Advanced book on Mathematics Olympiad

Geometry and Trigonometry 611

Since

−→

AB×

−→

ACand

−→

AE×

−→

ADare perpendicular to the plane of the triangle and oriented
the same way, this is equal to one-fourth of the area of the quadrilateralBCDE. Done.

589.We work in affine coordinates with the diagonals of the quadrilateral as axes. The
vertices areA(a, 0 ),B( 0 ,b),C(c, 0 ),D( 0 ,d). The midpoints of the sides areM(a 2 ,b 2 ),
N( 2 c 2 b),P( 2 c,d 2 ), andQ(a 2 ,d 2 ). The segmentsMPandNQhave the same midpoint,
namely, the centroid(a+ 4 c,b+ 4 d)of the quadrilateral. HenceMNPQis a parallelogram.

590.Choose a coordinate system that placesMat the origin and let the coordinates
ofA, B, C, respectively, be(xA,yA),(xB,yB),(xC,yC). Then the coordinates of the
centroids ofMAB,MAC, andMBCare, respectively,

GA=

(

xA+xB

3

,

yB+yB

3

)

,

GB=

(

xA+xC

3

,

yA+yC

3

)

,

GC=

(

xB+xC

3

,

yB+yC

3

)

.

The coordinates ofGA,GB,GCare obtained by subtracting the coordinates ofA, B,
andCfrom(xA+xB+xC,yA+yB+yC), then dividing by 3. Hence the triangle
GAGBGCis obtained by taking the reflection of triangleABCwith respect to the point
(xA+xB+xC,yA+yB+yC), then contracting with ratio^13 with respect to the origin
M. Consequently, the two triangles are similar.

591.Denote byδ(P, MN)the distance fromPto the lineMN. The problem asks for
the locus of pointsPfor which the inequalities

δ(P , AB) < δ(P , BC)+δ(P , CA),

δ(P , BC) < δ(P , CA)+δ(P , AB),

δ(P,CA)<δ(P,AB)+δ(P, BC)

are simultaneously satisfied.
Let us analyze the first inequality, written asf(P)=δ(P, BC)+δ(P, CA)−
δ(P, AB) >0. As a function of the coordinates(x, y)ofP, the distance fromPto a line
is of the formmx+ny+p. Combining three such functions, we see thatf(P)=f (x, y)
is of the same form,f (x, y)=αx+βy+γ. To solve the inequalityf (x, y) >0it
suffices to find the linef (x, y)=0 and determine on which side of the line the function
is positive. The line intersects the sideBCwhereδ(P, CA)=δ(P , AB), hence at the
pointEwhere the angle bisector fromAintersects this side. It intersects sideCAat the
pointFwhere the bisector fromBintersects the side. Also,f (x, y) >0 on sideAB,
hence on the same side of the lineEFas the segmentAB.