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

Revision as of 20:58, 28 August 2021

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

Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that $abc + abd + acd + bcd = -200$ can be factored into $ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).$ From the first equation, $c+d=18-a-b$ . Substituting it back into the equation, $-32(18-a-b)+62(a+b)=-200$ Expanding, $-576+32a+32b+62a+62b=-200 \implies 94a+94b=376$ So, $a+b=4$ and $c+d=14$ . Notice that $ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)$ Plugging all our values in, $-32+62+4(14)=\boxed{86}.$

~ kante314

Video Solution

https://youtu.be/5QdPQ3__a7I?t=589

~ pi_is_3.14

@@ 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>.
-==Solution==
+=== Solution 1 (ingenious)===
-Let the four roots be a, b, c, and d, so that ab=-32. From here we show two methods; the second is more slick, but harder to see.
-Solution #1
+Using Vieta's formulas, we have:
-Using Vieta's formulas, we have: \begin{align*} a+b+c+d &= 18, <br /> ab+ac+ad+bc+bd+cd &= k, <br /> abc+abd+acd+bcd &= -200, <br /> abcd &= -1984. <br /> \en... 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.
+<cmath>\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}</cmath>
-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.
+From the last of these equations, we see that <math>cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62</math>. Thus, the second equation becomes <math>-32+ac+ad+bc+bd+62=k</math>, and so <math>ac+ad+bc+bd=k-30</math>. The key insight is now to factor the left-hand side as a product of two binomials: <math>(a+b)(c+d)=k-30</math>, so that we now only need to determine <math>a+b</math> and <math>c+d</math> rather than all four of <math>a,b,c,d</math>.
-Therefore, we have (\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30, yielding k=4\cdot 14+30 = \boxed{86}.
+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>.
-Solution #2 (sketch)
+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>.
-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.
+=== Solution 2 (cool)===
-Now \begin{align*} x^4 - 18x^3 + kx^2 + 200x - 1984 &= (x^2 + rx - 32)(x^2 + sx + 62) <br /> & = x^4 + (r + s)x^3 + (62 - 32 ... 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}.
+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>.
+Now
+<cmath> \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*} </cmath>
+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>
+=== Solution 3 ===
+Let the roots of the equation be <math>a,b,c,</math> and <math>d</math>. By Vieta's,
+\begin{align*}
+a+b+c+d &= 18\\
+ab+ac+ad+bc+bd+cd &= k\\
+abc+abd+acd+bcd &=-200\\
+abcd &=-1984.\\
+\end{align*}
+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>
+~ kante314
+== 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

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