JHMT 05 Algeba Test Solutions 4 Febuay 05. In a Supe Smash Bothes tounament, of the contestants play as Fox, 3 of the contestants play as Falco, and 6 of the contestants play as Peach. Given that thee wee 40 moe people who played eithe Fox o Falco than who played Peach, how many contestants attended the tounament? Answe: 60 Solution: Let x denote the numbe of contestants in the tounament. Then x + 3 x 6 x = 40. Thus, 3x = 40 and hence x = 60 contestants attended the tounament.. Find all pais (x, y) that satisfy x + y = x +y = Answe: (0, ) and ( 4 5, 3 5 ) Solution: The second equation tells us that x = we have y. Substituting this into the fist equation, ( y) + y = 5y 8y +3=0 (5y 3)(y ) = 0 Thus, y = 3 5 o y =. Plugging in these values of y and solving fo x, we find that the possible 4 solutions to the system ae (0, ), 5, 3. 5 3. Find the unique x>0 such that p x + p x + p x =. Answe: 9 Solution: We solve p x + q x + p x = q x + p x = x + p x =( x + p x =+x 3 p x = p x p x) p x x = 9 4. Find the sum of all eal oots of x 5 +4x 4 + x 3 x 4x. Answe: 3
JHMT 05 Algeba Test Solutions 4 Febuay 05 Solution : Notice that x 5 +4x 4 + x 3 x 4x =x 3 (x +4x + ) (x +4x + ) = (x 3 )(x +4x + ). The only eal oot of x 3 is, and the eal oots of x +4x + ae ± p 3 by the quadatic fomula. Thus, the sum of all eal oots of the polynomial is +( + p p 3) + ( 3) = 3. Solution : By the ational oot theoem, we quickly discove that is a oot of the quintic polynomial. Factoing x out, we ae left with the quatic x 4 +5x 3 +6x +5x +. This polynomial is symmetic, so we may facto it as x ((x + x ) + 5(x + x ) + 4). Thus, it su ces to find the oots of (x + x ) + 5(x + x ) + 4. Witing y = x + x,wehavey +5y + 4 which has oots y =, 4. Thus, the oots of the quatic ae the solutions to the equations x + x = and x + x = 4. Solving the fist equation, we have x + x = x + x +=0 which has no eal oots. The second equation gives x + x = 4 x +4x +=0 which has the oots x = ± p 3. Theefoe, the sum of all eal oots of the polynomial is + ( + p 3) + ( p 3) = 3. 5. Let a, b, c, d be integes with no common diviso such that Compute a + b + c + d. Answe: 6 a 3p 4+b 3p +c d = 3p 4+ 3p + Solution: If we let a, b, c, d be ational numbes, then the solution is defined up to a scaling facto. Thus, we will fist solve fo ational a, b, c assuming d = and then scale the solution such that a, b, c, d ae all integes with no common diviso. Now, we wish to find ational a, b, c such that a 3p 4+b 3p +c = and simplifying, we obtain the equation: 3p 4+ 3p + (a + b +c) 3p 4+(4a + b + c) 3p +(a +4b + c ) = 0. Coss multiplying Since a, b, c ae to be ational, the coe cients of zeo. This gives the system of equations: 3p 4, 3 p, and the constant tem must all be a + b +c =0 4a + b + c =0 a +4b + c = Solving this system, we find that a = 3, b = 7 3, and c = 3 3. This was unde the assumption that d =, so if we scale eveything up by 3, we find that a =, b = 7, c = 3, and d = 3. Theefoe, a + b + c + d = 6.
