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"

m (→‎Solution)
m (→‎Solution)
Line 7: Line 7:
 
Let the four roots be <math>a, b, c,</math> and <math>d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see.  
 
Let the four roots be <math>a, b, c,</math> and <math>d</math>, so that <math>ab=-32</math>. From here we show two methods; the second is more slick, but harder to see.  
  
−
Solution #1  
+
=== Solution 1 ===
  
 
Using Vieta's formulas, we have:  
 
Using Vieta's formulas, we have:  
Line 20: Line 20:
 
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 (sketch)
+
=== Solution 2 ==
  
 
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>.  
Line 26: Line 26:
 
Now  
 
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>
+
<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>
 
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>

Revision as of 12:28, 18 July 2016

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

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:

\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

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 $k=\boxed{86}.$

See Also

1984 USAMO (Problems • Resources)
Preceded by
First
Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&oldid=79399"