i think you shuold put the solution and problem in different posts...

Your solution is so much shorter than mine. Instead of doing that adding, I substituted a+b for 18-c-d which made it take much longer.

oh, 2 more things. 1. take out "And again, how do I add this to the resources section?" 2. change the title. it shouldnt have vieta because it hints at the solution(regardless of how obvious it may be)

Nvm this doesn't work

Let $a,b,c,d$ be roots of the polynomial. By Vieta's formulas: $$a+b+c+d = 18$$ $$ab+ac+ad+bc+bd+cd = k$$ $$abc+abd+acd+bcd =-200$$ $$abcd =-1984.$$ Without loss of generality, assume that $ab=-32$, then by the last equation, $cd=62$. We can re-write the equations as follows, $$(a+b)+(c+d) = 18$$ $$ac+ad+bc+bd + 30 = (a+b)(c+d) + 30 = k$$ $$-32c-32d+62d+62d =-200$$ Letting $m=a+b$ and $n=c+d$, we see that $m+n = 18$, $-32n+62m=-200$, so $m=4$ and $n=14$. $k = mn + 30 = \boxed{86}$.

I wish USAMO was this easier nowadays

Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf

srisainandan6 wrote: Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf yup

Let $a$, $b$, $c$ and $d$ be the roots to this quartic equation. WLOG, let $a$ and $b$ be the roots to this quartic whose product is $-32$. By Vieta's Formula, we know that: $$\implies a + b + c + d = 18$$ $$\implies ab + ac + ad + bc + bd + cd = k$$ $$\implies abc + abd + bcd + acd = -200$$ $$\implies abcd = -1984$$ Note that: $$\implies \frac{abcd}{ab} = 62 = cd$$ $\implies ab + ac + ad + bc + bd + cd = -32 + ac + ad + bc + bd + 62 = k$ $$\implies ac + ad + bc + bd = k - 30$$ We can factor $ac + ad + bc + bd$ as $(a + b)(c + d)$. $$\implies (a + b)(c + d) = k - 30$$ Note that: $$\implies abc + abd + bcd + acd = ab(c + d) + cd(a + b) = -200$$ We know that $a + b + c + d = 18$ so $a + b = 18 - (c + d)$. We also figured $ab = -32$ and $cd = 62$. $$\implies -32(18 - (c + d)) + 62(c + d) = -200$$ $$\implies -576 + 32c + 32d + 62c + 62d = -200$$ $$\implies 94c + 94d = 376$$ $$\implies 94(c + d) = 376$$ $$\implies c + d = 4$$ $$\implies a + b = 14$$ (or vice versa). Hence: $$\implies (14)(4) = k - 30$$ $$\implies 56 = k - 30$$ $$\implies k = \boxed{86}$$ $\blacksquare$ credits to my blog

There is a much easier way (in my opinion)

Shoot this is the same as #6

@above i think you meant 86 but nice solution

OlympusHero wrote: There is a much easier way (in my opinion) We can factor the quartic as $(x^2+ax-32)(x^2+bx+62)$. Expanding this gives $x^4+(a+b)x^3+(ab+30)x^2+(62a-32b)x-1984$. We have $a+b=-18$, $62a-32b=200$. We know that $32a+32b=-576$, so $94a=-376$, which means that $a=-4$. So, $b=-14$, and we need to compute $ab+30=\boxed{84}$, the answer. Shoot this is the same as #6 nice solution but imo system of equations is kinda easier.

mathstats wrote: @above i think you meant 86 but nice solution Yeah, you're right, fixed. ATGY wrote: OlympusHero wrote: There is a much easier way (in my opinion) We can factor the quartic as $(x^2+ax-32)(x^2+bx+62)$. Expanding this gives $x^4+(a+b)x^3+(ab+30)x^2+(62a-32b)x-1984$. We have $a+b=-18$, $62a-32b=200$. We know that $32a+32b=-576$, so $94a=-376$, which means that $a=-4$. So, $b=-14$, and we need to compute $ab+30=\boxed{84}$, the answer. Shoot this is the same as #6 nice solution but imo system of equations is kinda easier. Maybe for you, but this method is easier than you think. It is called the Method of Undetermined Coefficients. In fact, you can use it to prove Vieta's Formulas.

Hello, OlympusHero's Method is faster, but ATGY's is easier to think of.

sshi wrote: Hello, OlympusHero's Method is faster, but ATGY's is easier to think of. Really? What first came to my mind was my method, I never even thought about ATGY's

OlympusHero wrote: sshi wrote: Hello, OlympusHero's Method is faster, but ATGY's is easier to think of. Really? What first came to my mind was my method, I never even thought about ATGY's oof lol ATGY's method is what I immediately thought of.

