More Circles!

Geometry Level 3

All three circles above are tangent to each other and tangent to the horizontal line. If the radius of the blue circle is $R_{1}$ and the radius of the green circle is $R_{2}$ , where $R_{1} > R_{2}$ , and the radius $x$ of the pink circle is $x = (\sqrt{R_{1}} - \sqrt{R_{2}})^2$ and $\phi$ is the golden ratio, then $\dfrac{R_{1}}{R_{2}} =$

\phi

\pi^2

\pi

\phi^2

1 solution

Rocco Dalto
May 25, 2020

Using the diagram above:

$l^2 = (R_{1} + x)^2 - (R_{1} - x)^2 = 4R_{1}x \implies l = 2\sqrt{R_{1}}\sqrt{x}$

$m^2 = (R_{2} + x)^2 - (R_{2} - x)^2 = 4R_{2}x \implies m = 2\sqrt{R_{2}}\sqrt{x}$

$p^2 = (R_{1} + R_{2})^2 - (R_{1} - R_{2})^2 \implies p = 2\sqrt{R_{1}R_{2}}$

$p = l + m \implies$

$\sqrt{R_{1}R_{2}} = (\sqrt{R_{1}} + \sqrt{R_{2}})\sqrt{x} \implies$

$x = \dfrac{R_{1}R_{2}}{(\sqrt{R_{1}} + \sqrt{R_{2}})^2} = (\sqrt{R_{1}} - \sqrt{R_{2}})^2 \implies$ $R_{1}R_{2} = (R_{1} - R_{2})^2 = R_{1}^2 - 2R_{2}R_{1} + R_{2}^2$

$\implies R_{1}^2 - 3R_{2}R_{1} + R_{2}^2 = 0 \implies R_{1} = (\dfrac{3 \pm \sqrt{5}}{2})R_{2}$

$R_{1} > R_{2} \therefore$ we drop $R_{1} = (\dfrac{3 - \sqrt{5}}{2})R_{2} \implies$

$\dfrac{R_{1}}{R_{2}} = \dfrac{3 + \sqrt{5}}{2} = 1 + \phi = \phi^2$ , where $\phi$ is the golden ratio.

More Circles!

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