# 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!

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This picture is not clear, nor is what is being asked. – Ron Gordon Mar 22 '13 at 19:27
why is it not whats beeing asked? i try to upload a clearer version – Sophia Mar 22 '13 at 19:30
I meant that I have no idea what you are asking us to do here. – Ron Gordon Mar 22 '13 at 19:31
its suppsed to find out r (radius) in relation to a (the length of the side of the square). something like r= g*a + k+a – Sophia Mar 22 '13 at 19:33
There is no relationship between the radius of the circle and the side of the square, except for the obvious fact side is $\ge 2a$. Presumably a sensible question is being asked. Maybe the centres of the circles are supposed to be collinear. If so, the radii are related. But the problem doesn't say the centres are collinear, and there is no drawn line that hints at it. Perhaps you could ask the original question, exactly as it was asked, with original picture, and in German. – André Nicolas Mar 22 '13 at 19:57

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 $r_1=\frac 13 r_3$.

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Thank you, you made my day :-) – Sophia Mar 23 '13 at 9:24
just one short question, because I can´t find on google. qhats homothetic? – Sophia Mar 23 '13 at 9:34
does is mean, that the cathetus are that size, 1:2:3? that would make sense to me. Thank you ;-) – Sophia Mar 23 '13 at 10:28
It means that the circles can all be superimposed with each other by a transformation which leaves $C$ fixed and moves any other point $P$ along the line $PC$, shrinking or expanding the distance $PC$ by a fixed ratio. In this case, it implies that the circles are in the ratio $1:2:3$, and that the right triangles circumscribed about them are similar in the same ratio, $1:2:3$. – David Moews Mar 23 '13 at 20:41
oki thank you for help! – Sophia Mar 24 '13 at 12:16