Difference between revisions of "1984 USAMO Problems/Problem 1"
Pi is 3.14 (talk | contribs) (→Solution 2) |
|||
Line 3: | Line 3: | ||
In the polynomial <math>x^4 - 18x^3 + kx^2 + 200x - 1984 = 0</math>, the product of <math>2</math> of its roots is <math>- 32</math>. Find <math>k</math>. | In the polynomial <math>x^4 - 18x^3 + kx^2 + 200x - 1984 = 0</math>, the product of <math>2</math> of its roots is <math>- 32</math>. Find <math>k</math>. | ||
− | === Solution 1 === | + | === Solution 1 (ingenious)=== |
Line 17: | Line 17: | ||
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 (cool)=== |
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>. |
Revision as of 05:44, 21 August 2021
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
Video Solution
https://youtu.be/5QdPQ3__a7I?t=589
~ 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.