Difference between revisions of "1984 USAMO Problems/Problem 1"
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=== Solution 3 === | === Solution 3 === | ||
Let the roots of the equatoon be <math>a,b,c,</math> and <math>d</math>. By Vieta's, | Let the roots of the equatoon be <math>a,b,c,</math> and <math>d</math>. By Vieta's, | ||
− | \begin{align*} | + | \begin{align*}a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} |
− | a+b+c+d=18 \\ | + | 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> |
− | 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 | ~ kante314 |
Revision as of 20:58, 28 August 2021
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 equatoon be and
. 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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.