2019 AIME I Problems/Problem 8
Contents
Problem
Let be a real number such that . Then where and are relatively prime positive integers. Find .
Solution 1
We can substitute . Since we know that , we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let , we can simplify the equation to: After using binomial theorem, this simplifies to: If we use the quadratic formula, we obtain , so (observe that either choice of doesn't matter). Substituting we get:
Therefore, the answer is .
-eric2020, inspired by Tommy2002
Motivation
The motivation to substitute comes so that after applying the binomial theorem to a lot of terms will cancel out. Note that all the terms with odd exponents in will cancel out, while the terms with even exponents will be doubled. mathboy282
Solution 2
First, for simplicity, let and . Note that . We then bash the rest of the problem out. Take the fifth power of this expression and get . Note that we also have . So, it suffices to compute . Let . We have from cubing that or . Next, using , we get or . Solving gives or . Clearly is extraneous, so . Now note that , and . Thus we finally get , giving .
- Emathmaster
Solution 3 (Newton Sums)
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in Solution . Let and be the roots of some polynomial . Then, by Vieta, for some .
Let . We want to find . Clearly and . Newton sums tells us that where for our polynomial .
Bashing, we have
Thus . Clearly, so .
Note . Solving for , we get . Finally, .
Solution 4
Factor the first equation. First of all, because We group the first, third, and fifth term and second and fourth term. The first group: The second group: Add the two together to make Because this equals , we have Let so we get Solving the quadratic gives us Because , we finally get .
Now from the second equation, Plug in to get which yields the answer
~ZericHang
Solution 5
Define the recursion We know that the characteristic equation of must have 2 roots, so we can recursively define as . is simply the sum of the roots of the characteristic equation, which is . is the product of the roots, which is . This value is not trivial and we have to solve for it. We know that , , . Solving the rest of the recursion gives
Solving for in the expression for gives us , so . Since , we know that the minimum value it can attain is by AM-GM, so cannot be .
Plugging in the value of into the expression for , we get . Our final answer is then
-Natmath
Solution 6
Let and , then and
Now factoring as solution 4 yields .
Since , .
Notice that can be rewritten as . Thus, and . As in solution 4, we get and
Substitute and , then , and the desired answer is
Solution 7 (Official MAA)
Let and let Then for Because and it follows that and Hence or and because the only possible value of is Therefore The requested sum is
Solution 8 (Recursion)
Let for non-negative integers . Then and . In addition,where . So we can compute \begin{align*} a_4&=1-2X\\ a_6&=1-3X\\ a_8&=1-4X+2X^2\\ a_{10}&=1-5X+5X^2=\frac{11}{36} \end{align*}so . But by the sin double angle formula, , so . Thenso the answer is as desired.
A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion.
Video Solution By The Power Of Logic
~ Hayabusa1
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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