Difference between revisions of "2021 AIME II Problems/Problem 4"
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There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math> | There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)== |
+ | By the Complex Conjugate Root Theorem, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are complex conjugates. Let <math>z=m+\sqrt{n}\cdot i</math> and <math>\overline{z}=m-\sqrt{n}\cdot i.</math> It follows that the roots of <math>x^3+ax+b</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math> | ||
− | + | We know that | |
+ | <cmath>\begin{align*} | ||
+ | z+\overline{z}&=2m, & (1) \\ | ||
+ | z\overline{z}&=m^2+n. & (2) | ||
+ | \end{align*}</cmath> | ||
+ | Applying Vieta's Formulas to <math>x^3+ax+b,</math> we have <math>-20+z+\overline{z}=0.</math> Substituting <math>(1)</math> into this equation, we get <math>m=10.</math> | ||
− | + | Applying Vieta's Formulas to <math>x^3+cx^2+d,</math> we have <math>-21z-21\overline{z}+z\overline{z}=0,</math> or <math>-21(z+\overline{z})+z\overline{z}=0.</math> Substituting <math>(1)</math> and <math>(2)</math> into this equation, we get <math>n=320.</math> | |
− | + | Finally, the answer is <math>m+n=\boxed{330}.</math> | |
− | ==Solution 2 | + | ~MRENTHUSIASM |
− | + | ||
+ | ==Solution 2 (Somewhat Bashy)== | ||
<math>(-20)^{3} + (-20)a + b = 0</math>, hence <math>-20a + b = 8000</math> | <math>(-20)^{3} + (-20)a + b = 0</math>, hence <math>-20a + b = 8000</math> | ||
Line 41: | Line 48: | ||
This simplifies to <math>m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261</math> | This simplifies to <math>m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261</math> | ||
− | Hence, <math>c|1261</math> | + | Hence, <math>c|1261 \Rightarrow c \in {1,13,97,1261}</math> |
+ | |||
+ | |||
+ | Consider case of <math>c = 1</math>: | ||
+ | |||
+ | <math>c = 1 \Rightarrow d = 8820</math> | ||
+ | Also, <math>a = 2m, b = 8000 + 40m</math> | ||
+ | |||
+ | <math>am + b = m^{2} - n + 8820</math> (because c = 1) | ||
+ | Also, <math>m^{2} + n + 40m = 820 \rightarrow (4)</math> | ||
+ | Also, Equation (2) gives <math>3m^{2} - n + 2m = 0 \rightarrow (5)</math> | ||
+ | |||
+ | Solving (4) and (5) simultaneously gives <math>m = 10, n = 320</math> | ||
+ | |||
+ | [AIME can not have more than one answer, so we can stop here ... Not suitable for objective exam] | ||
+ | |||
+ | Hence, <math>m + n = 10 + 320 = \boxed{330}</math> | ||
-Arnav Nigam | -Arnav Nigam | ||
− | ==See | + | ==Solution 3 (Heavy Calculation Solution)== |
+ | We start off by applying Vieta's, and we find that <math>a=m^2+n-40m</math>, <math>b=20m^2+20n</math>, <math>c=21-2m</math>, and <math>d=21m^2+21n</math>. After that, we use the fact that <math>-20</math> and <math>-21</math> are roots of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math>, respectively. Since substituting the roots back into the function returns zero, we have that <math>(-20)^3-20(a)+b=0</math> and <math>(-21)^3+c*(-21)^2+d=0</math>. Setting these two equations equal to each other while also substituting the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> above gives us <math>21m^2+21n-1682m+8000=0</math>. We then rearrange the equation into <math>21n = -21m^2+1682m-8000</math>. | ||
+ | |||
+ | With this property, we know that <math>-21m^2+1682m-8000</math> is divisible by <math>21</math>, so <math>1682m-8000=0 \pmod{21}</math>. This results in <math>2m-20=0 \pmod{21}</math>, which finally gives us <math>m=10 \pmod{21}</math>. We can test the first obvious value of <math>m</math>, which is <math>10</math>, to find that this works, and we get <math>m=10</math> and <math>n=320</math>. Therefore, the answer is <math>m + n = 10 + 320 = \boxed{330}</math> | ||
+ | |||
+ | ~Jske25 | ||
+ | |||
+ | ~sidkris (formatting edits) | ||
+ | |||
+ | ==Solution 4 (Synthetic Division)== | ||
+ | We note that <math>x^3 + ax + b = (x+20)P(x)</math> and <math>x^3 + cx^2 + d = (x+21)Q(x)</math> for some polynomials <math>P(x)</math> and <math>Q(x)</math>. | ||
+ | |||
+ | Through synthetic division (ignoring the remainder as we can set <math>b</math> and <math>d</math> to constant values such that the remainder is zero), <math>P(x) = x^2 - 20x + (400+a)</math>, and <math>Q(x) = x^2 + (c-21)x + (441 - 21c)</math>. | ||
+ | |||
+ | By the complex conjugate root theorem, we know that <math>P(x)</math> and <math>Q(x)</math> share the same roots, and they share the same leading coefficient, so <math>P(x) = Q(x)</math>. | ||
+ | |||
+ | Therefore, <math>c-21 = -20</math> and <math>441-21c = 400 + a</math>. Solving the system of equations, we get <math>a = 20</math> and <math>c = 1</math>, so <math>P(x) = Q(x) = x^2 - 20x + 420</math>. | ||
+ | |||
+ | Finally, by the quadratic formula, we have roots of <math>\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i</math>, so our final answer is <math>10 + 320 = \boxed{330}</math> | ||
+ | |||
+ | -faefeyfa | ||
+ | |||
+ | ==Solution 5 (Fast and Easy)== | ||
+ | We plug -20 into the equation obtaining <math>(-20)^3-20a+b</math>, likewise, plugging -21 into the second equation gets <math>(-21)^3+441c+d</math>. | ||
+ | |||
+ | Both equations must have 3 solutions exactly, so the other two solutions must be <math>m + \sqrt{n} \cdot i</math> and <math>m - \sqrt{n} \cdot i</math>. | ||
+ | |||
+ | By Vieta's, the sum of the roots in the first equation is <math>0</math>, so <math>m</math> must be <math>10</math>. | ||
