1999 IMO Problems/Problem 1
Contents
[hide]Problem
Determine all finite sets of at least three points in the plane which satisfy the following condition:
For any two distinct points and
in
, the perpendicular bisector of the line segment
is an axis of symmetry of
.
Solution
Upon reading this problem and drawing some points, one quickly realizes that the set consists of all the vertices of any regular polygon.
Now to prove it with some numbers:
Let , with
, where
is a vertex of a polygon which we can define their
coordinates as:
for
.
That defines the vertices of any regular polygon with being the radius of the circumcircle of the regular
-sided polygon.
Now we can pick any points and
of the set as:
and
, where
;
; and
Then,
and
Let be point
which is not part of
Then, , and
The perpendicular bisector of passes through
.
Let point , not in
be a point that passes through the perpendicular bisector of
at a distance
from
Then, and
CASE I: is even
and
is integer
Then
This means that the perpendicular bisector also passes through a point of
Let be any positive integer
and
Therefore, for any integer
.
Also, since for any integer
then this proves that the bisector of any points and
is an axis of symmetry for this case.
CASE II: is odd
and
is integer
and
is integer
Then
This means that the perpendicular bisector does not pass through any point of , but their closest points are
and
and that
Let be any positive integer
and
Therefore, for any integer
.
Since ,
Also, since for any integer
then this proves that the bisector of any points and
is an axis of symmetry for this case.
Having proven both cases, then the set of points that comply with the given condition is the set of the vertices of any regular polygon of any number of sides.
~Tomas Diaz. orders@tomasdiaz.com
Solution 2
First we prove no points can lie on a line. Say
were sequential points on a line. Considering the axis of symmetry between
and
one finds there lies a point
on the right side of
. Then considering the axis of symmetry between
and
one finds sequential points
and
that lie on the right side of
. One can continue this process ad infinitum to show
must have infinite points. A contradiction.
Let a line roll/rotate around the perimeter of such that at any time all points of
are on one side of the line and the line is always touching one point of
, but may touch two points of
at a time. Say the line sequentially touches the points
. We will now prove
form a regular polygon.
Identify with
,
with
, and
with
.
Considering the axis of symmetry between and
, a line which rolls counterclockwise around
with starting position passing through
and
will roll symmetrically to a line which rolls clockwise around
with starting position passing through
and
. Let
be the perpendicular bisector of
. There are two possibilities:
lies on the same side of
as
or
lies on
. In the latter case apparently
. In any event,
has the property all points of
are on one side of the line passing through
and
, and this is not true for any other point on the same side of
as
. A similar statement holds for
and
. It follows
must be symmetric to
about
. So
. Repeat for all other sequential
points of
to get
for all
. Now, if
didn't lie on the axis of symmetry between
and
there would exist another point
symmetric to
about that axis, and as the line rolled around
one would find it sequentially touched
or
. A contradiction to show
were defined. So
lies on the axis of symmetry, thus
. Repeat for all other sequential
points of
to get
for all
. We have shown
form a regular polygon.
Now, the regular polygon has a well-defined center
and there is certainly not another isometric polygon among
. Therefore any axis of symmetry about which we can reflect
must reflect
into itself. i.e. the axis of symmetry must intersect
.
Let be a point of
. We will show
is one of
. The axis of symmetry between
and
intersects
by the previous. So
is an isosceles triangle with
. Therefore
lies on the circle with center
and radius
. Note all points
lie on this circle. If
were not one of
, we could suppose
lies between
and
on this circle, whereas all points of
lie on one side of the line passing through
and
. A contradiction. So
is one of
. So
is the regular polygon
.
~not_detriti
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1999 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |