Problem:
Equilateral triangle β³ABC is inscribed in circle Ο with radius 18. Circle ΟAβ is tangent to sides AB and AC and is internally tangent to Ο. Circles ΟBβ and ΟCβ are defined analogously. Circles ΟAβ,ΟBβ, and ΟCβ meet in six points-two points for each pair of circles. The three intersection points closest to the vertices of β³ABC are the vertices of a large equilateral triangle in the interior of β³ABC, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of β³ABC. The side length of the smaller equilateral triangle can be written as aββbβ, where a and b are positive integers. Find a+b.
Solution:
Answer (378):
Let O be the center of Ο, and let D,E, and F be points diametrically opposite A,B, and C, respectively, in Ο. Denote the centers of circles ΟAβ,ΟBβ, and ΟCβ by OAβ,OBβ, and OCβ, respectively, and let r be the radius of these circles. Circle ΟBβ is the incircle of an equilateral triangle with altitude BE, so r=12 and OOAβ=ODβOAβD=18β12=6. Let circles ΟAβ and ΟBβ intersect at points P and Q, as shown in the figure, and let segments OAβOBββ and PQβ intersect at point M.
Note that OAβP=OAβQ=OBβP=OBβQ=12. Therefore POAβQOBβ is a rhombus and point M is the midpoint of its diagonals OAβOBββ and PQβ. Because β³OMOAβ is a 30β60β90β right triangle whose hypotenuse, OOAββ, has length 6, it follows that OM=3 and OAβM=33β. Then
PM2=OAβP2βOAβM2=144β27=117,
so PM=313β and OP=PMβOM=313ββ3. The circumradius of the smaller equilateral triangle is equal to 313ββ3, and therefore its side length is equal to 3ββ (313ββ3)=351ββ27β. The requested sum is 351+27=378.
Note: As long as abβ is irrational, as is the case with 351β 27β=2713β, the representation aββbβ is unique.
