second try: Find the radius in dependency of a (the side length) I am really hopeless with this task. I have been trying nearly all day now.It´s about finding a description for the three radius in dependency of a. somethin like r=3a for example.. My friend said it works with a formula for radius inside a triangle. ANd I tried to solve with pythagoras which led to: r²= (a²)² -4a³r+6a²r² -4ar²-4ar³+2(r²)² +a² -2ar+r²+2r².
can anyone please very urgent help me? I need it for my exam...Thank you!

 A: From the top picture, it appears that all three circles are meant to be tangent to $AC$.  Also, you can see gaps in the other segments approximately where the segment $AC$ should be.  Perhaps there was a line here that did not reproduce.
Adding this to other information from the picture:


*

*$AA_1D_1D$, $A_1A_2D_2D_1$, and $A_2BCD_2$ are all squares with side length $a$.

*Circle $M_1$, with radius $r_1$, is tangent to $CD_2$, $D_2A_2$, and $AC$.

*Circle $M_2$, with radius $r_2$, is tangent to $A_1D_1$, $D_1D_2$, and $AC$.

*Circle $M_3$, with radius $r_3$, is tangent to $AD$, $DD_1$, and $AC$.


$M_3$ is then the incircle of $\triangle ACD$.  This triangle is right, with legs $AD=a$ and $CD=3a$, so it has hypotenuse $AC=\sqrt{10}a$, area $\frac 32 a^2$, and perimeter $(4+\sqrt{10})a$.  A triangle with area $K$ and perimeter $p$ has inradius $2K/p$ (this can be seen by dissecting the triangle into three triangles by drawing segments from the incenter to the vertices.)  In this case then the inradius of $\triangle ACD$ is
$$
r_3=\frac{2\cdot\frac 32 a^2}{(4+\sqrt{10})a}=\frac{4-\sqrt{10}}{2}a.
$$
Since $M_1$, $M_2$ and $M_3$ are all homothetic with center $C$ in the ratio $1:2:3$, $r_2=\frac23 r_3$ and $r_1=\frac 13 r_3$.
