Difference between revisions of "2019 AIME I Problems/Problem 10"
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− | ==Problem | + | ==Problem== |
For distinct complex numbers <math>z_1,z_2,\dots,z_{673}</math>, the polynomial | For distinct complex numbers <math>z_1,z_2,\dots,z_{673}</math>, the polynomial | ||
− | <cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polynomial with complex coefficients and with degree at most <math>2016</math>. The | + | <cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polynomial with complex coefficients and with degree at most <math>2016</math>. The sum <cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath> can be expressed in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
− | <cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath>can be expressed in the form <math>\ | + | ==Video Solution & More by MegaMath== |
+ | Video #2 (Vieta's Formulas): | ||
+ | https://www.youtube.com/watch?v=6-kcdBsmCmc | ||
==Solution 1== | ==Solution 1== | ||
Line 46: | Line 48: | ||
<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)== | ||
+ | In the problem statement, we're given that the polynomial is <math>(x-z_1)^3(x-z_2)^3(x-z_3)^2\cdots (x-z_{673})^3</math> which can be expressed as <math>x^{2019}+20x^{2018}+19x^{2017}+g(x)</math>. So it has 673 roots of multiplicity 3 each, for a total of 2019 roots. | ||
+ | |||
+ | We start by calling <math> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| = S</math>. Next, let the sum of the roots of the polynomial be <math>P_1</math>, which equals <math>3(x_1+x_2+\cdots+x_{673})</math>, and the sum of the square of the roots be <math>P_2</math>, which equals <math>3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^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 that we missed no steps. | ||
+ | |||
+ | Now using [[Newton's Sums | Newton Sums]], where <math>P_2 = z_1^2+z_2^2+z_3^2+\dots+z_{673}^2</math>. We first have that <math>P_1\cdot a_n + 1\cdot a_{n-1}=0</math> and plugging in values, we get that <math>P_1\cdot 1 + 1\cdot 20 = 0 \implies P_1=-20.</math> Then, | ||
+ | <math>P_2\cdot a_n + P_1 \cdot a_{n-1} + 2\cdot a_{n-2}</math> and plugging in the values we know, we get that | ||
+ | |||
+ | <cmath>P_2 + -20\cdot20 + 19\cdot2 = 0 \implies P_2 = 362</cmath> | ||
+ | |||
+ | <cmath>19 = 362 + 9S \implies S = \frac{343}{9}</cmath> | ||
+ | Thus, <math>\frac{343}{9}</math> leads to the answer <math>\boxed{352}</math>. | ||
+ | |||
+ | ~YBSuburbanTea | ||
+ | |||
+ | ~minor edits by BakedPotato66 | ||
+ | |||
+ | ==Solution 6 (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> | ||
+ | |||
+ | Note that such a polynomial does exist. For example, let <math>z_{673}=-\tfrac{20}3,</math> and for <math>i=1,2,3,\dots,336,</math> let <cmath>z_i=\sqrt{\frac{343i}{9\sum_{j=1}^{336}j}}\qquad \text{and}\qquad z_{i+336}=-z_i.</cmath> Then <cmath>\sum_{i=1}^{673}z_i=-\frac{20}3\qquad\text{and}\qquad\sum_{i=1}^{673}z_i^2=2\sum_{i=1}^{336}\frac{343i}{9\sum_{i=1}^{336}j}+\left(\frac{20}3\right)^2=\frac{362}3,</cmath> as required. | ||
+ | |||
+ | ==Solution 7 (Official MAA 2)== | ||
+ | There are constants <math>a</math> and <math>b</math> such that <cmath>(x-z_1)(x-z_2)(x-z_3)\cdots(x-z_{673})=x^{673}+ax^{672}+bx^{671}+\cdots.</cmath> Then <cmath>(x^{673}+ax^{672}+bx^{671}+\cdots)^3=x^{2019}+20x^{2018}+19x^{2017}+\cdots.</cmath> Comparing the <math>x^{2018}</math> and <math>x^{2017}</math> coefficients shows that <math>3a=20</math> and <math>3a^2+3b=19.</math> Solving this system yields <math>a=\tfrac{20}3</math> and <math>b=-\tfrac{343}9.</math> Viète's Formulas then give <math>\left|\sum_{1\le j<k\le 673}z_jz_k\right|=|b|=\tfrac{343}9,</math> as above. | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | Note that we are trying to find the sum of the products of the pairwise distinct <math>z_k</math>. First, disregard the repeated roots and call the roots of the polynomial <math>x_1, x_2\ldots x_{2019}</math> for simplicity's sake. | ||
+ | |||
+ | Then, <math>(z_1+z_2\ldots z_{2019})^2=2\sum_{1\leq j<k\leq2019}{z_jz_k}+\sum_{1\leq k\leq 2019}{z_k^2}</math>. Notice that <math>\sum_{1\leq j<k\leq2019} | ||
+ | {z_jz_k}=19</math> and <math>(x_1+ x_2\ldots +x_{2019})^2=20^2=400</math> by Vietas so <math>\sum_{1\leq k\leq 2019}{z_k^2}=362</math>. Also, since every root is repeated 3 times, <math>\sum_{1\leq k\leq 2019}{z_k^2}=3\sum_{1\leq k\leq673}{z_k^2} \Longrightarrow \sum_{1\leq k\leq673}{z_k^2}=\frac{362}{3}</math> where the <math>z_k</math> in the second sum are all distinct. | ||
+ | |||
+ | Call <math>|\sum_{1 \le j <k \le 673} z_jz_k|=|s|</math>. Taking into account repeated roots, if we pair any two roots together, there will be <math>\binom{3}{1}=3</math> copies for each <math>z_k^2, 1\leq z\leq 673</math> and <math>3\cdot 3=9</math> copies for each <math>z_jz_k</math> where <math>z_j</math> and <math>z_k</math> are distinct. | ||
+ | |||
+ | Since the sum of any two pairs of roots is 19, <math>9s+3\sum_{1\leq k\leq673}{z_k^2}=19</math> or <math>9s+362=19 \Longrightarrow s=\frac{-343}{9}, |s|=\frac{343}{9}</math>. <math>m+n=343+9=352</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ||
+ | |||
+ | == 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== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=9|num-a=11}} | {{AIME box|year=2019|n=I|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:02, 8 February 2024
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
.
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 | |
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.