+ | |||
+ | Next, using Vieta's theorem on the second equation, you get <math>x1x2+x2x3+x1x3 = 0</math>. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so <math>-21*(20)+(m^2+n)=0</math>. | ||
+ | |||
+ | Given that <math>m</math> is <math>10</math>, then <math>n</math> is equal to <math>320</math>. | ||
+ | |||
+ | Therefore, the answer to the equation is <math>\boxed{330}</math> | ||
+ | |||
+ | ==Solution 6 (solution by '''integralarefun''')== | ||
+ | Since <math>m+i\sqrt{n}</math> is a common root and all the coefficients are real, <math>m-i\sqrt{n}</math> must be a common root, too. | ||
+ | |||
+ | Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients). | ||
+ | |||
+ | First, however, the product of the two common roots is: | ||
+ | <cmath>\begin{align*} | ||
+ | &&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\ | ||
+ | &=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\ | ||
+ | &=&&x^2-2xm+(m^2-i^2n)\\ | ||
+ | &=&&x^2-2xm+m^2+n | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Now, let's equate the two forms of both the polynomials: | ||
+ | <cmath>x^3+ax+b=(x^2-2xm+m^2+n)(x+20)</cmath> | ||
+ | <cmath>x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)</cmath> | ||
+ | Now we can match the zero coefficients. | ||
+ | <cmath>-2m+20=0\to m=10\text{ and}</cmath> | ||
+ | <cmath>-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}</cmath> | ||
+ | Thus, <math>m+n=10+320=\boxed{330}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=sYRWWQayNyQ | ||
+ | |||
+ | ==Video Solution by TheCALT== | ||
+ | https://www.youtube.com/watch?v=HJ0EldshLuE | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=II|num-b=3|num-a=5}} | {{AIME box|year=2021|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:52, 28 June 2024
Contents
- 1 Problem
- 2 Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)
- 3 Solution 2 (Somewhat Bashy)
- 4 Solution 3 (Heavy Calculation Solution)
- 5 Solution 4 (Synthetic Division)
- 6 Solution 5 (Fast and Easy)
- 7 Solution 6 (solution by integralarefun)
- 8 Video Solution
- 9 Video Solution by TheCALT
- 10 See Also
Problem
There are real numbers and
such that
is a root of
and
is a root of
These two polynomials share a complex root
where
and
are positive integers and
Find
Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)
By the Complex Conjugate Root Theorem, the imaginary roots for each of and
are complex conjugates. Let
and
It follows that the roots of
are
and the roots of
are
We know that
Applying Vieta's Formulas to
we have
Substituting
into this equation, we get
Applying Vieta's Formulas to we have
or
Substituting
and
into this equation, we get
Finally, the answer is
~MRENTHUSIASM
Solution 2 (Somewhat Bashy)
, hence
Also, , hence
satisfies both
we can put it in both equations and equate to 0.
In the first equation, we get
Simplifying this further, we get
Hence, and
In the second equation, we get
Simplifying this further, we get
Hence, and
Comparing (1) and (2),
and
;
Substituting these in gives,
This simplifies to
Hence,
Consider case of :
Also,
(because c = 1)
Also,
Also, Equation (2) gives
Solving (4) and (5) simultaneously gives
[AIME can not have more than one answer, so we can stop here ... Not suitable for objective exam]
Hence,
-Arnav Nigam
Solution 3 (Heavy Calculation Solution)
We start off by applying Vieta's, and we find that ,
,
, and
. After that, we use the fact that
and
are roots of
and
, respectively. Since substituting the roots back into the function returns zero, we have that
and
. Setting these two equations equal to each other while also substituting the values of
,
,
, and
above gives us
. We then rearrange the equation into
.
With this property, we know that is divisible by
, so
. This results in
, which finally gives us
. We can test the first obvious value of
, which is
, to find that this works, and we get
and
. Therefore, the answer is
~Jske25
~sidkris (formatting edits)
Solution 4 (Synthetic Division)
We note that and
for some polynomials
and
.
Through synthetic division (ignoring the remainder as we can set and
to constant values such that the remainder is zero),
, and
.
By the complex conjugate root theorem, we know that and
share the same roots, and they share the same leading coefficient, so
.
Therefore, and
. Solving the system of equations, we get
and
, so
.
Finally, by the quadratic formula, we have roots of , so our final answer is
-faefeyfa
Solution 5 (Fast and Easy)
We plug -20 into the equation obtaining , likewise, plugging -21 into the second equation gets
.
Both equations must have 3 solutions exactly, so the other two solutions must be and
.
By Vieta's, the sum of the roots in the first equation is , so
must be
.
Next, using Vieta's theorem on the second equation, you get . However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so
.
Given that is
, then
is equal to
.
Therefore, the answer to the equation is
Solution 6 (solution by integralarefun)
Since is a common root and all the coefficients are real,
must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).
First, however, the product of the two common roots is:
Now, let's equate the two forms of both the polynomials:
Now we can match the zero coefficients.
Thus,
.
Video Solution
https://www.youtube.com/watch?v=sYRWWQayNyQ
Video Solution by TheCALT
https://www.youtube.com/watch?v=HJ0EldshLuE
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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