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)^2</math> with two of the vertices sharing the same 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 y-coordinate. | ||
− | + | ||
+ | BASE CASE: | ||
If <math>n = 0</math>, consider the parabolic triangle <math>ABC</math> with <math>A(0, 0), B(1, 1), C(-1, 1)</math> that has area <math>1/2 \cdot 1 \cdot 2 = 1</math>, so that <math>n = 0</math> and <math>m = 1</math>. | If <math>n = 0</math>, consider the parabolic triangle <math>ABC</math> with <math>A(0, 0), B(1, 1), C(-1, 1)</math> that has area <math>1/2 \cdot 1 \cdot 2 = 1</math>, so that <math>n = 0</math> and <math>m = 1</math>. | ||
If <math>n = 1</math>, let <math>ABC = A(5, 25), B(4, 16), C(-4, 16)</math>. Because <math>ABC</math> has area <math>1/2 \cdot 8 \cdot 9 = 36</math>, we set <math>n = 1</math> and <math>m = 3</math>. | If <math>n = 1</math>, let <math>ABC = A(5, 25), B(4, 16), C(-4, 16)</math>. Because <math>ABC</math> has area <math>1/2 \cdot 8 \cdot 9 = 36</math>, we set <math>n = 1</math> and <math>m = 3</math>. | ||
− | If <math>n = 2</math>, consider the triangle formed by <math>A(21, 441), B(3, 9), C(-3, 9)</math>. It is parabolic and has area <math>1/2 \cdot 6 \cdot 432 = 1296 = | + | If <math>n = 2</math>, consider the triangle formed by <math>A(21, 441), B(3, 9), C(-3, 9)</math>. It is parabolic and has area <math>1/2 \cdot 6 \cdot 432 = 1296 = 36^2</math>, so <math>n = 2</math> and <math>m = 9</math>. |
− | + | INDUCTIVE STEP: | |
− | If n = k produces parabolic triangle ABC</math> with <math>A(a, a^2), B(b, b^2),</math> and <math>C(-b, b^2)</math>, consider <math>A'B'C | + | If <math>n = k</math> produces parabolic triangle <math>ABC</math> with <math>A(a, a^2), B(b, b^2),</math> and <math>C(-b, b^2)</math>, consider <math>A</math>'<math>B</math>'<math>C</math>' with vertices <math>A(4a, 16a^2)</math>, <math>B(4b, 16b^2)</math>, and <math>C(-4b, 16b^2)</math>. If <math>ABC</math> has area <math>(2^km)^2</math>, then <math>A</math>'<math>B</math>'<math>C</math>' has area <math>(2^{k+3}m)^2</math>, which is easily verified using the <math>1/2 \cdot\text{base} \cdot \text{height}</math> formula for triangle area. This completes the inductive step for <math>k \implies k+3</math>. |
− | 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. | + | Hence, for every nonnegative integer <math>n</math>, there exists an odd <math>m</math> 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. -MathGenius_ |
− | -MathGenius_ | ||
==Solution 3 (without induction)== | ==Solution 3 (without induction)== |
Revision as of 16:06, 25 February 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.