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Here is a geometry problem that is driving me nuts! I feel like I am missing something (the author glosses over proving the part I am having trouble with - really makes me feel dumb), and the problem is very easy to state.

Problem: Start with two small central circles of unit diameter. Then find the radius $R$ of the two circles on their left and right. The requirement is that there exists a pair of congruent circles (dotted) that are simultaneously tangent to all the other circles.

The problem also states that repeated use of the Pythagorean theorem reveals that $R_1$ is the golden ratio. I have come up with three equations (one of them is not from the Pythagorean theorem) in three unknowns:

  1. $\quad 2 \cdot R_1+1 = R_2 + h$
  2. $\quad (R_1+1)^2 + h^2 = (R_1+R_2)^2$
  3. $\quad (1/2)^2 + h^2 = (R_2+1/2)^2$

A linear system of three equations in three unknowns would be easy to solve, but this isn’t that unfortunately. Any help would be greatly appreciated.

Diagram of setup

dxiv
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Phil
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1 Answers1

3

The equations are correct, and all that's missing is the last step to solve them for $R_1$. It is not a linear system, but similar techniques of elimination work in this case, too.

  1. $\quad 2 R_1+1 = R_2 + h$
  2. $\quad (R_1+1)^2 + h^2 = (R_1+R_2)^2$
  3. $\quad (1/2)^2 + h^2 = (R_2+1/2)^2$

Eliminating $h$ between 1. and 3. by rearranging 1. and squaring then adding the equations:

$$ \require{cancel} \begin{align} (2 R_1+1-R_2)^2 + 1/4 + \cancel{h^2} &= \cancel{h^2} + (R_2+1/2)^2 \\ (2R_1+1)^2 - 2(2R_1+1)R_2 + \bcancel{R_2^2} + \xcancel{1/4} &= \bcancel{R_2^2} +R_2 + \xcancel{1/4} \\ (2R_1+1)^2 &= (4R_1+3) R_2 \tag{I} \end{align} $$

Eliminating $h$ between 2. and 3. by subtracting the equations:

$$ \begin{align} (R_1+1)^2 + \cancel{h^2} - 1/4 - \cancel{h^2} &= (R_1+R_2)^2 - (R_2+1/2)^2 \\ \cancel{R_1^2} + 2 R_1 + 1 - \xcancel{1/4} &= \cancel{R_1^2} + 2 R_1 R_2 + \bcancel{R_2^2} - \bcancel{R_2^2} - R_2 - \xcancel{1/4} \\ 2R_1 + 1 &= (2R_1-1)R_2\tag{II} \end{align} $$

Eliminating $R_2$ between $(I)$ and $(II)$ by cross-multiplying:

$$ \begin{align} (2R_1+1)^\bcancel{2}(2R_1-1)\cancel{R_2} &= \bcancel{(2R_1+1)}(4R_1+3)\cancel{R_2} \\ 4 R_1^2 - 4 R_1 - 4 &= 0 \tag{III} \end{align} $$

The latter is indeed the equation that defines the golden ratio, so $\,R_1 = \varphi\,$.

dxiv
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