# Tangential circles 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.

• What led you to assume the triangles were 30, 60, 90. I'm pretty sure they aren't and that is where your problem lies. Jun 3, 2020 at 15:34
• "The length is larger than the width by a factor of two."? what does that mean? what length and width?
– Oray
Jun 3, 2020 at 15:39
• @Oray it means the length & width of the full figure with all 9 circles. Jun 3, 2020 at 15:45
• I assumed that they are 30,60,90 because one of the angles is 90 and the hypotenuse is twice larger than one of the legs (2005 and 4010). Maybe this assumption is wrong. Jun 3, 2020 at 15:48
• @Displaymaths Yeah I'm not sure where you got the assumption that hypotenuse is twice larger Jun 3, 2020 at 15:57

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.

• You’re correct! Thank you for solving it. Jun 3, 2020 at 16:38
• @Displaymaths well it is pure math/geometry, not a puzzle though... good luck :)
– Oray
Jun 3, 2020 at 16:39

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 $$b=a+\frac{c}{2}=\frac{4c}{5}$$.