Advanced book on Mathematics Olympiad

218 4 Geometry and Trigonometry

Separate the variables

dx=−

√

1 −y^2

y

dy,

and then integrate to obtain

x=−

√

1 −y^2 −lny−ln( 1 +

√

1 −y^2 )+C.

The initial condition givesC=0. The answer to the problem is therefore the curve

x=−

√

1 −y^2 −lny−ln( 1 +

√

1 −y^2 ),

depicted in Figure 28. 

0

3

2.5

2

1.5

5

1

0.5

4

0

321

Figure 28

This curve is called atractrix, a name given by Ch. Huygens. Clearly, it has thex-
axis as an asymptote. E. Beltrami has shown that the surface of revolution of the tractrix
around its asymptote provides a partial model for hyperbolic geometry. This surface has
been used in recent years for the shape of loudspeakers.
A variety of other curves show up in the problems below. In some of the solutions,
polar coordinates might be useful. Recall the formulas for changing between Cartesian
and polar coordinates:x=rcosθ,y=rsinθ.

616.Find the points where the tangent to the cardioidr= 1 +cosθis vertical.

617.Given a circle of diameterAB, a variable secant throughAintersects the circle at
Cand the tangent throughBatD. On the half-lineACa pointMis chosen such
thatAM=CD. Find the locus ofM.