Advanced book on Mathematics Olympiad

620 Geometry and Trigonometry

thereforeQP=OQ. We conclude thatQlies on the parabola of focusOand directrix

t. A continuity argument shows that the locus is the entire parabola.

(A. Myller,Geometrie Analitic ̆a(Analytical Geometry), 3rd ed., Editura Didactic ̆a

̧si Pedagogica, Bucharest, 1972, solutions found by the students from the Mathematical ̆

Olympiad Summer Program, 2004)

605.We will use the equation of the tangent with prescribed slope. Write the parabola in

standard form

y^2 = 4 px.

The tangent of slopemto this parabola is given by

y=mx+

p

m

.

IfA(p+a, 0 )andB(p−a, 0 )are the two fixed points,(p, 0 )being the focus, then the

distances to the tangent are

∣

∣

∣∣m(p±a)+

p

√ m

1 +m^2

∣

∣∣

∣.

The difference of their squares is

(

m^2 (p+a)^2 + 2 p(p+a)+p

2

m^2

)

−

(

m^2 (p−a)^2 + 2 p(p−a)+p

2

m^2

)

1 +m^2

.

An easy computation shows that this is equal to 4pa, which does not depend onm,
meaning that it does not depend on the tangent.
(A. Myller,Geometrie Analitica ̆(Analytical Geometry), 3rd ed., Editura Didactica ̧ ̆si
Pedagogica, Bucharest, 1972) ̆
606.The statement of the problem is invariant under affine transformations, so we can
assume the hyperbola to have the equationxy=1, such that the asymptotes are the
coordinate axes. IfP(x 1 ,y 1 )andQ(x 2 ,y 2 )are two of the vertices, then the other two
vertices of the parallelogram are(x 1 ,y 2 )and(x 2 ,y 1 ). The line they determine has the
equation

y−y 1 =

y 2 −y 1

x 1 −x 2

(x−x 2 ).

Substituting the coordinates of the origin in this equation yields−y 1 =yx^21 −−xy^12 (−x 2 ),or

x 1 y 1 −x 2 y 1 =x 2 y 2 −x 2 y 1. This clearly holds, sincex 1 y 1 =x 2 y 2 =1, and the property

is proved.

(A. Myller,Geometrie Analitica ̆(Analytical Geometry), 3rd ed., Editura Didactica ̧ ̆si

Pedagogica, Bucharest, 1972) ̆