As seen in my solution, both methods ultimately boil down to the same system.

Let the roots be $a,b,c,d$. Then $ab=-32,cd=62$. Next, note $$(a+b)+(c+d) = 18$$ $$-200 = abc+abd+acd+bcd = -32(c+d)+ 62(a+b)$$ Thus, $$94(a+b) = 18\cdot 32 - 200 = 376$$ Thus, $a+b=4,c+d=14$. Now, we may answer extract $$k = ab+cd+(a+b)(c+d) = 30+56 = 86$$

My solution from WOOT: Let the roots of the quartic be $a,b,c,d$ with $ab=-32$. By Vieta's, we have: $$\begin{cases}a+b+c+d=18\\ab+ac+ad+bc+bd+cd=k\\abc+bcd+cda+dab=-200\\abcd=-1984\end{cases}$$ Using $ab=-32$, this is: $$\begin{cases}a+b+c+d=18\\ac+ad+bc+bd+cd=k+32\\bcd+cda-32c-32d=-200\\cd=62\end{cases}$$ $$\Rightarrow\begin{cases}a+b+c+d=18\\(a+b)(c+d)=k-30\\62a+62b-32c-32d=-200\\cd=62\end{cases}$$ Let $x=a+b$ and $y=c+d$. We have $x+y=18$ and $62x-32y=-200$. Solving, $x=4$ and $y=14$. Substituting into $xy=k-30$, we will find that $\boxed{k=86}$.

The answer is $86$. By Vieta we have $$\begin{align*} \sum r_i&=18\tag{$1$} \\ \sum r_ir_j&=k\tag{$2$} \\ \sum r_ir_jr_k&=-200\tag{$3$} \\ r_1r_2r_3r_4&=-1984\tag{$4$} \end{align*}$$ Let WLOG $r_1r_2=-32$. Using $(2)$ we have $r_3r_4=-1984/-32=62$. Using $r_1r_2$, $r_3r_4$ and equation $(1)$ in equation $(3)$ we have $$62(r_1+r_2)-32(r_3+r_4)=-200\tag{$5$}$$ Rewriting the equation $(2)$ we have $$(r_1+r_2)(r_3+r_4)+30=k\tag{$6$}$$ Solving the equation $(1)$ and $(5)$ in terms of variable $r_1+r_2$ and $r_3+r_4$ we get $4, 14$. Putting these value in the equation $(6)$ we get out answer $k=86$.

Let $$x^4-18x^3+kx^2+200x-1984 = (x^2+ax-32)(x^2+bx+32)$$ for some $a, b$. Equating the $x^3$ and $x$ coefficients yields $$\begin{align*} a+b &= -18 \\ 62a-32b &= 200 \end{align*}$$ so $a = -4, b = -14$, and $k = 56-32+62=\boxed{86}$.

By Veita the product of the other 2 roots are 62, let the 4 roots be $a,b,c,d$ by vieta's we have $a+b+c+d=18,ab+ac+Ad+bc+bd+cd$, and $abc+acd+abd+bcd=-200$ And then we can do some calculator from here to get that $k=\boxed{86}$

Assume the roots of the equation $x^4-18x^3+kx^2+200x-1984=0$ are $\alpha ,\beta ,\gamma ,\delta$ Given product of any two roots of this equation is -32. By Vietas Relation we get $$\alpha+\beta+\gamma+\delta=18$$ $$\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\delta+\beta\delta+\gamma\delta=k$$ $$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta=-200$$ $$\alpha\beta\gamma\delta=-1984$$ Lets assume $\alpha\beta=-32$ from product of roots we get $\alpha\beta\gamma\delta=-1984\implies\gamma\delta=62$ $$\alpha\beta\gamma+\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta=-200\implies62(\alpha+\beta)-32(\gamma+\delta)\implies62(\alpha+\beta)-32(18-(\alpha+\beta))=-200\implies94(\alpha+\beta)=376\implies\boxed{\alpha+\beta=4}$$ From these we also get $\boxed{\gamma+\delta=14}$ Therefore, $$\alpha\beta+\beta\gamma+\gamma\alpha+\delta\alpha+\beta\delta+\gamma\delta=k\implies(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=14.4+62-32=56+30\implies86.$$ Is the correct answer... Vietas Relation

