Difference between revisions of "2010 USAJMO Problems/Problem 4"
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Then, consider triangle with vertices <math>(a,a^2), (-a,a^2), (b,b^2)</math>, and set <math>a=2^{2n}</math> and <math>b=2^{4n-2}+1</math>. | Then, consider triangle with vertices <math>(a,a^2), (-a,a^2), (b,b^2)</math>, and set <math>a=2^{2n}</math> and <math>b=2^{4n-2}+1</math>. | ||
− | The area of this triangle is <math>\frac{1}{2} | + | The area of this triangle is <math>\frac{1}{2} \cdot base \cdot height=a(b^2-a^2)</math>. |
We have that <math>b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2</math> | We have that <math>b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2</math> | ||
We desire <math>A=a(b^2-a^2)=2^{2n}m^2</math>, or <math>2^{4n-1}-1=m</math>, and <math>m</math> is clearly always odd for positive <math>n</math>, completing the proof. | We desire <math>A=a(b^2-a^2)=2^{2n}m^2</math>, or <math>2^{4n-1}-1=m</math>, and <math>m</math> is clearly always odd for positive <math>n</math>, completing the proof. |
Revision as of 16:37, 5 June 2020
Contents
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
.
A Small Hint
Before you read the solution, try using induction on n. (And don't step by one!)
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.
Solution 2
We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area with two of the vertices sharing the same y-coordinate.
BASE CASE:
If , consider the parabolic triangle
with
that has area
, so that
and
.
If
, let
. Because
has area
, we set
and
.
If
, consider the triangle formed by
. It is parabolic and has area
, so
and
.
INDUCTIVE STEP:
If produces parabolic triangle
with
and
, consider
'
'
' with vertices
,
, and
. If
has area
, then
'
'
' has area
, which is easily verified using the
formula for triangle area. This completes the inductive step for
.
Hence, for every nonnegative integer , there exists an odd
and a parabolic triangle with area
with two vertices sharing the same ordinate. The problem statement is a direct result of this result. -MathGenius_
Solution 3 (without induction)
First, consider triangle with vertices ,
,
. This has area
so
case is satisfied.
Then, consider triangle with vertices , and set
and
.
The area of this triangle is
.
We have that
We desire
, or
, and
is clearly always odd for positive
, completing the proof.
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.