Prove a geometry question about angles and radii in five collinear circles?

Question

Our teacher give us this question and I worked on it but I couldn't find a way to prove that. Is it possible to help me to prove that?

Thanks.

Answer 1

Assumptions

I'm assuming that the following are true:

• $A$, $B$, $G$, $H$, $I$, $K$ are collinear
• $B$, $G$, $H$, $I$, $K$ are the centres of their respective circles
• $A$, $C$, $D$, $E$, $F$, $J$ are collinear and the line on which they lie is tangent to all the circles.

Definitions

We know these angles:

• $\angle CAB$ = $\angle DAG$ = $\angle EAH$ = $\angle FAI$ = $\angle JAK$. Call this angle $\theta$.
• $\angle ACB$ = $\angle ADG$ = $\angle AEH$ = $\angle AFI$ = $\angle AJK$ = $90^{\circ}$.

And we can define some lengths. Let:

• $R_1$, $R_2$, etc. be the lengths of $CB$, $DG$, etc.
• $H_1$, $H_2$, etc. be the lengths of $AB$, $AG$, etc.

Known Information

• For right angled trianges, we know that $\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypoteneuse}}$, so:

$$\forall n \in \{1, \dots, 5\}, \sin\theta = \frac{R_n}{H_n} \Rightarrow H_n = \frac{R_n}{\sin\theta}$$

• Looking at the diagram, we can also see that:

$$\forall n \in \{1, \dots, 4 \}, H_{n+1} = H_n + (R_n + R_{n+1})$$

Forming a Rule

We can substitute for $H_n$ from the first equation into the second and rearrange:

\begin{align} && \frac{R_{n+1}}{\sin\theta} &= \frac{R_n}{\sin\theta} + R_n + R_{n+1} \\ \Rightarrow && R_{n+1} \left(\frac{1}{\sin\theta} - 1\right) &= R_n \left(\frac{1}{\sin\theta} + 1\right) \\ \Rightarrow && R_{n+1} \left(1 - \sin\theta\right) &= R_n \left(1 + \sin\theta \right) \\ \Rightarrow && R_{n+1} &= R_n \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right) \\ \Rightarrow &&R_{n+k} &= R_n {\left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)}^k \\ \end{align}

The result

This means that:

\begin{align} R_1 \cdot R_5 &= R_1 \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^4\right) \\ &= \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \\ &= R_3 \cdot R_3 \end{align}

Extra results

This rule, coupled with the fact that there's nothing special about $R_1$ (it could just as easily have been called $R_0$ or $R_{1000}$) means that if we take any two collections of integers $A = \{A_1, A_2, \dots\}$ and $B = \{B_1, B_2, \dots\}$ and define $S_A = \sum_{a \in A}a$ and $S_B = \sum_{b \in B}b$ we get:

\begin{align} \prod_{a \in A} R_{a} &= R_0^{|A|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A} \\ &= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} R_0^{|B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_B} \\ &= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} \prod_{b \in B} R_{b} \end{align}

If $|A| \neq |B|$ and $z = \frac{S_A - S_B}{|A|-|B|}$ is an integer, this simplifies to:

$$ \prod_{a \in A} R_{a} = R_z^{|A| - |B|} \prod_{b \in B} R_{b} $$

Or if $|A| = |B|$ and $S_A = S_B$ (i.e. two equal sized collections with equal sums), we end up with:

$$ \prod_{a \in A} R_{a} = \prod_{b \in B} R_{b} $$

and your question is a special case of this where $A = \{1,5\}$ and $B = \{3,3\}$.

Answer 2

Hint: Consider the uniform scaling (i.e., similarity transformation) of the entire plane about point $A$ that takes $K$ to $H$. Argue that this transformation takes $H$ to $B$, $J$ to $E$, and $E$ to $C$. Once you see this, it's clear that $$ \frac{R_5}{R_3} = \frac{R_3}{R_1} $$

If the language of transformations isn't familiar to you, try instead noting that the segments labelled $R_1, R_3, R_5$ are all parallel (being orthogonal to the upper ray), and hence that $AKJ$ is similar to $AHE$, and work from there.

Answer 3

Hint. Start by proving that $R_{i-1} R_{i+1}=R_{i}^2$ by noting that $$\frac{x}{R_i}=\frac{x-R_{i}-R_{i-1}}{R_{i-1}}=\frac{x+R_{i}+R_{i+1}}{R_{i+1}}.$$ where $x$ is distance of the center of the $i$-th circle from $A$.

Answer 4

That is a immediate consequence of Thales Theorem.

$$\frac{BG}{GH}=\frac{CD}{DE} \Rightarrow \frac{R_1+R_2}{R_2+R_3}=\frac{\sqrt{4R_1R_2}}{\sqrt{4R_2R_3}} \Rightarrow R_1R_3=(R_2)^2 (*)$$

We get $CD=\sqrt{4R_1R_2}$ by a Pythagoras Theorem at $BCDG$. The relation $(*)$ means that those radius are in a geometric sequence and so

$$R_1R_5=(R_3)^2$$

Answer 5

Let $c_n$ denote the distance of $A$ to the center of the circle with radius $R_n$.

Since the triangles formed by the radius and $A$ contain the same angles, you have

$R_n = k \cdot c_n$ for some constant $k$.

Now since the neighboring circles intersect in exactly one point on the line connecting the centers and $A$, we can come up with the following recursive formula for $c_n$:

$c_{n+1} = c_n + R_n + R_{n+1} = c_n + k \cdot c_n + k \cdot c_{n+1}$

Solve for $c_{n+1}$ to get

$c_{n+1} = c_n \cdot \frac{k+1}{1-k}$ which is the recursive form of a geometric sequence and hence

$c_{n} = c_1 \cdot \left(\frac{k+1}{1-k}\right)^{n-1}$

Proving the claim now is trivial. (use $R_n = k \cdot c_n$)

Answer 6

By the similarity $R_{i+1}/R_i = \mathrm{const.}$ , hence $R_5/R_3 = R_3/R_1$.