Advanced book on Mathematics Olympiad

216 4 Geometry and Trigonometry

inside the triangle) such that the cevians on the linesAP,BP, andCPhave equal

lengths. LetSBCbe an equilateral triangle, and letAbe chosen in the interior of

SBC, on the altitude dropped fromS.

(a) Show thatABChas two equicevian points.

(b) Show that the common length of the cevians through either of the equicevian

points is constant, independent of the choice ofA.

(c) Show that the equicevian points divide the cevian throughAin a constant ratio,

which is independent of the choice ofA.

(d) Find the locus of the equicevian points asAvaries.

(e) LetS′be the reflection ofSin the lineBC. Show that (a), (b), and (c) hold ifA

varies on any ellipse withSandS′as its foci. Find the locus of the equicevian

points asAvaries on the ellipse.

A planar curve is called rational if it can be parametrized as(x(t), y(t))withx(t)
andy(t)rational functions of the real variablet. Here we have to pass to the closed real
line, sotis allowed to be infinite, while the plane is understood as the projective plane,
zero denominators giving rise to points on the line at infinity.

Theorem.All conics are rational curves.

Proof.The case of degenerate conics (i.e., pairs of lines) is trivial. The parabolay^2 = 4 px
is parametrized by(t
2
4 p,t), the ellipse

x^2

a^2 +

y^2

b^2 =1by(a

1 −t^2

1 +t^2 ,b

2 t

1 +t^2 ), and the hyperbola

x^2

a^2 −

y^2

b^2 =1by(a

t+t−^1

2 ,b

t−t−^1
2 ).The general case follows from the fact that coordinate
changes are rational (in fact, linear) transformations. 

Compare the standard parametrization of the circle(cosx,sinx)to the rational
parametrization(^1 −t
2
1 +t^2 ,

2 t

1 +t^2 ). This gives rise to the trigonometric substitution tan

x
2 =t
and explains why integrals of the form
∫
R(cosx,sinx)dx,

withRa two-variable rational function, can be reduced to integrals of rational functions.
Let us change slightly our point of view and take a look at the conic

y^2 =ax^2 +bx+c.

If we fix a point(x 0 ,y 0 )on this conic, the liney−y 0 =t(x−x 0 )intersects the conic in
exactly one more point(x, y). Writing the conditions that this point is both on the line
and on the conic and eliminatingy, we obtain the equation

[y 0 +t(x−x 0 )]^2 =ax^2 +bx+c.