Difference between revisions of "2010 USAJMO Problems/Problem 4"
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<math>A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2</math>, | <math>A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2</math>, | ||
which yields the <math>n=2</math> case. This completes the construction. | which yields the <math>n=2</math> 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 <math>2^nm</math> 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 m = 1. | ||
+ | If n = 1, let ABC = A(1, 1), B(2, 4), C(-2, 4). Because ABC has area 1/2 * 3 * 4 = 6, we set n = 1 and m = 3. | ||
+ | If n = 2, consider the triangle formed by A(3, 9), B(4, 16), C(-4, 16). It is parabolic and has area 1/2 * 7 * 8 = 28, so n = 2 and m = 7. | ||
+ | |||
+ | 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(2a, <math>4a^2</math>), B(2b, <math>4b^2</math>), and C(-2b, <math>4b^2</math>). If ABC has area <math>2^km</math>, then A'B'C' has area <math>2^{k+3}m</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. | ||
== See Also == | == See Also == |
Revision as of 22:13, 11 April 2014
Contents
[hide]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.
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 m = 1. If n = 1, let ABC = A(1, 1), B(2, 4), C(-2, 4). Because ABC has area 1/2 * 3 * 4 = 6, we set n = 1 and m = 3. If n = 2, consider the triangle formed by A(3, 9), B(4, 16), C(-4, 16). It is parabolic and has area 1/2 * 7 * 8 = 28, so n = 2 and m = 7.
Inductive step: If n = k produces parabolic triangle ABC with A(a, ), B(b, ), and C(-b, ), consider A'B'C' with vertices A(2a, ), B(2b, ), and C(-2b, ). 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.