612 Geometry and Trigonometry
Arguing similarly for the other two inequalities, we deduce that the locus is the interior
of the triangle formed by the points where the angle bisectors meet the opposite sides.
592.Consider an affine system of coordinates such that none of the segments determined
by thenpoints is parallel to thex-axis. If the coordinates of the midpoints are(xi,yi),
i= 1 , 2 ,...,m, thenxi =xjfori =j. Thus we have reduced the problem to the one-
dimensional situation. So letA 1 ,A 2 ,...,Anlie on a line in this order. The midpoints of
A 1 A 2 ,A 1 A 3 ,..., A 1 Anare all distinct and different from the (also distinct) midpoints
ofA 2 An,A 3 An,..., An− 1 An. Hence there are at least(n− 1 )+(n− 2 )= 2 n− 3
midpoints. This bound can be achieved forA 1 ,A 2 ,...,Anthe points 1, 2 ,...,non the
real axis.
(K ̋ozépiskolai Matematikai Lapok(Mathematics Magazine for High Schools, Bu-
dapest), proposed by M. Salát)
593.We consider a Cartesian system of coordinates withBCandADas thex- andy-
axes, respectively (the origin is atD). LetA( 0 ,a),B(b, 0 ),C(c, 0 ),M( 0 ,m). Because
the triangle is acute,a, c >0 andb<0. Also,m>0. The equation ofBMis
mx+by=bm, and the equation ofACisax+cy=ac. Their intersection is
E
(
bc(a−m)
ab−cm,
am(b−c)
ab−cm)
.
Note that the denominator is strictly negative, hence nonzero. The pointEtherefore
exists.
The slope of the lineDEis the ratio of the coordinates ofE, namely,
am(b−c)
bc(a−m).
Interchangingbandc, we find that the slope ofDFis
am(c−b)
bc(a−m),
which is the negative of the slope ofDE. It follows that the linesDEandDFare
symmetric with respect to they-axis, i.e., the angles∠ADEand∠ADFare equal.
(18th W.L. Putnam Mathematical Competition, 1958)
594.We refer everything to Figure 74. LetA(c, 0 ),cbeing the parameter that determines
the variable line. BecauseBhas the coordinates(a 2 ,b 2 ), the lineABis given by the
equation
y=
b
a− 2 cx+
bc
2 c−a