Difference between revisions of "1984 USAMO Problems/Problem 1"
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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 1 (ingenious)== | |
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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 (cool)== | |
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>. | ||
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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> | ||
− | + | == 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== |
Latest revision as of 19:25, 28 March 2024
Contents
Problem
In the polynomial , the product of
of its roots is
. Find
.
Solution 1 (ingenious)
Using Vieta's formulas, we have:
From the last of these equations, we see that . Thus, the second equation becomes
, and so
. The key insight is now to factor the left-hand side as a product of two binomials:
, so that we now only need to determine
and
rather than all four of
.
Let and
. Plugging our known values for
and
into the third Vieta equation,
, we have
. Moreover, the first Vieta equation,
, gives
. Thus we have two linear equations in
and
, which we solve to obtain
and
.
Therefore, we have , yielding
.
Solution 2 (cool)
We start as before: and
. We now observe that a and b must be the roots of a quadratic,
, 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
.
Now
Equating the coefficients of and
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
and get
Solution 3
Let the roots of the equation be and
. By Vieta's,
Since
and
, then,
. Notice that
can be factored into
From the first equation,
. Substituting it back into the equation,
Expanding,
So,
and
. Notice that
Plugging all our values in,
~ kante314
Solution 4 (Alcumus)
Since two of the roots have product the equation can be factored in the form
Expanding, we get
Matching coefficients, we get
\begin{align*}
a + b &= -18, \\
ab + c - 32 &= k, \\
ac - 32b &= 200, \\
-32c &= -1984.
\end{align*}Then
so
With
we can solve to find
and
Then
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.