Inscribed Rhombus

Let the radius of each red circle be $1$ and let the radius of the yellow circle be $r$ , so that the ratio of the radius of the yellow circle to the radius of a red circle is $r$ , and label the diagram as follows:

$\triangle HOI \cong \triangle JOI$ by HL congruency, so $\angle HOI = \angle JOI$ . Let $\theta = \angle HOI = \angle JOI$ .

Since the diagonals of a rhombus are perpendicular, $\angle DOC = 90°$ , and $\angle KOH = \angle DOC - \angle HOI - \angle JOI = 90° - 2\theta$ , and from the angle sum of $\triangle KHO$ makes $\angle KHO = 2\theta$ .

Since $OH = OL + LH = r + 1$ and $OI = OF - IF = OF - HE = r - 1$ , from $\triangle HOI$ we have $\cos \angle HOI = \cos \theta = \cfrac{r - 1}{r + 1}$ .

Also, from $\triangle KHO$ we have $\cos \angle KHO = \cos 2\theta = \cfrac{1}{r + 1}$ .

Substituting $\cos \theta = \cfrac{r - 1}{r + 1}$ and $\cos 2\theta = \cfrac{1}{r + 1}$ into the double angle identity $\cos 2\theta = 2 \cos^2 \theta - 1$ gives $\cfrac{1}{r + 1} = 2\bigg(\cfrac{r - 1}{r + 1}\bigg)^2 - 1$ , which rearranges to $r(r - 7) = 0$ , and solves to $r = \boxed{7}$ for $r > 0$ .

Let radius of the yellow circle be $1$ and its center $O$ ; the radius of the red circles be $r$ and their centers be $P$ , $Q$ , and $R$ ; and $\angle D = 4 \theta$ . Then $\angle C = 180^\circ - 4\theta$ . Let $EG$ and $FH$ through $O$ be perpendicular to the sides of the $ABCD$ as shown. Then $OE=OF=OG=OH=1$ .

From the figure:

$\begin{aligned} CT+TG & = CG \\ TR \cdot \cot \frac C2 + TG & = OG \cdot \cot \frac C2 \\ r \cot (90^\circ - 2\theta) + SR & = 1 \cdot \cot (90^\circ - 2\theta) \\ r \tan 2\theta + \sqrt{OR^2-OS^2} & = \tan 2\theta \\ r \tan 2\theta + \sqrt{(1+r)^2-(1-r)^2} & = \tan 2\theta \\ r \tan 2\theta + 2\sqrt r & = \tan 2 \theta \\ r \tan 2\theta + 2\sqrt r - \tan 2 \theta & = 0 \\ \implies \sqrt r & = \frac {\sqrt{4+4 \tan^2 2\theta}-2}{2\tan 2\theta} \\ &= \frac {\sec 2 \theta -1}{\tan 2\theta} \\ & = \frac {1-\cos 2 \theta}{\sin 2\theta} \\ & = \frac {2\sin^2 \theta}{2 \sin \theta \cos \theta} \\ \implies \sqrt r & = \tan \theta \quad \pink{\rm Beautiful!} \end{aligned}$

Similarly,

$\begin{aligned} DU+UG & = DG \\ DU + QS & = DG \\ UQ \cdot \cot \theta + \sqrt{OQ^2-OS^2} & = OG \cdot \cot 2\theta \\ r \cot \theta + 2 \sqrt r & = \cot 2\theta & \small \blue{\text{Since }\sqrt r = \tan \theta} \\ \frac r{\sqrt r} + 2 \sqrt r & = \frac {1-r}{2\sqrt r} \\ 6r & = 1 - r \\ \implies r & = \frac 17 \quad \pink{\rm Beautiful!} \end{aligned}$

Therefore the ratio of the radius of the yellow circle to the radius of the red circle is $\dfrac 1{\frac 17} = \boxed 7$ .