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1984 USAMO Problems/Problem 1

Problem

In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$.

Solution 1 (ingenious)

Using Vieta's formulas, we have:

\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}


From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$. Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$, and so $ac+ad+bc+bd=k-30$. The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$, so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$.

Let $p=a+b$ and $q=c+d$. Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$, we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$. Moreover, the first Vieta equation, $a+b+c+d=18$, gives $p+q=18$. Thus we have two linear equations in $p$ and $q$, which we solve to obtain $p=4$ and $q=14$.

Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$, yielding $k=4\cdot 14+30 = \boxed{86}$.

Solution 2 (cool)

We start as before: $ab=-32$ and $cd=62$. We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$, 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 $x^2+sx+62$.

Now

\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*}

Equating the coefficients of $x^3$ and $x$ 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 $x^2$ and get $k=\boxed{86}.$

Solution 3 AM-GM

Let $r_1, r_2, r_3, r_4$ be the roots of the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$. We are given that $r_1 r_2 = -32$.

By Vieta's formulas, we have:

$r_1+r_2+r_3+r_4 = 18$

$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4 = k$

$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -200$

$r_1r_2r_3r_4 = -1984$

Since $r_1r_2 = -32$, we have $(-32)r_3r_4 = -1984$, so $r_3r_4 = \frac{-1984}{-32} = 62$.

Also, $r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -200$, so $r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2) = -200$. Substituting $r_1r_2 = -32$ and $r_3r_4 = 62$, we have $-32(r_3+r_4)+62(r_1+r_2) = -200$.

Let $A = r_1+r_2$ and $B = r_3+r_4$. Then $A+B = 18$, so $B = 18-A$. Substituting this into the equation, we have $-32B+62A = -200$, so $-32(18-A)+62A = -200$. $-576+32A+62A = -200$, so $94A = 376$, which means $A = \frac{376}{94} = 4$. Then $B = 18-A = 18-4 = 14$.

So we have $r_1+r_2 = 4$ and $r_1r_2 = -32$. Then $r_1$ and $r_2$ are roots of $x^2-4x-32 = 0$, so $(x-8)(x+4) = 0$, which means $r_1 = 8$ and $r_2 = -4$ (or vice versa).

Also we have $r_3+r_4 = 14$ and $r_3r_4 = 62$. Then $r_3$ and $r_4$ are roots of $x^2-14x+62 = 0$. Using the quadratic formula, $x = \frac{14 \pm \sqrt{14^2-4(62)}}{2} = \frac{14 \pm \sqrt{196-248}}{2} = \frac{14 \pm \sqrt{-52}}{2} = 7 \pm i\sqrt{13}$. Then $r_3 = 7+i\sqrt{13}$ and $r_4 = 7-i\sqrt{13}$.

We want to find $k = r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4$. $k = r_1r_2+r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86$.

Final Answer: The final answer is $\boxed{86}$

~avm2023

Solution 4 (Alcumus)

Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\]Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\]Matching coefficients, we get a+b=−18,ab+c−32=k,ac−32b=200,−32c=−1984.Then $c = \frac{-1984}{-32} = 62,$ so $62a - 32b = 200.$ With $a + b = -18,$ we can solve to find $a = -4$ and $b = -14.$ Then \[k = ab + c - 32 = \boxed{86}.\]

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

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