2008 USAMO Problems/Problem 4
Problem
(Gregory Galparin) Let be a convex polygon with
sides,
. Any set of
diagonals of
that do not intersect in the interior of the polygon determine a triangulation of
into
triangles. If
is regular and there is a triangulation of
consisting of only isosceles triangles, find all the possible values of
.
Contents
Solution
We label the vertices of as
. Consider a diagonal
in the triangulation. This diagonal partitions
into two regions
, and is the side of an isosceles triangle in both regions. Without loss of generality suppose the area of
is less than the area of
(so the center of
does not lie in the interior of
); it follows that the lengths of the edges and diagonals in
are smaller than
. Thus
must the be the base of the isosceles triangle in
, from which it follows that the isosceles triangle is
, and so
. Repeating this process on the legs of isosceles triangle (
), it follows that
for some positive integer
.
Now take the isosceles triangle in the triangulation that contains the center of
in its interior; if a diagonal passes through the center, selecting either of the isosceles triangles with that diagonal is an edge will suffice. Without loss of generality, suppose
. From our previous result, it follows that (sum of powers of
), eg
We now claim that this condition is sufficient. Suppose WLOG that ; then we rewrite this as
- Lemma 1: All
and
have triangulations that work.
- Lemma 2: If
-sided polygon works, then
-sided polygon works.
By induction, it follows that we can cover all the .
Proof 1: For , this is trivial. For
, construct
and
, then make midpoint isosceles triangles, repeat.
Proof 2: Construct every other edge, then reduce etc.
Template:Incomplete
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2008 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
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