Question

Three circles, each of diameter $1$, are drawn each tangential to the others. A square enclosing the three circles is drawn so that two adjacent sides of the square are tangents to one of the circles and the square is as small as possible. The side length of this square is $a+\frac{\sqrt{b}+\sqrt{c}}{12}$ where $a,b,c$ are integers that are unique (except for swapping $b$ and $c.$) Find $a+b+c.$
(correct answer + 5, wrong answer 0)

Answer 1

Let $WXYZ$ be a square that encloses the three circles and is as small as possible.
Let the centres of the three given circles be $A,B,C.$
Then $ABC$ is an equilateral triangle of side length $1$.
We may assume that $A,B,C$ are arranged anticlockwise and that the circle with centre $A$ touches $WX$ and $WZ.$ We may also assume that $WX$ is horizontal.

Note that if neither $YX$ nor $YZ$ touch a circle, then the square can be contracted by moving $Y$ along the diagonal $WY$ towards $W.$ So at least one of $YX$ and $YZ$ must touch a circle and it can't be the circle with centre $A.$
We may assume that $XY$ touches the circle with centre $B.$


If $YZ$ does not touch a circle, then the $3$-circle cluster can be rotated anticlockwise, allowing neither $YX$ nor $YZ$ to touch a circle. So $YZ$ touches the circle with centre $C.$

Let $ADEF$ be the rectangle with sides through $C$ and $B$ parallel to $WX$ and $WZ$ respectively.


Since $AF=WZ-1=WX-1=AD,$
$ADEF$ is a square.
Since $AC=1=AB,$ triangles $AFC$ and $ADB$ are congruent.
So $FC=DB$ and $CE=BE.$
Let $x=AD$.
Since $AB=1$ and triangle $ADB$ is right-angled,
$DB=\sqrt{1-x^2}$

Since triangle $CBE$ isosceles with $BC=1,$ we have $BE=\frac{1}{\sqrt{2}}.$
So, $x=DE=\sqrt{1-x^2}+\frac{1}{\sqrt{2}}$
$\Rightarrow x-\frac{1}{\sqrt{2}}=\sqrt{1-x^2}$
$\Rightarrow 1-x^2={(x-\frac{1}{\sqrt{2}})}^2$
$\Rightarrow 1-x^2=x^2-\sqrt{2}x+\frac{1}{2}$
$\Rightarrow 2x^2-\sqrt{2}x-\frac{1}{2}=0$
$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{2+4}}{4}$

Since $x>0$, we have
$x=\frac{\sqrt{2}+\sqrt{6}}{4}=\frac{\sqrt{18}+\sqrt{54}}{12}$
Hence, $WX=1+\frac{\sqrt{18}+\sqrt{54}}{12}$
We are told that $WX=a+\frac{\sqrt{b}+\sqrt{c}}{12}$
where $a,b,c$ are unique integers.
This gives $a+b+c=1+18+54=73.$