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enter image description here

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:enter image description here

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:

enter image description here

And I didn’t get the right answer.

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    $\begingroup$ 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. $\endgroup$ – Ankit Jun 3 at 15:34
  • $\begingroup$ "The length is larger than the width by a factor of two."? what does that mean? what length and width? $\endgroup$ – Oray Jun 3 at 15:39
  • $\begingroup$ @Oray it means the length & width of the full figure with all 9 circles. $\endgroup$ – Ankit Jun 3 at 15:45
  • $\begingroup$ 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. $\endgroup$ – Display maths Jun 3 at 15:48
  • $\begingroup$ @Ankit thanks :) $\endgroup$ – Oray Jun 3 at 15:53
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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;

enter image description here

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.

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  • $\begingroup$ You’re correct! Thank you for solving it. $\endgroup$ – Display maths Jun 3 at 16:38
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    $\begingroup$ @Displaymaths well it is pure math/geometry, not a puzzle though... good luck :) $\endgroup$ – Oray Jun 3 at 16:39
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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:enter image description here
$ADC$ 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 $b=a+\frac{c}{2}=\frac{4c}{5}$.

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