Difference between revisions of "2010 USAJMO Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
− | We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area <math>2^nm</math> with two of the vertices sharing the same ordinate (y-coordinate). | + | We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area <math>(2^nm)^2</math> with two of the vertices sharing the same ordinate (y-coordinate). |
Base case: | Base case: | ||
− | If n = 0, consider the parabolic triangle ABC with A(0, 0), B(1, 1), C(-1, 1) that has area 1/2 * 1 * 2 = 1, so that m = 1. | + | If n = 0, consider the parabolic triangle ABC with A(0, 0), B(1, 1), C(-1, 1) that has area 1/2 * 1 * 2 = 1, so that n = 0 and m = 1. |
− | If n = 1, let ABC = A( | + | If n = 1, let ABC = A(5, 25), B(4, 16), C(-4, 16). Because ABC has area 1/2 * 8 * 9 = 36, we set n = 1 and m = 3. |
− | If n = 2, consider the triangle formed by A( | + | If n = 2, consider the triangle formed by A(21, 441), B(3, 9), C(-3, 9). It is parabolic and has area 1/2 * 6 * 432 = 1296 = <math>36^2</math>, so n = 2 and m = 9. |
Inductive step: | Inductive step: | ||
− | If n = k produces parabolic triangle ABC with A(a, <math>a^2</math>), B(b, <math>b^2</math>), and C(-b, <math>b^2</math>), consider A'B'C' with vertices A( | + | If n = k produces parabolic triangle ABC with A(a, <math>a^2</math>), B(b, <math>b^2</math>), and C(-b, <math>b^2</math>), consider A'B'C' with vertices A(4a, <math>16a^2</math>), B(4b, <math>16b^2</math>), and C(-4b, <math>16b^2</math>). If ABC has area <math>(2^km)^2</math>, then A'B'C' has area <math>(2^{k+3}m)^2</math>, which is easily verified using the 1/2 * base * height formula for triangle area. This completes the inductive step for k -> k+3. |
− | Hence, for every nonnegative integer n, there exists an odd m and a parabolic triangle with area <math>2^nm</math> with two vertices sharing the same ordinate. The problem statement is a direct result of this result. | + | Hence, for every nonnegative integer n, there exists an odd m and a parabolic triangle with area <math>(2^nm)^2</math> with two vertices sharing the same ordinate. The problem statement is a direct result of this result. |
== See Also == | == See Also == |
Revision as of 21:54, 14 April 2014
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 ordinate (y-coordinate).
Base case:
If n = 0, consider the parabolic triangle ABC with A(0, 0), B(1, 1), C(-1, 1) that has area 1/2 * 1 * 2 = 1, so that n = 0 and m = 1.
If n = 1, let ABC = A(5, 25), B(4, 16), C(-4, 16). Because ABC has area 1/2 * 8 * 9 = 36, we set n = 1 and m = 3.
If n = 2, consider the triangle formed by A(21, 441), B(3, 9), C(-3, 9). It is parabolic and has area 1/2 * 6 * 432 = 1296 = , so n = 2 and m = 9.
Inductive step:
If n = k produces parabolic triangle ABC with A(a, ), B(b,
), and C(-b,
), consider A'B'C' with vertices A(4a,
), B(4b,
), and C(-4b,
). If ABC has area
, then A'B'C' has area
, which is easily verified using the 1/2 * base * height formula for triangle area. This completes the inductive step for k -> k+3.
Hence, for every nonnegative integer n, there exists an odd m and a parabolic triangle with area with two vertices sharing the same ordinate. The problem statement is a direct result of this result.
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.