Difference between revisions of "1996 AIME Problems/Problem 5"
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Suppose that the [[root]]s of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</math>. Find <math>t</math>. | Suppose that the [[root]]s of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</math>. Find <math>t</math>. | ||
− | == Solution == | + | == Solution 1 == |
By [[Vieta's formulas]] on the polynomial <math>P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0</math>, we have <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math>. Then | By [[Vieta's formulas]] on the polynomial <math>P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0</math>, we have <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math>. Then | ||
<center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center> | <center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center> | ||
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t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</cmath> | t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</cmath> | ||
+ | == Solution 2== | ||
− | + | Each term in the expansion of <math>(a+b)(b+c)(c+a)</math> has a total degree of 3. Another way to get terms with degree 3 is to multiply out <math>(a+b+c)(ab+bc+ca)</math>. Expanding both of these expressions and comparing them shows that: | |
<math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math> | <math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math> |
Latest revision as of 21:34, 30 January 2018
Contents
Problem
Suppose that the roots of are , , and , and that the roots of are , , and . Find .
Solution 1
By Vieta's formulas on the polynomial , we have , , and . Then
This is just the definition for .
Alternatively, we can expand the expression to get
Solution 2
Each term in the expansion of has a total degree of 3. Another way to get terms with degree 3 is to multiply out . Expanding both of these expressions and comparing them shows that:
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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