2010 USAJMO Problems/Problem 4
Problem
A triangle is called a parabolic triangle if its vertices lie on a
parabola . Prove that for every nonnegative integer
, there
is an odd number
and a parabolic triangle with vertices at three
distinct points with integer coordinates with area
.
Solution
Let the vertices of the triangle be .
The area of the triangle is the absolute value of
in the equation:
If we choose ,
and gives the actual area. Furthermore,
we clearly see that the area does not change when we subtract the same
constant value from each of
,
and
. Thus, all possible areas
can be obtained with
, in which case
.
If a particular choice of and
gives an area
,
with
a positive integer and
a positive odd integer, then setting
,
gives an area
.
Therefore, if we can find solutions for ,
and
,
all other solutions can be generated by repeated multiplication
of
and
by a factor of
.
Setting and
, we get
, which yields
the
case.
Setting and
, we get
, which yields
the
case.
Setting and
, we get
. Multiplying these
values of
and
by
, we get
,
,
,
which yields the
case. This completes the construction.
See Also
2010 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.