2019 AIME I Problems/Problem 10
Contents
Problem
For distinct complex numbers , the polynomial can be expressed as , where is a polynomial with complex coefficients and with degree at most . The sum can be expressed in the form , where and are relatively prime positive integers. Find .
Video Solution & More by MegaMath
Video #2 (Vieta's Formulas): https://www.youtube.com/watch?v=6-kcdBsmCmc
Solution 1
In order to begin this problem, we must first understand what it is asking for. The notation simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or Call this sum .
Now we can begin the problem. Rewrite the polynomial as . Then we have that the roots of are .
By Vieta's formulas, we have that the sum of the roots of is . Thus,
Similarly, we also have that the the sum of the roots of taken two at a time is This is equal to
Now we need to find and expression for in terms of . We note that Thus, .
Plugging this into our other Vieta equation, we have . This gives . Since 343 is relatively prime to 9, .
Solution 2
This is a quick fake solve using where and only .
By Vieta's, and Rearranging gives and giving .
Substituting gives which simplifies to .
So, , , ,
Solution 3
Let . By Vieta's, Then, consider the term. To produce the product of two roots, the two roots can either be either for some , or for some . In the former case, this can happen in ways, and in the latter case, this can happen in ways. Hence, and the requested sum is .
Solution 4
Let Therefore, . This is also equivalent to for some real coefficients and and some polynomial with degree . We can see that the big summation expression is simply summing the product of the roots of taken two at a time. By Vieta's, this is just the coefficient . The first three terms of can be bashed in terms of and to get Thus, and . That is .
Solution 5 (Newton's Sums)
In the problem statement, we're given that the polynomial is which can be expressed as . So it has 673 roots of multiplicity 3 each, for a total of 2019 roots.
We start by calling . Next, let the sum of the roots of the polynomial be , which equals , and the sum of the square of the roots be , which equals .
By Vieta's,
The latter can be easily verified by using a combinatorics approach. is the sum of all the possible pairs of two roots of the polynomial. Which has without simplification. Now looking at the latter above, there are terms in the first part and .
With some computation, we see . This step was simply done to check that we missed no steps.
Now using Newton Sums, where . We first have that and plugging in values, we get that Then, and plugging in the values we know, we get that
Thus, leads to the answer .
~YBSuburbanTea
~minor edits by BakedPotato66
Solution 6 (Official MAA 1)
Because each root of the polynomial appears with multiplicity Viète's Formulas show that and Then the identity shows that The requested sum is
Note that such a polynomial does exist. For example, let and for let Then as required.
Solution 7 (Official MAA 2)
There are constants and such that Then Comparing the and coefficients shows that and Solving this system yields and Viète's Formulas then give as above.
Solution 8
Note that we are trying to find the sum of the products of the pairwise distinct . First, disregard the repeated roots and call the roots of the polynomial for simplicity's sake.
Then, . Notice that and by Vietas so . Also, since every root is repeated 3 times, where the in the second sum are all distinct.
Call . Taking into account repeated roots, if we pair any two roots together, there will be copies for each and copies for each where and are distinct.
Since the sum of any two pairs of roots is 19, or . .
Video Solution by OmegaLearn
https://youtu.be/Dp-pw6NNKRo?t=776
~ pi_is_3.14
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=7SFKuEdgwMA
~The Power of Logic
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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