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Difference between revisions of "1984 USAMO Problems/Problem 1"

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Equating the coefficients of <math>x^3</math> and <math>x</math> 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 <math>x^2</math> and get <math>k=\boxed{86}.</math>
 
Equating the coefficients of <math>x^3</math> and <math>x</math> 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 <math>x^2</math> and get <math>k=\boxed{86}.</math>
  
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== Solution 3 ==
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== Solution 3 AM-GM==
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Let the roots of the equation be <math>a,b,c,</math> and <math>d</math>. By Vieta's,
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Let <math>r_1, r_2, r_3, r_4</math> be the roots of the polynomial <math>x^4 - 18x^3 + kx^2 + 200x - 1984 = 0</math>. We are given that <math>r_1 r_2 = -32</math>. By Vieta's formulas, we have: \begin{align*} \label{eq:1} 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\end{align*} Since <math>r_1r_2 = -32</math>, we have <math>(-32)r_3r_4 = -1984</math>, so <math>r_3r_4 = \frac{-1984}{-32} = 62</math>. Also, <math>r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -200</math>, so <math>r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2) = -200</math>. Substituting <math>r_1r_2 = -32</math> and <math>r_3r_4 = 62</math>, we have <math>-32(r_3+r_4)+62(r_1+r_2) = -200</math>. Let <math>A = r_1+r_2</math> and <math>B = r_3+r_4</math>. Then <math>A+B = 18</math>, so <math>B = 18-A</math>. Substituting this into the equation, we have <math>-32B+62A = -200</math>, so <math>-32(18-A)+62A = -200</math>. <math>-576+32A+62A = -200</math>, so <math>94A = 376</math>, which means <math>A = \frac{376}{94} = 4</math>. Then <math>B = 18-A = 18-4 = 14</math>. So we have <math>r_1+r_2 = 4</math> and <math>r_1r_2 = -32</math>. Then <math>r_1</math> and <math>r_2</math> are roots of <math>x^2-4x-32 = 0</math>, so <math>(x-8)(x+4) = 0</math>, which means <math>r_1 = 8</math> and <math>r_2 = -4</math> (or vice versa). Also we have <math>r_3+r_4 = 14</math> and <math>r_3r_4 = 62</math>. Then <math>r_3</math> and <math>r_4</math> are roots of <math>x^2-14x+62 = 0</math>. Using the quadratic formula, <math>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}</math>. Then <math>r_3 = 7+i\sqrt{13}</math> and <math>r_4 = 7-i\sqrt{13}</math>. We want to find <math>k = r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4</math>. <math>k = r_1r_2+r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86</math>.
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<cmath>\begin{align*}
 
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a+b+c+d &= 18\\
 
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ab+ac+ad+bc+bd+cd &= k\\
 
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abc+abd+acd+bcd &=-200\\
 
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abcd &=-1984.\\
 
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\end{align*}</cmath>
 
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Since <math>abcd=-1984</math> and <math>ab=-32</math>, then, <math>cd=62</math>. Notice that<cmath>abc + abd + acd + bcd = -200</cmath>can be factored into<cmath>ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).</cmath>From the first equation, <math>c+d=18-a-b</math>. Substituting it back into the equation,<cmath>-32(18-a-b)+62(a+b)=-200</cmath>Expanding,<cmath>-576+32a+32b+62a+62b=-200 \implies 94a+94b=376</cmath>So, <math>a+b=4</math> and <math>c+d=14</math>. Notice that<cmath>ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)</cmath>Plugging all our values in,<cmath>-32+62+4(14)=\boxed{86}.</cmath>
 
  
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~ kante314
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Then <math>k = \sum_{i < j} r_i r_j = r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 = r_1r_2 + r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86</math>.
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~ avm2023
  
 
== Solution 4 (Alcumus)==
 
== Solution 4 (Alcumus)==

Revision as of 15:46, 28 July 2025

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: \begin{align*} \label{eq:1} 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\end{align*} 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 $k = r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4$. $k = r_1r_2+r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86$.

Then $k = \sum_{i < j} r_i r_j = r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 = r_1r_2 + r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86$.

~ 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 \begin{align*} a + b &= -18, \\ ab + c - 32 &= k, \\ ac - 32b &= 200, \\ -32c &= -1984. \end{align*}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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