2010 USAJMO Problems/Problem 4
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.
Solution 4
We simply need to provide an example for all that satisfies the condition, and we do so.
Let . Then consider the triangle with coordinates .
By the shoelace formula, this triangle has area which clearly can be written in the form , where or . Now, we just have to prove that is always odd. This is true because is even (because it's a power of ), so is odd, and since an odd number squared is odd, the whole thing is odd as well. And we are done, since we have proved that for all , we can show that there exists such a triangle by merely providing an example.
~thinker123
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 |
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