Difference between revisions of "1984 USAMO Problems/Problem 1"
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− | Using Vieta's formulas, we have: \begin{align*} a+b+c+d &= 18, | + | Using Vieta's formulas, we have: <cmath> \begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*} </cmath> |
− | Let p=a+b and q=c+d. Plugging our known values for ab and cd into the third Vieta equation, -200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b), we have -200 = -32(c+d) + 62(a+b) = 62p-32q. Moreover, the first Vieta equation, a+b+c+d=18, gives p+q=18. Thus we have two linear equations in p and q, which we solve to obtain p=4 and q=14. | + | From the last of these equations, we see that <math>cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62</math>. Thus, the second equation becomes <math>-32+ac+ad+bc+bd+62=k</math>, and so <math>ac+ad+bc+bd=k-30</math>. The key insight is now to factor the left-hand side as a product of two binomials: <math>(a+b)(c+d)=k-30</math>, so that we now only need to determine <math>a+b</math> and <math>c+d</math> rather than all four of <math>a,b,c,d</math>. |
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+ | Let <math>p=a+b</math> and <math>q=c+d</math>. Plugging our known values for <math>ab</math> and <math>cd</math> into the third Vieta equation, <math>-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)</math>, we have <math>-200 = -32(c+d) + 62(a+b) = 62p-32q</math>. Moreover, the first Vieta equation, a+b+c+d=18, gives p+q=18. Thus we have two linear equations in p and q, which we solve to obtain p=4 and q=14. | ||
Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}. | Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}. |
Revision as of 21:24, 27 April 2014
Problem
In the polynomial , the product of of its roots is . Find .
Solution
Let the four roots be a, b, c, and d, so that ab=-32. From here we show two methods; the second is more slick, but harder to see.
Solution #1
Using Vieta's formulas, we have:
From the last of these equations, we see that . Thus, the second equation becomes , and so . The key insight is now to factor the left-hand side as a product of two binomials: , so that we now only need to determine and rather than all four of .
Let and . Plugging our known values for and into the third Vieta equation, , we have . Moreover, the first Vieta equation, a+b+c+d=18, gives p+q=18. Thus we have two linear equations in p and q, which we solve to obtain p=4 and q=14.
Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}.
Solution #2 (sketch)
We start as before: ab=-32 and cd=62. We now observe that a and b must be the roots of a quadratic, x^2+rx-32, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic x^2+sx+62.
Now \begin{align*} x^4 - 18x^3 + kx^2 + 200x - 1984 &= (x^2 + rx - 32)(x^2 + sx + 62)
& = x^4 + (r + s)x^3 + (62 - 32 ... Equating the coefficients of x^3 and x with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of x^2 and get k=\boxed{86}.
See Also
1984 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.