Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Theory of Equations | AIME I, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Theory of Equations.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

Theory of Equations – AIME 2015


Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

  • is 107
  • is 72
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Algebra

Check the Answer


Answer: is 72.

AIME, 2015, Question 10.

Polynomials by Barbeau.

Try with Hints


First hint

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Second Hint

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Final Step

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.