderTom replied onSat, 01/15/2022 - 05:22 Permalink
free DIABLO IMMORTAL gold
spiderTom replied onSat, 01/15/2022 - 06:00 Permalink
DIABLO IMMORTAL Gold 2022
opaltm4 replied onSat, 01/15/2022 - 06:34 Permalink
Sexy photo galleries, daily updated pics
spiderTom replied onSat, 01/15/2022 - 06:38 Permalink
How To Get DIABLO IMMORTAL Gold For Free? 2022
spiderTom replied onSat, 01/15/2022 - 07:15 Permalink
How To Get DIABLO IMMORTAL Eternal Orbs Free 2022
Jannakank replied onSat, 01/15/2022 - 07:36 Permalink
Para comprar generico Cardarone 200 mg de forma segura
spiderTom replied onSat, 01/15/2022 - 08:00 Permalink
DIABLO IMMORTAL Free Hack Updated 2022
pa60 replied onSat, 01/15/2022 - 08:11 Permalink
Hot new pictures each day
spiderTom replied onSat, 01/15/2022 - 08:53 Permalink
how to get DIABLO IMMORTAL Eternal Orbs free
ShanaRex replied onSat, 01/15/2022 - 09:17 Permalink
Tetracycline 500 mg donde comprar sin receta en farmacia online
spiderTom replied onSat, 01/15/2022 - 09:40 Permalink
DIABLO IMMORTAL hack 2022
vq60 replied onSat, 01/15/2022 - 10:03 Permalink
enjoy new website
spiderTom replied onSat, 01/15/2022 - 10:12 Permalink
DIABLO IMMORTAL honor hack
Jannakank replied onSat, 01/15/2022 - 10:44 Permalink
Spedra gГјnstig bestellen online Schweiz
spiderTom replied onSat, 01/15/2022 - 10:47 Permalink
DIABLO IMMORTAL eternals generator
neldahn4 replied onSat, 01/15/2022 - 10:58 Permalink
enjoy new website
spiderTom replied onSat, 01/15/2022 - 11:22 Permalink
DIABLO IMMORTAL hack no survey
Jannakank replied onSat, 01/15/2022 - 11:29 Permalink
Billig Isoptin Sr in online apotheke ohne rezept bestellen
Jannakank replied onSat, 01/15/2022 - 11:46 Permalink
Amlodipine Norvasc bestellen online
spiderTom replied onSat, 01/15/2022 - 11:56 Permalink
DIABLO IMMORTAL gold free
madgeck1 replied onSat, 01/15/2022 - 11:58 Permalink
Sexy teen photo galleries
spiderTom replied onSat, 01/15/2022 - 12:32 Permalink
DIABLO IMMORTAL Hack Android 2022
spiderTom replied onSat, 01/15/2022 - 13:05 Permalink
DIABLO IMMORTAL Hack 2022
deannsa3 replied onSat, 01/15/2022 - 13:20 Permalink
Hot galleries, daily updated collections
spiderTom replied onSat, 01/15/2022 - 13:39 Permalink
DIABLO IMMORTAL hack ios
spiderTom replied onSat, 01/15/2022 - 13:58 Permalink
DIABLO IMMORTAL Hack No Survey 2022
spiderTom replied onSat, 01/15/2022 - 14:32 Permalink
DIABLO IMMORTAL Eternal Orbs 2022
spiderTom replied onSat, 01/15/2022 - 15:05 Permalink
DIABLO IMMORTAL hack no offer
spiderTom replied onSat, 01/15/2022 - 15:43 Permalink
How To Get DIABLO IMMORTAL Eternal Orbs Free 2022
spiderTom replied onSat, 01/15/2022 - 16:17 Permalink
DIABLO IMMORTAL gold generator
ShanaRex replied onSat, 01/15/2022 - 16:57 Permalink
Megalis Tadalafil in online apotheke bestellen Schweiz
spiderTom replied onSat, 01/15/2022 - 17:06 Permalink
DIABLO IMMORTAL Platinum Hack 2022
spiderTom replied onSat, 01/15/2022 - 17:57 Permalink
DIABLO IMMORTAL gold free
ShanaRex replied onSat, 01/15/2022 - 18:22 Permalink
Donde para ordenar Geodon 20 mg sin receta en línea
Jannakank replied onSat, 01/15/2022 - 18:38 Permalink
Cómo realizar un pedido Manforce sin receta online Panamá
spiderTom replied onSat, 01/15/2022 - 18:51 Permalink
DIABLO IMMORTAL hack no survey
ShanaRex replied onSat, 01/15/2022 - 19:09 Permalink
Farmacia online donde comprar Diltiazem 180 mg entrega rápida
spiderTom replied onSat, 01/15/2022 - 19:44 Permalink
DIABLO IMMORTAL eternals hack
spiderTom replied onSat, 01/15/2022 - 20:36 Permalink
DIABLO IMMORTAL hack no offer
spiderTom replied onSat, 01/15/2022 - 21:29 Permalink
DIABLO IMMORTAL Gold 2022
ShanaRex replied onSat, 01/15/2022 - 22:02 Permalink
Wo kann ich Dutasteride sicher kaufen in der Schweiz?
coreyey1 replied onSat, 01/15/2022 - 22:15 Permalink
Enjoy our scandal amateur galleries that looks incredibly dirty
spiderTom replied onSat, 01/15/2022 - 22:18 Permalink
DIABLO IMMORTAL Free Hack Ios 2022
spiderTom replied onSat, 01/15/2022 - 23:07 Permalink
how to get DIABLO IMMORTAL gold for free?
Pages