000RM.dvi

924 Some geometry problems from recent journals

Crux 2836, proposed by G. Tsintsifas, Thessaloniki, Greece

Suppose that triangleABCis equilateral and thatPis an interior point.
The linesAP,BP,CPintersect the opposite sides atD,E,Frespec-
tively. Suppose thatPD=PE=PF. Determine the locus ofP.
SupposePhas homogeneous barycentric coordinates(x:y:
z)with respect to triangleABC. Then,DdividesBCin the ratio
BP:PC=z:y, andPD=x+xy+z·AD. By Stewart’s theorem,

AD^2 =

z

y+z

a^2 +

y

y+z

a^2 −

yz

(y+z)^2

a^2 =

y^2 +yz+z^2

(y+z)^2

·a^2.

Therefore,

PD^2 =

x^2 (y^2 +yz+z^2 )

(x+y+z)^2 (y+z)^2

·a^2.

Similarly, we obtain

PE^2 =

y^2 (z^2 +zx+x^2 )

(x+y+z)^2 (z+x)^2

·a^2 ,

PF^2 =

z^2 (x^2 +xy+y^2 )

(x+y+z)^2 (x+y)^2

·a^2.

Therefore,PE=PFif and only if

(x+y)^2 y^2 (z^2 +zx+x^2 )=(z+x)^2 z^2 (x^2 +xy+y^2 ),

or

(y−z)(x+y+z)(x^3 (y+z)+x^2 (y^2 +yz+z^2 )+xyz(y+z)+y^2 z^2 )=0.

SincePis an interior point,x,y,zare positive. We conclude that
y=z. Similarly,PF=PDif and only ifz=x.
The pointPhas coordinatesx:y:z=1:1:1; it is the center
of the equilateral triangle.