Advanced book on Mathematics Olympiad

632 Geometry and Trigonometry

this segment. ThenOA=cosrθandOA=ABsinθ, which yield the equation of the
locus

r=asinθcosθ=

a

2

sin 2θ.

This is a four-leaf rose.

621.Choosing a Cartesian system of coordinates whose axes are the asymptotes, we can
bring the equation of the hyperbola into the formxy=a^2. The equation of the tangent
to the hyperbola at a point(x 0 ,y 0 )isx 0 y+y 0 x− 2 a^2 =0. Sincea^2 =x 0 y 0 , thexand
yintercepts of this line are 2x 0 and 2y 0 , respectively.
Let(r, θ )be the polar coordinates of the foot of the perpendicular from the origin to
the tangent. In the right triangle determined by the center of the hyperbola and the two
intercepts we have 2x 0 cosθ=rand 2y 0 sinθ =r. Multiplying, we obtain the polar
equation of the locus

r^2 = 2 a^2 sin 2θ.

This is the lemniscate of Bernoulli, shown in Figure 82.
(1st W.L. Putnam Mathematical Competition, 1938)

0.8

0.8

00 0.4

0.4

-0.4

-0.8

-0.8 -0.4

Figure 82

622.The solution uses complex and polar coordinates. Our goal is to map the circle onto
a cardioid of the form

r=a( 1 +cosθ), a > 0.

Because this cardioid passes through the origin, it is natural to work with a circle that
itself passes through the origin, for example|z− 1 |=1. Ifφ:C→Cmaps this circle
into the cardioid, then the equation of the cardioid will have the form

|φ−^1 (z)− 1 |= 1.