Difference between revisions of "1984 USAMO Problems/Problem 1"
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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, <math>a+b+c+d=18</math>, gives <math>p+q=18</math>. Thus we have two linear equations in <math>p</math> and <math>q</math>, which we solve to obtain <math>p=4</math> and <math>q=14</math>. | 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, <math>a+b+c+d=18</math>, gives <math>p+q=18</math>. Thus we have two linear equations in <math>p</math> and <math>q</math>, which we solve to obtain <math>p=4</math> and <math>q=14</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>. | + | 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 2 === | === Solution 2 === |
Revision as of 12:01, 19 July 2020
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.