The product of two of the four roots of the quartic equation x4−18x3+kx2+200x−1984=0 is −32. Determine the value of k.
Problem
Source: 1984 USAMO #1
Tags: algebra, polynomial, AMC, USAMO
11.08.2009 20:04
i think you shuold put the solution and problem in different posts...
11.08.2009 20:10
11.08.2009 20:24
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.
11.08.2009 23:28
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)
11.09.2009 06:12
16.02.2012 00:50
18.10.2016 01:09
Nvm this doesn't work
26.11.2019 19:24
Let a,b,c,d be roots of the polynomial. By Vieta's formulas: a+b+c+d=18ab+ac+ad+bc+bd+cd=kabc+abd+acd+bcd=−200abcd=−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)=18ac+ad+bc+bd+30=(a+b)(c+d)+30=k−32c−32d+62d+62d=−200Letting 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=86.
11.05.2020 16:37
I wish USAMO was this easier nowadays
11.05.2020 16:38
Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf
11.05.2020 16:40
srisainandan6 wrote: Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf yup
20.07.2020 17:16
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: ⟹a+b+c+d=18⟹ab+ac+ad+bc+bd+cd=k⟹abc+abd+bcd+acd=−200⟹abcd=−1984Note that: ⟹abcdab=62=cd⟹ab+ac+ad+bc+bd+cd=−32+ac+ad+bc+bd+62=k ⟹ac+ad+bc+bd=k−30We can factor ac+ad+bc+bd as (a+b)(c+d). ⟹(a+b)(c+d)=k−30Note that: ⟹abc+abd+bcd+acd=ab(c+d)+cd(a+b)=−200We know that a+b+c+d=18 so a+b=18−(c+d). We also figured ab=−32 and cd=62. ⟹−32(18−(c+d))+62(c+d)=−200⟹−576+32c+32d+62c+62d=−200⟹94c+94d=376⟹94(c+d)=376⟹c+d=4⟹a+b=14(or vice versa). Hence: ⟹(14)(4)=k−30⟹56=k−30⟹k=86◼ credits to my blog
20.07.2020 18:21
There is a much easier way (in my opinion)
Shoot this is the same as #6
20.07.2020 18:24
@above i think you meant 86 but nice solution
20.07.2020 19:04
OlympusHero wrote: There is a much easier way (in my opinion) We can factor the quartic as (x2+ax−32)(x2+bx+62). Expanding this gives x4+(a+b)x3+(ab+30)x2+(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=84, the answer. Shoot this is the same as #6 nice solution but imo system of equations is kinda easier.
20.07.2020 19:25
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 (x2+ax−32)(x2+bx+62). Expanding this gives x4+(a+b)x3+(ab+30)x2+(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=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.
20.07.2020 19:31
Hello, OlympusHero's Method is faster, but ATGY's is easier to think of.
20.07.2020 19:38
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
20.07.2020 19:43
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.
15.08.2021 02:40
As seen in my solution, both methods ultimately boil down to the same system.
15.08.2021 03:18
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⋅32−200=376Thus, a+b=4,c+d=14. Now, we may answer extract k=ab+cd+(a+b)(c+d)=30+56=86
24.08.2021 03:42
My solution from WOOT: Let the roots of the quartic be a,b,c,d with ab=−32. By Vieta's, we have: {a+b+c+d=18ab+ac+ad+bc+bd+cd=kabc+bcd+cda+dab=−200abcd=−1984Using ab=−32, this is: {a+b+c+d=18ac+ad+bc+bd+cd=k+32bcd+cda−32c−32d=−200cd=62⇒{a+b+c+d=18(a+b)(c+d)=k−3062a+62b−32c−32d=−200cd=62Let 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 k=86.
18.10.2022 22:50
The answer is 86. By Vieta we have ∑ri=18∑rirj=k∑rirjrk=−200r1r2r3r4=−1984Let WLOG r1r2=−32. Using (2) we have r3r4=−1984/−32=62. Using r1r2, r3r4 and equation (1) in equation (3) we have 62(r1+r2)−32(r3+r4)=−200Rewriting the equation (2) we have (r1+r2)(r3+r4)+30=kSolving the equation (1) and (5) in terms of variable r1+r2 and r3+r4 we get 4,14. Putting these value in the equation (6) we get out answer k=86.
19.12.2022 05:49
Let x4−18x3+kx2+200x−1984=(x2+ax−32)(x2+bx+32)for some a,b. Equating the x3 and x coefficients yields a+b=−1862a−32b=200so a=−4,b=−14, and k=56−32+62=86.
20.04.2023 07:52
21.04.2023 16:08
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=86
31.05.2023 14:49
Assume the roots of the equation x4−18x3+kx2+200x−1984=0 are α,β,γ,δ Given product of any two roots of this equation is -32. By Vietas Relation we get α+β+γ+δ=18αβ+βγ+γα+αδ+βδ+γδ=kαβγ+βγδ+αβδ+αγδ=−200αβγδ=−1984Lets assume αβ=−32 from product of roots we get αβγδ=−1984⟹γδ=62 αβγ+βγδ+αγδ+αβδ=−200⟹62(α+β)−32(γ+δ)⟹62(α+β)−32(18−(α+β))=−200⟹94(α+β)=376⟹α+β=4From these we also get γ+δ=14 Therefore, αβ+βγ+γα+δα+βδ+γδ=k⟹(α+β)(γ+δ)+αβ+γδ=14.4+62−32=56+30⟹86.Is the correct answer... Vietas Relation
01.08.2023 05:59
Let the roots be r1,r2,r3,r4. By Vieta, if r1r2=−32, then r3r4=62. Additionally, Vieta gives r1r2r3+r1r2r4+r1r3r4+r2r3r4=−200⟹−32(r3+r4)+62(r1+r2).But notice that r3+r4=18−(r1+r2) again by Vieta. Then, solving we get that r1+r2=4 and r3+r4=14. Finally, by Vieta, we see have k=r1r2+r1r3+r1r4+r2r3+r2r4+r3r4=(r1r2+r3r4)+(r1r3+r1r4+r2r3+r2r4)=−32+62+(r1+r2)(r3+r4)=30+56=86,and we're done.
02.08.2023 00:22
easiest usamo ever
13.10.2023 19:59
24.12.2023 21:55
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.
05.05.2024 16:21
By Vieta's we have that 1) x1+x2+x3+x4=18, 2) x1x2+x1x3+x1x4+x2x3+x2x4+x3x4=k, 3) x1x2x3+x1x2x4+x1x3x4+x2x3x4=−200, 4) x1x2x3x4=−1984. Now we can assume x1x2 = -32 ⇒ x3x4 = 62. We now plug this into 3) and we get −32x3−32x4+62x1+62x2=62(x1+x2)−32(x3+x4) and after using 1) we get 62(x1+x2)−32(18−x1−x2)=−200. Let x1+x2=a, we get 62a−32(18−a)+200=94a−576+200=94a−376=0 ⇒ a = 4 ⇒ x1+x2=4 and x3+x4=14. Now we plug in everything we know in 2) so x1x2+x1x3+x1x4+x2x3+x2x4+x3x4=−32+x1(x3+x4)+x2(x3+x4)+62=30+(x1+x2)(x3+x4)=30+4.14=30+56=86=k ⇒ k = 86 ⇒ the answer is 86.
18.08.2024 23:46
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: x4−18x3+kx2+200x−1984=(x2+ax−32)(x2+bx+62) By matching coefficients, we find the following relationships: a+b=−18,ab+30=k,62a−32b=200. 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=86.
02.01.2025 00:39
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.