Difference between revisions of "1984 USAMO Problems/Problem 1"
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+ | == Solution 4 (Alcumus)== | ||
+ | Since two of the roots have product <math>-32,</math> the equation can be factored in the form | ||
+ | <cmath>x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).</cmath>Expanding, we get | ||
+ | <cmath>x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.</cmath>Matching coefficients, we get | ||
+ | \begin{align*} | ||
+ | a + b &= -18, \ | ||
+ | ab + c - 32 &= k, \ | ||
+ | ac - 32b &= 200, \ | ||
+ | -32c &= -1984. | ||
+ | \end{align*}Then <math>c = \frac{-1984}{-32} = 62,</math> so <math>62a - 32b = 200.</math> With <math>a + b = -18,</math> we can solve to find <math>a = -4</math> and <math>b = -14.</math> Then | ||
+ | <cmath>k = ab + c - 32 = \boxed{86}.</cmath> | ||
== Video Solution by Omega Learn == | == Video Solution by Omega Learn == |
Latest revision as of 18:25, 28 March 2024
Contents
[hide]Problem
In the polynomial , the product of of its roots is . Find .
Solution 1 (ingenious)
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 (cool)
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
Solution 3
Let the roots of the equation be and . By Vieta's, Since and , then, . Notice thatcan be factored intoFrom the first equation, . Substituting it back into the equation,Expanding,So, and . Notice thatPlugging all our values in,
~ kante314
Solution 4 (Alcumus)
Since two of the roots have product the equation can be factored in the form
Expanding, we get
Matching coefficients, we get
Video Solution by Omega Learn
https://youtu.be/Dp-pw6NNKRo?t=316
~ pi_is_3.14
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.