Difference between revisions of "2019 AIME I Problems/Problem 10"
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<cmath> 19 = 3a^2+3b </cmath> | <cmath> 19 = 3a^2+3b </cmath> | ||
Thus, <math>a=\frac{20}{3}</math> and <math>b=\frac{1}{3}\left(19-3\left(\frac{20}{3} \right)^2 \right)</math>. That is <math>|b|=\frac{343}{9}=\frac{m}{n}</math>. <math>m+n=343+9=\boxed{352}</math> | Thus, <math>a=\frac{20}{3}</math> and <math>b=\frac{1}{3}\left(19-3\left(\frac{20}{3} \right)^2 \right)</math>. That is <math>|b|=\frac{343}{9}=\frac{m}{n}</math>. <math>m+n=343+9=\boxed{352}</math> | ||
+ | |||
+ | ==Solution 5(Newton's Sums)== | ||
+ | We start by calling <math> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| = S</math>, the sum of the roots of the polynomial <math>P_1</math> and the sum of the square of the roots <math>P_2</math> | ||
+ | |||
+ | By Vieta's, | ||
+ | <cmath> -20 = 3(z_1+z_2+z_3+z_4 \dots+z_{673}) </cmath> | ||
+ | <cmath>19 = 3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2) + 9S </cmath> | ||
+ | |||
+ | The latter can be easily verified by using a combinatorics approach. <math>19</math> is the sum of all the possible pairs of two roots of the polynomial. Which has <math>\binom{2019}{2}</math> without simplification. Now looking at the latter above, there are <math>3\cdot673</math> terms in the first part and <math>9\cdot\binom{673}{2}</math>. | ||
+ | |||
+ | With some computation, we see | ||
+ | <math>\binom{2019}{2}</math> <math>= 3\cdot673+</math> <math>9\cdot\binom{673}{2}</math>. | ||
+ | This step was simply done to check I missed no steps | ||
+ | |||
+ | Now using Newton Sums, where <math>P_2 = z_1^2+z_2^2+z_3^2+\dots+z_{673}^2</math> | ||
+ | |||
+ | <cmath>P_2 + -20\cdot20 + 19\cdot2 \Rightarrow P_2 = 362</cmath> | ||
+ | |||
+ | <cmath>19 = 362 + 9S \Rightarrow S = \frac{343}{9}</cmath> | ||
+ | Thus, <math>\frac{343}{9}</math> leads to the answer <math>\boxed{352}</math>. | ||
+ | ~YBSuburbanTea | ||
+ | |||
==Solution 5 (Official MAA 1)== | ==Solution 5 (Official MAA 1)== | ||
Because each root of the polynomial appears with multiplicity <math>3,</math> Viète's Formulas show that <cmath>z_1+z_2+\cdots+z_{673}=-\frac{20}3</cmath> and <cmath>z_1^2+z_2^2+\cdots+z_{673}^2 =\frac13\left((-20)^2-2\cdot19\right)=\frac{362}3.</cmath> Then the identity <cmath>\left(\sum_{i=1}^{673}z_i\right)^2=\sum_{i=1}^{673}z_i^2+2\left(\sum_{1\le j<k\le 673}z_jz_k\right)</cmath> shows that <cmath>\sum_{1\le j<k\le 673}z_jz_k=\frac{\left(-\frac{20}3\right)^2-\frac{362}3}{2}=-\frac{343}9.</cmath> The requested sum is <math>343+9=352.</math> | Because each root of the polynomial appears with multiplicity <math>3,</math> Viète's Formulas show that <cmath>z_1+z_2+\cdots+z_{673}=-\frac{20}3</cmath> and <cmath>z_1^2+z_2^2+\cdots+z_{673}^2 =\frac13\left((-20)^2-2\cdot19\right)=\frac{362}3.</cmath> Then the identity <cmath>\left(\sum_{i=1}^{673}z_i\right)^2=\sum_{i=1}^{673}z_i^2+2\left(\sum_{1\le j<k\le 673}z_jz_k\right)</cmath> shows that <cmath>\sum_{1\le j<k\le 673}z_jz_k=\frac{\left(-\frac{20}3\right)^2-\frac{362}3}{2}=-\frac{343}9.</cmath> The requested sum is <math>343+9=352.</math> |
Revision as of 12:10, 3 January 2023
Contents
[hide]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
.
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)
We start by calling , the sum of the roots of the polynomial
and the sum of the square of the roots
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 I missed no steps
Now using Newton Sums, where
Thus,
leads to the answer
.
~YBSuburbanTea
Solution 5 (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 6 (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.
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.