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+√b+√c12 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)

Solution

## 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=√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 Since x>0, we have x=√2+√64=√18+√5412 Hence, WX=1+√18+√5412  We are told that WX=a+√b+√c12 where a,b,c are unique integers. This gives a+b+c=1+18+54=73.

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