Tangential circles

Question

The following figure has two axes of symmetry which define its width and length. The length (horizontal distance) is twice the width (vertical distance). The largest circle has a radius of 2005 and all the contacts with small circles are tangent. If the six smallest circles have the same radii, then what’s the radius of the medium circle? Swiss competition: https://fsjm.ethz.ch/static/oldwebsite/documents/FI_19e_1.pdf

My attempt: I’ve tried solving it with simple algebra, but I failed and didn’t know where was my mistake. Here’s my attempt:

y:distance between the center of the medium circle and the large circle

x:distance between the center of the large circle and the small circle

r:the radius of the small circle

I then drew two 30,60,90 triangles and set up a system of equations:

And I didn’t get the right answer.

Answer 1

Let's draw our problem and I used the same notations as shown in the question except $x$ since $x=2y$ and it was too obvious;

Simply,

we calculate $|AH|$ from $|AD|$ and $|DH|$ as $\sqrt{4y^2+4yr}$ then we know that $|DI|=|AH|$ since $DI$ is parallel to $|AH|$ and perpendicular to $|AC|$.

then we need to calculate |CI|, so to do that;

we know $|CG|=y$,

and

$FI=2y-r$ since $|DH|=r$ and $|AI|=r$

then we can find $|CI|$ as

$|CI|=|CG|+2r+|FI|=3y+r$

then

$|CI|^2+DI^2=|CD|^2=9y^2+6yr+r^2+4y^2+4yr=9r^2+6yr+y^2$

then we find another relation between y and r;

$y=2r/3$

and we also know that

$2y+2r=2005$

then

we find y as

$401$

Note that I did not get into much detail of how I construct it since it was too obvious for some cases, if you ask, I may add details and drawing is not perfect. To be honest it does not sound like a puzzle to me, it was pure geometry problem without any logic deduction etc.

Answer 2

The radius of the medium circle is

$\frac{4}{5} \cdot 2005 = 1604$

Proof:

Let the circles, in order of increasing size, be $A$, $B$, and $C$ with radii $a$,$b$, and $c$ respectively. From the problem, we have that $c+2b-2a=2c$, so $b=a+\frac{c}{2}$. For now, let $c=2$.
Consider the following diagram:
$ADC$ is a right triangle with hypotenuse $2-a$ and leg $a$, so the other leg $CD$ is $2\sqrt{1-a}$.
The height of $ABC$ is the same as $CD$, so we have $AB^2 = CD^2 + (BC-AD)^2$.
But $AB = a+b=2a+1$, $CD=2\sqrt{1-a}$, and $BC-AD = 4-b-a=3-2a$, so $4a^2+4a+1=4-4a+9-12a+4a^2$. The $4a^2$ cancels, and transposing and dividing by 20 yields $a = \frac{3}{5} = \frac{3c}{10}$ and thus $b=a+\frac{c}{2}=\frac{4c}{5}$.