Advanced book on Mathematics Olympiad

Geometry and Trigonometry 625

x^2 +y^2 +cx+dy+e= 0 ,i= 1 , 2 ,...,n.

Writing the fact that the vertices also satisfy the equation of the ellipse and subtracting,
we obtain( 1 −a)y^2 i+cxi+dyi+(e+b)=0. Hence

y^2 i=−

c

1 −a

xi−

d

1 −a

yi−

e+b

1 −a

.

The numberccannot be 0, for otherwise the quadratic equation would have two solutions
yiand each of these would yield two solutionsxi, so the polygon would have four or fewer
sides, a contradiction. This means that the regular polygon is inscribed in a parabola.
Change the coordinates so that the parabola has the standard equationy^2 = 4 px. Let
the new coordinates of the vertices be(ξi,ηi)and the new equation of the circumcircle
bex^2 +y^2 +c′x+d′y+e′=0. That the vertices belong to both the parabola and the
circle translates to

η^2 i = 4 pξi and ξi^2 +ηi^2 +c′ξ+d′η+e′= 0 , fori= 1 , 2 ,...,n.

So theηi’s satisfy the fourth-degree equation

1

16 p^2

ηi^4 +η^2 i+

c′

4 p

ηi^2 +d′ηi+e′= 0.

This equation has at most four solutions, and each solution yields a uniquexi. So the
regular polygon can have at most four vertices, a contradiction. We conclude that no
regular polygon with five or more vertices can be inscribed in an ellipse that is not also
a circle.

612.SetFBk=tk,k= 1 , 2 ,...,n. Also, letαbe the angle made by the half-line|FB 1
with thex-axis andαk=α+^2 (k−n^1 )π,k= 2 ,...,n. The coordinates of the focusFare
(p 2 , 0 ).
In general, the coordinates of the points on a ray that originates inFand makes an
angleβwith thexaxis are(p 2 +tcosβ, tsinβ),t>0 (just draw a ray from the origin
of the coordinate system that makes an angleβwith thex-axis; then translate it toF). It
follows that the coordinates ofBkare(p 2 +tkcosαk,tksinαk),k= 1 , 2 ,...,n.
The condition thatBkbelongs to the parabola is written astk^2 sin^2 αk = p^2 +
2 ptkcosαk. The positive root of this equation istk=p/( 1 −cosαk). We are sup-
posed to prove thatt 1 +t 2 +···+tk>np, which translates to

1

1 −cosα 1

+

1

1 −cosα 2

+···+

1

1 −cosαn

>n.

To prove this inequality, note that

( 1 −cosα 1 )+( 1 −cosα 2 )+···+( 1 −cosαn)