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 | + | == 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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Now | Now | ||
− | <cmath> \begin{align*}x^4-18x^3+kx^2+200x-1984 & | + | <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> | ||
+ | |||
+ | == 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 thatcan be factored intoFrom the first equation, . Substituting it back into the equation,Expanding,So, and . Notice thatPlugging 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
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.