Geometry and Trigonometry 623±
1
2
∣∣
∣∣
∣
∣
4 pα^214 pα 11
4 pα^224 pα 21
4 pα^234 pα 31∣∣
∣∣
∣
∣
=± 8 p^2∣∣
∣∣
∣
∣
α^21 α 11
α^22 α 21
α^23 α 31∣∣
∣∣
∣
∣
= 8 p^2 |(α 1 −α 2 )(α 1 −α 3 )(α 2 −α 3 )|.The area of the triangleP 12 P 23 P 31 is given by
±
1
2
∣∣
∣∣
∣∣
4 pα 1 α 22 p(α 1 +α 2 ) 1
4 pα 2 α 32 p(α 2 +α 3 ) 1
4 pα 3 α 12 p(α 3 +α 1 ) 1∣∣
∣∣
∣∣
=± 4 p^2∣
∣∣
∣∣
∣
α 1 α 2 (α 1 +α 2 ) 1
α 2 α 3 (α 2 +α 3 ) 1
α 3 α 1 (α 3 +α 1 ) 1∣
∣∣
∣∣
∣
=± 4 p^2∣
∣∣
∣∣
∣
(α 1 −α 3 )α 2 (α 1 −α 3 ) 0
(α 2 −α 1 )α 3 (α 2 −α 1 ) 0
α 3 α 1 (α 3 +α 1 ) 1∣
∣∣
∣∣
∣
= 4 p^2 |(α 1 −α 3 )(α 1 −α 2 )(α 2 −α 3 )|.We conclude that the ratio of the two areas is 2, regardless of the location of the three
points or the shape of the parabola.
(Gh. Calug ̆ ari ̧ta, V. Mangu, ̆ Probleme de Matematic ̆a pentru TreaptaI ̧si aII-ade
Liceu(Mathematics Problems for High School), Editura Albatros, Bucharest, 1977)
610.Choose a Cartesian system of coordinates such that the focus isF(p, 0 )and the
directrix isx=−p, in which case the equation of the parabola isy^2 = 4 px. Let the
three points beA(a
2
4 p,a),B(
b^2
4 p,b),C(c^2
4 p,c).
(a) The tangentsNP,PM, andMNto the parabola are given, respectively, byay= 2 px+
a^2
2,by= 2 px+
b^2
2,cy= 2 px+
c^2
2,
from which we deduce the coordinates of the vertices
M(
bc
4 p,
b+c
2)
,N
(
ca
4 p,
c+a
2)
,P
(
ab
4 p,
a+b
2)
.
The intersection of the lineACof equation 4px−(c+a)y+ca=0 with the parallel
to the symmetry axis throughB, which has equationy =b,isL(ab+ 4 bcp−ca,b).Itis
straightforward to verify that the segmentsMPandLNhave the same midpoint, the
point with coordinates(b(c 8 +pa),a+^24 b+c). Consequently,LMN Pis a parallelogram.
(b) Writing that the equation of the circlex^2 +y^2 + 2 αx+ 2 βy+γ=0 is satisfied by
the pointsM, N, Phelps us determine the parametersα, β, γ. We obtain the equation
of the circumcircle ofMNP,
x^2 +y^2 −
ab+bc+ca+ 4 p^2
4 px+
abc− 4 p^2 (a+b+c)
8 p^2y+
ab+bc+ca
4= 0.
This equation is satisfied by(p, 0 ), showing that the focusFis on the circle.