Difference between revisions of "1984 USAMO Problems/Problem 1"
(→Solution 3) |
|||
Line 30: | Line 30: | ||
=== 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 19:58, 28 August 2021
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 equatoon be and . By Vieta's,
~ 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.