American Mathematical Monthly Problem 10678

    Problem. (Proposed by Clark Kimberling, University of Evansville, Evansville, IN, and Peter Yff, Muncie, IN.) Let the incircle K of any triangle ABC touch BC, CA, AB at A', B', C' respectively and have I as center. Let A'IF be a diameter. Let A" be a point on K such that the circle A"BC touches K at A". Define B", C" similarly. Prove that AA", BB", CC" are concurrent.

    Solution. We shall extend the problem, and prove that
    (i) FA", C'B', BC concur at some point E such that IE is perpendicular to AA', and (B,C;A',E) form a harmonic quadruple (i.e., A'B/A'C = EB/EC.) (ii) Let the lines AF, AA" cut BC at M,N respectively. Then (M,N;A',E) is also a harmonic quadruple, and BM = A'C. (iii) AA", BB" CC" concur at some point X. (iv) The trilinear coordinates of X, i.e., the ratios of the distances of X from the sides of DABC, are 1/a(s-a)³ : 1/b(s-b)³ : 1/c(s-c)³.
    (We avoided using trigonometry.) Let s=(a+b+c)/2. Using Ceva's theorem and the fact that BA' = BC' = s-b, we can prove that AA', BB', CC' concur (at what is called the Gergonne center). Let C'B' intersect BC at E, then Menelaus' theorem gives BE/CE = (BC'/CA) (AB'/B'C) = BA'/A'C, hence (B,C;A',E) form a harmonic quadruple.
    Let the tangent to K at A" intersect BC at D. Then A'D = A"D = Ö(BD×CD) (because A"D also touches the circle A"BC). From BA'/CA' = BE/CE we can deduce that D is the midpoint of A'E, so that A'D = A"D = ED. \ A'A" ^ A"E. Hence FA"E is a str. line.
     C'B'E is the polar of A relative to K, hence AA' is the polar of E, so that IE ^ AA'. (This can be proved by more elementary means.) (i) is now proved.
     Let the line FA"E cut AA' at V, then (F,A";V,E) is a harmonic quadruple because AA' is the polar of E. Hence (M,N;A',E) is also a harmonic quadruple.
    To find BM, Let the tangent to K at F intersect AB at B₂ and AC at C₂. Then DAB₂C₂ is similar to DABC. Since K is one of the three excircles of DAB₂C₂, and AFM is a str. line, M is then the point of contact of the excircle of DABC inside ÐA with the side BC. Hence BM = s-c = CA'. This proved (ii).
    It follows that MA'=c-b. Let x = A'N, y = A'E.
First, EB/EC = (y + s-b)/(y-s+c) = A'B/A'C = (s-b)/(s-c), yielding y = 2 (s-b) (s-c)/(c-b).
Next, x/(c-b) = A'N/A'M = EM/EN = (y-x) / (y+(c-b)), which yields
A'N = x = (s-b)(s-c)a(c-b) / a((c-b)²+(s-b)(s-c)).
From this, we get NC = (s-c)-x = (s-c)³/((c-b)²+(s-b)(s-c))
and NB = (s-b)³/((c-b)²+(s-b)(s-c)).
Thus NB/NC = (s-b)³/(s-c)³, a very elegant result, proving (iii)
     Similarly, if BB" intersects AC at N', and CC" intersects AB at N", (see the small diagram), then
(BN/NC) (CN'/N'A) (AN"/N"B) = (s-b)³/(s-c)³×(s-c)³/(s-a)³ ×(s-a)³/(s-b)³ = 1.
\By Ceva's theorem, AA", BB", CC" are concurrent at some point X.
     Let a= (s-a)³, b= (s-b)³, g=(s-c)³.
From Menelaus' theorem, AX/XN = (AN"/N"B) (BC/NC) = (a/b) (b+g)/g.
It follows that XN/AN = bg /(gb+ag+ab).
\The homogeneous coordinates of X are XN/AN : XN'/BN' : XN"/CN"
= 1/(s-a)³ : 1/(s-b)³ : 1/(s-c)³.
\The trilinear coordinates of X are 1/a(s-a)³ : 1/b(s-b)³ : 1/c(s-c)³.