Problem:
In △ABC, the sides have integers lengths and AB=AC. Circle ω has its center at the incenter of △ABC. An excircle of △ABC is a circle in the exterior of △ABC 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 ω, and the other two excircles are both externally tangent to ω. Find the minimum possible value of the perimeter of △ABC.
Solution:
Rescale the triangle so that BC=1 and AB=AC=x. Then [ABC]=21x2−41. Let the incircle of △ABC have center I and radius r. Let the excircles opposite A and B have centers IA and IB and radii rA and rB, respectively. For any triangle ABC with a=BC,b=AC,c=AB, inradius r, and the radius of the excircle opposite A,rA, the area of △ABC is given by r⋅2a+b+c and by rA⋅2b+c−a. It follows that
r=21x+21x−21,rA=21x−21x+21, and rB=x2−41
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 IBD so that IX⊥IBD. Note that the radius of ω is equal to r+2rA. Note also that BD is the semiperimeter of △ABC; that is, BD=x+21, and so IX=MD=BD−BM=x. The Pythagorean Theorem applied to △IXIB yields
x2+(rB−r)2=(r+2rA+rB)2
Expressing each term in the above equation in terms in x gives
implying that x=29. Thus the minimum possible perimeter with integer side lengths occurs when BC=2 and AB=AC=9, giving a perimeter of 20 .
OR
Set TA and TB to be the tangency points of ω with the excircle opposite A, ωA, and excircle opposite B,ωB, respectively. Note that the homothety HA sending ωA to ω has positive scale factor and is centered at TA, while the homothety HB sending ω to ωB has negative scale factor and is centered at TB. Thus the composition HB∘HA is another homothety with negative scale factor and sends ωA to ωB. Because ωA and ωB have common tangent lines\ AC and BC, the center of this homothety is C and therefore TA,TB, and C are collinear. In turn,
∠BTAI=90∘−2∠TBITA=90∘−2∠AIC=45∘−4∠B.
Now note that BIA is a median of △BTAM, where M is the midpoint of BC; combining this with ∠IABM=90∘−2∠B yields the trigonometric equation
cot(45∘−4∠B)=2tan(90∘−2∠B)=tan2∠B2
Let β=4∠B. Then
cot(45∘−β)=tan(45∘+β)=1−tanβ1+tanβ
and
tan(2β)2=tanβ1−tan2β
This shows that tanβ=23−5 and tan(4β)=tan(∠B)=45. It follows that AB=29BC as above.
