Geometry easy-med

Let the center of the blue circle be the origin $(O(0,0)$ of the $xy$ -plane, the radius of the yellow and purple circles be $r_1$ and $r_2$ respectively. Then the center of the top and bottom green circles are $P(0,1)$ and $P'(0,-1)$ . Due to symmetry, the center of the purple circle is $R(2-r_2,0)$ and that by Pythagorean theorem on $\triangle PRO$ :

$\begin{aligned} (2-r_2)^2 + 1^2 & = (1+r_2)^2 \\ r_2^2 - 4r_2 + 4 + 1 & = r_2^2 + 2r_2+1 \\ \implies r_2 & = \frac 23 \end{aligned}$

Now we note that the point $Q(x,y)$ is the intersection of three circles centered at $O(0,0)$ with radius $2-r_1$ , at $P(0,1)$ with radius $1+r_1$ , and at $R\left(\frac 43, 0\right)$ with radius $\frac 23 + r_1$ . Or in other words, $Q(x,y)$ satisfy the system of equations below:

$\begin{cases} OQ^2: & (x-0)^2 + (y-0)^2 = (2-r_1)^2 & \implies x^2 + y^2 = r_1^2 - 4r_1 + 4 & ...(1) \\ PQ^2: & (x-0)^2 + (y-1)^2 = (1+r_1)^2 & \implies x^2 + y^2 - 2y = r_1^2 + 2r_1 & ...(2) \\ RQ^2: & \left(x-\frac 43\right)^2 + (y-0)^2 = \left(\frac 23 +r_1\right)^2 & \implies x^2 + y^2 - \frac 83x = r_1^2 + \frac 43 r_1 - \frac 43 & ...(3) \end{cases}$

$\begin{cases} (1)-(2): & 2y = - 6r_1 + 4 & \implies y = 2-3r_1 \\ (1)-(3): & \frac 83 x = - \frac {16}3 r_1 + \frac {16}3 & \implies x = 2-2r_1 \end{cases}$

From $(1)$ :

$\begin{aligned} x^2 + y^2 & = r_1^2 - 4r_1 + 4 \\ (2-2r_1)^2 + (2-3r_1)^2 & = r_1^2 - 4r_1 + 4 \\ 13r_1^2 - 20r_1 + 8 & = r_1^2 - 4r_1 + 4 \\ 12r_1^2 - 16r_1 + 4 & = 0 \\ 3r_1^2 - 4r_1 + 1 & = 0 \\ (3r_1-1)(r_1 - 1) & = 0 \\ \implies r_1 & = \frac 13 \approx \boxed{0.333} & \small \color{#3D99F6} \text{Note that }r_1 \text{ cannot be 1.} \end{aligned}$

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$\color{#D61F06}\text{Previous solution: }$ The solution below has assumed that the centers of four circles form a rectangle without justifications.

Let the radii of the yellow circle and purple circle be $r_1$ and $r_2$ respectively. We note that

$\begin{cases} r_1 + r_2 = 1 & ...(1) \\ (1+r_1)^2 + 1^2 = (1+r_2)^2 & ...(2) \end{cases}$

From $(1): r_2 = 1-r_1$ , then

$\begin{aligned} (2): \quad (1+r_1)^2 + 1 & = (2-r_1)^2 \\ r_1^2 + 2r_1+2 & = r_1^2 - 4r_1+4 \\ 6r_1 & = 2 \\ \implies r_1 & = \frac 13 \approx \boxed{0.333} \end{aligned}$