Difference between revisions of "2008 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
Let <math>r</math>, <math>s</math>, and <math>t</math> be the three roots of the equation | Let <math>r</math>, <math>s</math>, and <math>t</math> be the three roots of the equation | ||
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Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>. | Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>. | ||
− | + | ==Solution 1 == | |
− | == Solution == | + | By [[Vieta's formulas]], we have <math>r + s + t = 0</math> so <math>t = -r - s.</math> Substituting this into our problem statement, our desired quantity is <cmath>(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).</cmath> Also by [[Vieta's formulas]] we have <cmath>rst = -rs(r + s) = -\dfrac{2008}{8} = -251</cmath> so negating both sides and multiplying through by 3 gives our answer of <math>\boxed{753}.</math> |
− | === Solution | + | |
+ | == Solution 2 == | ||
By [[Vieta's formulas]], we have <math>r+s+t = 0</math>, and so the desired answer is <math>(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)</math>. Additionally, using the factorization | By [[Vieta's formulas]], we have <math>r+s+t = 0</math>, and so the desired answer is <math>(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)</math>. Additionally, using the factorization | ||
<cmath>r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0</cmath> | <cmath>r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0</cmath> | ||
we have that <math>r^3 + s^3 + t^3 = 3rst</math>. By Vieta's again, <math>rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.</math> | we have that <math>r^3 + s^3 + t^3 = 3rst</math>. By Vieta's again, <math>rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.</math> | ||
− | + | == Solution 3 == | |
Vieta's formulas gives <math>r + s + t = 0</math>. Since <math>r</math> is a root of the polynomial, <math>8r^3 + 1001r + 2008 = 0\Longleftrightarrow - 8r^3 = 1001r + 2008</math>, and the same can be done with <math>s,\ t</math>. Therefore, we have | Vieta's formulas gives <math>r + s + t = 0</math>. Since <math>r</math> is a root of the polynomial, <math>8r^3 + 1001r + 2008 = 0\Longleftrightarrow - 8r^3 = 1001r + 2008</math>, and the same can be done with <math>s,\ t</math>. Therefore, we have | ||
<cmath>\begin{align*}8\{(r + s)^3 + (s + t)^3 + (t + r)^3\} &= - 8(r^3 + s^3 + t^3)\\ | <cmath>\begin{align*}8\{(r + s)^3 + (s + t)^3 + (t + r)^3\} &= - 8(r^3 + s^3 + t^3)\\ | ||
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http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums | http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums | ||
− | + | == Solution 4 == | |
Expanding, you get: | Expanding, you get: | ||
<cmath>r^3 + 3r^2s + 3s^2r +s^3 +</cmath> | <cmath>r^3 + 3r^2s + 3s^2r +s^3 +</cmath> | ||
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<cmath>r^3 + 3r^2t + 3t^2r +t^3</cmath> | <cmath>r^3 + 3r^2t + 3t^2r +t^3</cmath> | ||
<cmath>= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r </cmath> | <cmath>= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r </cmath> | ||
− | This looks similar to <math>(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + | + | This looks similar to <math>(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + 6rst</math> |
Substituting: | Substituting: | ||
<cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath> | <cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath> | ||
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We are trying to find <math> -(r^3 + s^3 + t^3)</math>. | We are trying to find <math> -(r^3 + s^3 + t^3)</math>. | ||
Substituting: | Substituting: | ||
− | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}</cmath> | + | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}.</cmath> |
− | + | == Solution 5 == | |
Write <math>(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)</math> and let <math>f(x)=8x^3+1001x+2008</math>. Then <cmath>f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.</cmath> Solving for <math>r^3+s^3+t^3</math> and negating the result yields the answer <math>\boxed{753}.</math> | Write <math>(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)</math> and let <math>f(x)=8x^3+1001x+2008</math>. Then <cmath>f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.</cmath> Solving for <math>r^3+s^3+t^3</math> and negating the result yields the answer <math>\boxed{753}.</math> | ||
+ | |||
+ | == Solution 6 == | ||
+ | Here by [[Vieta's formulas]]: | ||
+ | <math>r+s+t = 0</math> --(1) | ||
+ | |||
+ | <math>rst = \frac{-2008}{8} = -251</math> --(2) | ||
+ | |||
+ | By the factorisation formula: | ||
+ | Let <math>a = r+s</math>, <math>b = s+t</math>, <math>c = t+r</math>, | ||
+ | <math>a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0</math> (By (1)) | ||
+ | |||
+ | So <cmath>a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \boxed{753}.</cmath> | ||
+ | |||
+ | == Solution 7 == | ||
+ | Let's construct a polynomial with the roots <math>(r+s), (s+t),</math> and <math>(t+r)</math>. | ||
+ | |||
+ | sum of the roots: | ||
+ | |||
+ | <math>=2(r+s+t)=2\cdot0=0</math> | ||
+ | |||
+ | pairwise product of the roots: | ||
+ | |||
+ | <math>(r+s)(s+t)+(s+t)(t+r)+(t+r)(r+s)=r^2+s^2+t^2+3(rs+st+tr)</math> | ||
+ | |||
+ | <math>=(r+s+t)^2+rs+st+tr=0+\frac{1001}{8}</math> | ||
+ | |||
+ | product of the roots: | ||
+ | |||
+ | <math>(r+s)(s+t)(t+r)=r^2t+r^2s+s^2r+s^2t+t^2r+t^2s+3rst</math> | ||
+ | |||
+ | <math>=(rs+st+tr)(r+s+t)-3rst+2rst=-rst=-\frac{2008}{8}</math> | ||
+ | |||
+ | thus, the polynomial we get is | ||
+ | |||
+ | <math>x^3+\frac{1001}{8}x+-\frac{2008}{8}=0</math> | ||
+ | |||
+ | as <math>(r+s), (s+t),</math> and <math>(t+r)</math> are roots of this polynomial, we know that (using power reduction) | ||
+ | |||
+ | <math>(r+s)^3+\frac{1001}{8}(r+s)-\frac{2008}{8}=0</math> | ||
+ | |||
+ | <math>(s+t)^3+\frac{1001}{8}(s+t)-\frac{2008}{8}=0</math> | ||
+ | |||
+ | <math>(t+r)^3+\frac{1001}{8}(t+r)-\frac{2008}{8}=0</math> | ||
+ | |||
+ | adding all of the equations up, we see that | ||
+ | |||
+ | <math>(r+s)^3+(s+t)^3+(t+r)^3=3\cdot\frac{2008}{8}-\frac{1001}{8}(2r+2s+2t)=251(3)+0=\boxed{753}</math> | ||
== See also == | == See also == |
Latest revision as of 23:49, 15 August 2020
Contents
Problem
Let , , and be the three roots of the equation Find .
Solution 1
By Vieta's formulas, we have so Substituting this into our problem statement, our desired quantity is Also by Vieta's formulas we have so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 3
Vieta's formulas gives . Since is a root of the polynomial, , and the same can be done with . Therefore, we have yielding the answer .
Also, Newton's Sums yields an answer through the application. http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums
Solution 4
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 5
Write and let . Then Solving for and negating the result yields the answer
Solution 6
Here by Vieta's formulas: --(1)
--(2)
By the factorisation formula: Let , , , (By (1))
So
Solution 7
Let's construct a polynomial with the roots and .
sum of the roots:
pairwise product of the roots:
product of the roots:
thus, the polynomial we get is
as and are roots of this polynomial, we know that (using power reduction)
adding all of the equations up, we see that
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.