Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1984 USAMO Problems/Problem 1"
(→‎Solution 3)
 
(24 intermediate revisions by 11 users not shown)
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==
+
== Solution 1 (ingenious)==
  
−
Let the four roots be <math>a,b,c,d</math>. By Vieta's Formulas, we have the following:
 
  
−
*<math>ab = - 32</math>
+
Using Vieta's formulas, we have:
−
*<math>abcd = - 1984</math>
 
−
*<math>cd = 62</math>
 
−
*<math>a + b + c + d = 18</math>*
 
−
*<math>ab + ac + ad + bc + bd + cd = k</math>
 
−
*<math>abc + abd + acd + bcd = - 200</math>
 
  
−
Substituting given values obtains the following:
+
<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>
  
−
*<math>ac + ad + bc + bd = k - 62 + 32 = k - 30</math>
+
−
*<math>(a + b)(c + d) = k - 30</math>
+
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>.
−
*<math>- 32c + 62b + 62a - 32d = - 200</math>
 
−
*<math>31(a + b) - 16(c + d) = - 100</math>
 
  
−
Multiplying the * equation by 16 gives
+
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>.
  
−
<math>16(a + b) + 16(c + d) = 288</math>. Adding it to the previous equation gives
+
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>.
  
−
*<math>47(a + b) = 188</math>
+
== Solution 2 (cool)==
−
*<math>a + b = 4</math>
+
 
−
*<math>c + d = 18 - 4 = 14</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>.
−
*<math>(a + b)(c + d) = 4(14) = 56 = k - 30</math>
+
 
−
*<math>k = \boxed{86}</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,
 +
<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>
 +
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
 +
 
 +
== 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 ==
 +
https://youtu.be/Dp-pw6NNKRo?t=316
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
  
 
{{USAMO box|year=1984|before=First<br>Problem|num-a=2}}
 
{{USAMO box|year=1984|before=First<br>Problem|num-a=2}}
 +
{{MAA Notice}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 18:25, 28 March 2024

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&oldid=217550"