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.
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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