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.
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
We may also assume that WX
Note that if neither YX
touch a circle, then the square can be contracted by moving Y
along the diagonal WY
So at least one of YX
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.
does not touch a circle, then the 3
-circle cluster can be rotated anticlockwise, allowing neither YX
to touch a circle. So YZ
touches the circle with centre C.
be the rectangle with sides through C
parallel to WX
Since AF=WZ−1=WX−1=AD, ADEF
is a square.
and triangle ADB
is right-angled, DB=√1−x2
Since triangle CBE
isosceles with BC=1,
we have BE=1√2.
So, x=DE=√1−x2+1√2 ⇒x−1√2=√1−x2 ⇒1−x2=(x−1√2)2 ⇒1−x2=x2−√2x+12 ⇒2x2−√2x−12=0 ⇒x=√2±√2+44
, we have x=√2+√64=√18+√5412
We are told that WX=a+√b+√c12
are unique integers.
This gives a+b+c=1+18+54=73.