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. 
