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American Mathematical Monthly
Problem 10763
by Peter Y. Woo
Biola University, La Mirada, Calif.
Problem. (Proposed by Jean Anglesio, Garches, France.) Let ABC be a triangle. Let O be its circumcenter, H its orthocenter, I its incenter, N its Nagel point, and X,Y,Z its excenters. Let S be the point such that O is the midpoint of HS. Let T be the midpoint of SN. It is known that the orthocenter and the nine-point center of triangle XYZ are I and O respectively. Prove that
(a) T is the circumcenter of triangle XYZ, and
(b) the centroids of triangles XYZ and SIN coincide.
Solution. (Avoiding trigonometric ratios is my quest for elegance.) Let a, b, c denote BC, CA, AB, and s = (a+b+c)/2. Let I", X", Y" be the projections of I, X, Y on line AC. For each point V in the plane, let V' denote its projection on the line BC. Then CX' = BI' = s-b.
The Nagel point N lies on AX' and BY", hence Menelaus' theorem gives
X'N/NA = X'B/BC × CY"/Y"A = (s-c)/a × (s-a)/(s-c) = (s-a)/a.
\X'N' : X'H' = X'N : X'A = s-a : s = AI" : AX" = AI : AX = H'I' : H'X'. \ H'I' = N'X'
On the other hand, HO = OS implies H'O' = O'S'.
\S'X' = S'O'-X'O' = H'O'-I'O' = H'I' = N'X'.
Hence the midpoint T of NS lies on the line XX'. Now ÐTXC = p/2-ÐBCX = C/2.
Similarly, we can prove ÐTYC = C/2, so that ÐXTY = p-C = A+B.
But ÐXZY = p-(p/2-B/2)-(p/2-A/2) = (A+B)/2. Hence ÐXTY = 2ÐXZY.
Let T" be the circumcenter of DXYZ, then T lies on the circular arc XT"Y. Similarly, T lies on the circular arcs XT"Z and YT"Z, so that T and T" must coincide, proving that T is the circumcenter of XYZ.
From the properties of Euler lines, we know that the centroid G" of DXYZ lies on the line IT, because I and T are the orthocenter and circumcenter of DXYZ, and that IT = 3 G"T. Thus G" is also the centroid of DSIN because IT is a median of DSIN. QED