Let the roots be $r_1, r_2, r_3, r_4$. By Vieta, if $r_1r_2=-32$, then $r_3r_4=62$. Additionally, Vieta gives $$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4=-200 \implies -32(r_3+r_4)+62(r_1+r_2).$$ But notice that $r_3+r_4=18-(r_1+r_2)$ again by Vieta. Then, solving we get that $r_1+r_2=4$ and $r_3+r_4=14$. Finally, by Vieta, we see have $$k=r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=(r_1r_2+r_3r_4)+(r_1r_3+r_1r_4+r_2r_3+r_2r_4)=-32+62+(r_1+r_2)(r_3+r_4)=30+56=86,$$ and we're done.

easiest usamo ever

Let the roots of the quartic be $a$, $b$, $c$ and $d$. WLOG, let $ab=-32$, and by Vieta's, we have that $abcd=-1984$, meaning that $cd=-62$. Let $x=a+b$ and $y=c+d$. By Vieta's, we have that $$18=a+b+c+d=(a+b)+(c+d)=x+y,$$ and $$-200=abc+abd+acd+bcd=cd(a+b)+ab(c+d)=62x-32y.$$ Solving the system of equations gives us that $x=4$ and $y=14$. Again using Vieta's, we have that $$k=ac+ad+bc+bd+ab+cd=(a+b)(c+d)+ab+cd=xy-32+62,$$ or $xy+30$, giving us that $k$ is $86$, finishing the problem.

By Vieta's we have that 1) $x_1 + x_2 + x_3 + x_4 = 18$, 2) $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = k$, 3) $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -200$, 4) $x_1x_2x_3x_4 = -1984$. Now we can assume $x_1x_2$ = -32 $\Rightarrow$ $x_3x_4$ = 62. We now plug this into 3) and we get $-32x_3 - 32x_4 + 62x_1 + 62x_2 = 62(x_1 + x_2) - 32(x_3 + x_4)$ and after using 1) we get $62(x_1 + x_2) - 32(18 - x_1 - x_2) = -200$. Let $x_1 + x_2 = a$, we get $62a - 32(18 - a) + 200 = 94a - 576 + 200 = 94a - 376 = 0$ $\Rightarrow$ a = 4 $\Rightarrow$ $x_1 + x_2 = 4$ and $x_3 + x_4 = 14$. Now we plug in everything we know in 2) so $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = - 32 + x_1(x_3 + x_4) + x_2(x_3 + x_4) + 62 = 30 + (x_1 + x_2)(x_3 + x_4) = 30 + 4.14 = 30 + 56 = 86 = k$ $\Rightarrow$ k = 86 $\Rightarrow$ the answer is 86.

The product of all the roots in this polynomial is $-1984$. Since we know the product of one pair of roots, we can find the product of the other pair of roots as $62$. We can split the original polynomial into two quadratic polynomials: $$x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + 62)$$ By matching coefficients, we find the following relationships: $$\begin{align*} a + b &= -18, \\ ab + 30 &= k, \\ 62a - 32b &= 200. \end{align*}$$ Using the first and last relationships, we solve for $a$ and $b$: $$(a, b) = (-4, -14).$$ Substituting into the equation $ab + 30 = k$, we find: $$k = \boxed{86}.$$

Let's assume that the biquadratic equation x⁴ - 18x³ + kx² + 200x - 1984 = 0 has roots a,b,c and d respectively such that ab = -32. By Vieta's Relation we've abcd = -1984 —————(1) abc + bcd + cda + dab = -200 ab(c + d) + cd(a + b) = -200 —————(2) ab + ac + ad + bc + bd + cd = k ab + cd + (a + b)c + (a + b)d = k ab + cd + (a + b)(c + d) = k —————(3) and, (a + b) + (c + d) = 18 or, (a + b) = 18 - (c + d) —————(4) Substituting ab = -32 in equation (1), we get (-32)cd = -1984 Therefore, cd = 62 On substituting ab = -32, cd = 62 and (a + b) = 18 - (c + d) in equation (2), we get -32(c + d) + 62[18 - (c + d)] = -200 1116 - 62(c + d) - 32(c + d) = -200 94(c + d) = 1316 Therefore, (c + d) = 14 Similarly, (a + b) = 18 - (c + d) = 4 Now, ab + cd + (a + b)(c + d) = k On substituting the values of ab,cd,(a + b) and (c + d), we get k = -32 + 62 + 4(14) k = 30 + 56 = 86 Therefore, the required value of k is 86.