Difference between revisions of "1984 USAMO Problems/Problem 1"
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Let the four roots be <math>a, b, c,</math> and <math>d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see. | Let the four roots be <math>a, b, c,</math> and <math>d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see. | ||
− | Solution | + | === Solution 1 === |
Using Vieta's formulas, we have: | Using Vieta's formulas, we have: | ||
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Therefore, we have <math>(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30</math>, yielding <math>k=4\cdot 14+30 = \boxed{86}</math>. | Therefore, we have <math>(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30</math>, yielding <math>k=4\cdot 14+30 = \boxed{86}</math>. | ||
− | Solution | + | === Solution 2 == |
We start as before: <math>ab=-32</math> and <math>cd=62</math>. We now observe that a and b must be the roots of a quadratic, <math>x^2+rx-32</math>, 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 <math>x^2+sx+62</math>. | We start as before: <math>ab=-32</math> and <math>cd=62</math>. We now observe that a and b must be the roots of a quadratic, <math>x^2+rx-32</math>, 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 <math>x^2+sx+62</math>. | ||
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Now | Now | ||
− | <cmath> \begin{align*}x^4-18x^3+kx^2+200x-1984 & | + | <cmath> \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+rs)x^2\ |
+ | &+(62s-32r)x-1984.\end{align*} </cmath> | ||
Equating the coefficients of <math>x^3</math> and <math>x</math> 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 <math>x^2</math> and get <math>k=\boxed{86}.</math> | Equating the coefficients of <math>x^3</math> and <math>x</math> 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 <math>x^2</math> and get <math>k=\boxed{86}.</math> |
Revision as of 11:28, 18 July 2016
Contents
[hide]Problem
In the polynomial , the product of of its roots is . Find .
Solution
Let the four roots be and , so that . 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, , gives . Thus we have two linear equations in and , which we solve to obtain and .
Therefore, we have , yielding .
= Solution 2
We start as before: and . We now observe that a and b must be the roots of a quadratic, , 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 .
Now
Equating the coefficients of and 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 and get
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.