Problem:
In β³ABC, the sides have integers lengths and AB=AC. Circle Ο has its center at the incenter of β³ABC. An excircle of β³ABC is a circle in the exterior of β³ABC that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to BC is internally tangent to Ο, and the other two excircles are both externally tangent to Ο. Find the minimum possible value of the perimeter of β³ABC.
Solution:
Rescale the triangle so that BC=1 and AB=AC=x. Then [ABC]=21βx2β41ββ. Let the incircle of β³ABC have center I and radius r. Let the excircles opposite A and B have centers IAβ and IBβ and radii rAβ and rBβ, respectively. For any triangle ABC with a=BC,b=AC,c=AB, inradius r, and the radius of the excircle opposite A,rAβ, the area of β³ABC is given by rβ 2a+b+cβ and by rAββ 2b+cβaβ. It follows that
r=21βx+21βxβ21βββ,rAβ=21βxβ21βx+21βββ, and rBβ=x2β41ββ
Let M be the midpoint of BC, and let D be the point of tangency of the excircle opposite B with line BC. Let point X lie on line IBβD so that IXβ₯IBβDβ. Note that the radius of Ο is equal to r+2rAβ. Note also that BD is the semiperimeter of β³ABC; that is, BD=x+21β, and so IX=MD=BDβBM=x. The Pythagorean Theorem applied to β³IXIBβ yields
x2+(rBββr)2=(r+2rAβ+rBβ)2
Expressing each term in the above equation in terms in x gives
implying that x=29β. Thus the minimum possible perimeter with integer side lengths occurs when BC=2 and AB=AC=9, giving a perimeter of 20 .
OR
Set TAβ and TBβ to be the tangency points of Ο with the excircle opposite A, ΟAβ, and excircle opposite B,ΟBβ, respectively. Note that the homothety HAβ sending ΟAβ to Ο has positive scale factor and is centered at TAβ, while the homothety HBβ sending Ο to ΟBβ has negative scale factor and is centered at TBβ. Thus the composition HBββHAβ is another homothety with negative scale factor and sends ΟAβ to ΟBβ. Because ΟAβ and ΟBβ have common tangent lines\ AC and BC, the center of this homothety is C and therefore TAβ,TBβ, and C are collinear. In turn,
Now note that BIAββ is a median of β³BTAβM, where M is the midpoint of BC; combining this with β IAβBM=90ββ2β Bβ yields the trigonometric equation
