Problem:
Let ABC be an acute triangle with ω,Ω, and R being its incircle, circumcircle, and circumradius, respectively. Circle ωA is tangent internally to Ω at A and tangent externally to ω. Circle ΩA is tangent internally to Ω at A and tangent internally to ω. Let PA and QA denote the centers of ωA and ΩA, respectively. Define points PB,QB,PC,QC analogously. Prove that
8PAQA⋅PBQB⋅PCQC≤R3
with equality if and only if triangle ABC is equilateral.
Solution:
Let the incircle touch the sides AB,BC, and CA at C1,A1, and B1, respectively. Set AB=c,BC=a,CA=b. By equal tangents, we may assume that AB1=AC1=x, BC1=BA1=y, and CA1=CB1=z. Then a=y+z,b=z+x,c=x+y. By the AM-GM inequality, we have a≥2yz,b≥2zx, and c≥2xy. Multiplying the last three inequalities yields
abc≥8xyz
with equality if and only if x=y=z; that is, triangle ABC is equilateral.
Let k denote the area of triangle ABC. By the Extended Law of Sines, c=2Rsin∠C. Hence
k=2absin∠C=4Rabc or R=4kabc
We are going to show that
PAQA=4kxa2(*)
In exactly the same way, we can also establish its cyclic analogous forms
PBQB=4kyb2 and PCQC=4kzc2
Multiplying the last three equations together gives
with equality if and only if triangle ABC is equilateral.
Hence it suffices to show (∗). Let r,rA,rA′ denote the radii of ω,ωA,ΩA, respectively. We consider the inversion I with center A and radius x. Clearly, I(B1)=B1,I(C1)=C1, and I(ω)=ω. Let ray AO intersect ωA and ΩA at S and T, respectively. It is not difficult to see that AT>AS, because ω is tangent to ωA and ΩA externally and internally, respectively. Set S1=I(S) and T1=I(T). Let ℓ denote the line tangent to Ω at A. Then the image of ωA (under the inversion) is the line (denoted by ℓ1 ) passing through S1 and parallel to ℓ, and the image of ΩA is the line (denoted by ℓ2 ) passing through T1 and parallel to ℓ. Furthermore, since ω is tangent to both ωA and ΩA,ℓ1 and ℓ2 are also tangent to the image of ω, which is ω itself. Thus the distance between these two lines is 2r; that is, S1T1=2r. Hence we can consider the following configuration. (The darkened circle is ωA, and its image is the darkened line ℓ1.)
By the definition of inversion, we have AS1⋅AS=AT1⋅AT=x2. Note that AS=2rA, AT=2rA′, and S1T1=2r. We have
rA=2AS1x2. and rA′=2AT1x2=2(AS1−2r)x2
Hence
PAQA=AQA−APA=rA′−rA=2x2(AS1−2r1+AS11)
Let HA be the foot of the perpendicular from A to side BC. It is well known that ∠BAS1=∠BAO=90∘−∠C=∠CAHA. Since ray AI bisects ∠BAC, it follows that rays AS1 and AHA are symmetric with respect to ray AI. Further note that both line ℓ1 (passing through S1 ) and line BC (passing through HA ) are tangent to ω. We conclude that AS1=AHA. In light of this observation and using the fact 2k=AHA⋅BC=(AB+BC+CA)r, we can compute PAQA as follows:
