2015 AIME I Problems/Problem 10
Contents
Problem
Let be a third-degree polynomial with real coefficients satisfying Find .
Solution
Let = . Since 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 and , it is easy to see that , and ; otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that , and . This provides the following system of equations. Using any four of these functions as a system of equations yields
Solution 2 (fast)
By drawing the function, and similar to Solution 1, WLOG let . Then, . Set . Then the roots of are . So, . Plug in to find a. We know . So, . Thus, , and then .
Solution 3
Without loss of generality, let . (If , then take as the polynomial, which leaves unchanged.) Because is third-degree, write where clearly must be a permutation of from the given condition. Thus However, subtracting the two equations gives , so comparing coefficients gives and thus both values equal to . As a result, . As a result, and so . Now, we easily deduce that and so removing the without loss of generality gives , which is our answer.
Solution 4
The following solution is similar to solution 3, but assumes nothing. Let . Since has degree 3, has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, for some . Hence . Note that . Since has degree 3, so do and ; and both have the same leading coefficient. Hence and for some (else is not cubic) where is the same as the set . Subtracting the second equation from the first, expanding, and collecting like terms, we have that which must hold for all . Since we have that (1) , (2) and (3) . Since is the sum of 1,2,3,5,6, and 7, we have so that by (1) we have and . We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be and . Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that . Since is the leading coefficient of , the leading coefficient of is . Thus the leading coefficient of is 4, i.e. . Then from earlier, so that the answer is .
Solution 5
Express in terms of powers of : By the same argument as in the first Solution, we see that is an odd function about the line , so its coefficients and are 0. From there it is relatively simple to solve (as in the above solution, but with a smaller system of equations): and
Solution 6 (Finite differences)
Because a cubic must come in a "wave form" with two "humps" (Called points of inflections) , we can see that , and . By symmetry, . Now, WLOG let , and . Then, we can use finite differences to get that the third (constant) difference is , and therefore .
Solution 7 (Like solution 1 without annoying systems)
We can rewrite our function as two different cubics, . Note that is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing , , , , , yields: For the constant terms to have a difference of 24 (), we must have , so the constant term of our polynomial is , the absolute value of which is . -- Solution by eiis1000
Solution 8(godspeed)
Trivially, assume the function can be written as , where is quadratic. Then we set up the following equations using the coefficients of .
Subtract the first equation from the second and the second from the third to get a system of two variables. Solve and plug it in to get .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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