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== | + | == Solution 1 (ingenious)== |
− | |||
− | |||
− | |||
Using Vieta's formulas, we have: | Using Vieta's formulas, we have: | ||
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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>. | 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>. | ||
− | 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 18:25, 28 March 2024
Contents
[hide]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
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.