JHMT 05 Algeba Test Solutions 4 Febuay 05 Solution: Thee must exist polynomials p(x) = ax + bx + c and q(x) = dx + e so that p(x)(x + x + ) + q(x)(x 3 ) = fo all x, as the polynomials ae elatively pime. We expand the expession to obtain (a + d)x 4 +(a +b + e)x 3 +(a + b +c)x +(b + c d)x +(c e) =, whee all of these coe cients ae equal to 0 except fo the last one which is equal to. Thus, we have a system of linea equations. Howeve we solve the system, we find that a = 3,b= 7 3,c= 3 3.Whenwepluginx = 3p into the polynomial expession, we get 3p 4+ 3p = 3p 4+7 3p 3 + 3 so a =, b = 7, c = 3, and d = 3. Theefoe, a + b + c + d = 6. 6. Let f(a, b) = a+b. Suppose that x, y, z ae distinct integes such that x + y + z = 05 and f(f(x, y),z)=f(x, f(y, z)). Compute y. Answe: 05 Solution: Since f(f(x, y),z)=f(x, f(y, z)), we have that Simplifying, we have that x+y + z = x + y+z x + y + z = x + y + z x(y + z)(x + y)+(x + y) =(y + z)+z(x + y)(y + z) x ((y + z)(x + y) + ) = z ((x + y)(y + z) + ) (x z)((y + z)(x + y) + ) = 0 Since x and z ae distinct, x z 6= 0 so we may divide though by x z to obtain (y + z)(x + y)+=0 (y + z)(x + y) = Since x + y + z = 05, y + z = 05 x and x + y = 05 z so (05 x)(05 z) = Since x, y, z ae integes, we have that x = 04 and z = 06 (o the othe way aound). In eithe case, y = 05 x z = 05 04 06 = 05. 7. Compute all pais (a, b) such that (x + ax + b) + a(x + ax + b) b =(x ) 4 fo some eal numbe. Answe: (0, 0),, 4 Solution: Let P (x) =x + ax + b and Q(x) =x + ax b. Then we ae looking fo a, b such that Q(P (x)) has only a single eal epeated oot. The oots of Q(P (x)) ae the solutions to the equations P (x) = and P (x) = whee, ae the oots of Q. Thus, a necessay condition is fo Q(x) to have a epeated oot, so we equie that the disciminant a +4b = 0. Then the epeated oot of Q is = a.
JHMT 05 Algeba Test Solutions 4 Febuay 05 Now, we equie that P (x) = a have a epeated oot, so the disciminant of P (x)+ a must be zeo. Theefoe, we equie that a 4(b + a ) = 0. Fom ealie, we know that a +4b = 0, so we may substitute in b = a 4. Hence, we have the equation a 4( a 4 + a )=0 a a =0 a(a ) = 0 So a =0, with coesponding b =0, 4. Thus, the desied pais of a and b ae (0, 0),, 8. Let a, b, c, d satisfy Find the geatest possible value of a. Answe: p 30 ab + cd = ac + bd = 3 ad + bc = 7 abcd = 30. 4 Solution: Note that ab, cd ae oots of the quadatic equation x x+30 because ab+cd = and ab cd = abcd = 30. But this clealy has oots 5, 6, thus {ab, cd} = {5, 6}. Similaly, we must have that {ab, cd} = {5, 6}, {ac, bd} = {3, 0}, {ad, bc} = {, 5}. But we have that a = ab ac ad abcd is maximized when we maximize ab, ac, ad. Using ou pevious esult, we set (ab, cd) =(6, 5), (ac, bd) = (0, 3), (ad, bc) = (5, ) and conclude that the maximum of a is 6 0 5 30 = 30, thus a apple p p30, q q q 6 30. Note that (a, b, c, d) = 5, 0 3, 5 satisfies the equation (b,c,d wee found in analogous ways to a ), thus we have a = p 30 is the geatest possible value of a. 9. Given that eal numbes x, y satisfy the equation x 4 + x y + y 4 = 7, what is the minimum possible value of x + xy +y? Answe: 6 p 6 Solution : Let a = x + xy + y and b = x xy + y.thenab = x 4 + x y + y 4 = 7 and the expession we ae tying to minimize is a+b 7 + a. Substitutingb = a, the expession wo want to q minimize becomes 3a + 36 a. By AM-GM, this is geate than o equal to 3a 36 a =p 54 = 6 p 6. Finally, we show 6 p 6 is achievable. In AM-GM, equality is only achieved when 3a = 36 a so a = p 6. If we let x = p p 6 and y = p p 6, then a = p 6so6 p 6 is achievable. Hence, the minimum possible value is 6 p 6. Solution : Let a = x +y and b = xy. Then we ae tying to minimize a+b. The expession x 4 +x y +y 4 can be factoed as (x +y ) (xy) =(a b)(a+b). Thus, we have the condition
JHMT 05 Algeba Test Solutions 4 Febuay 05 (a b)(a + b) = 7. Now, notice that x + y > xy fo any choice of x, y, so both a b and a + b ae always positive. Now, suppose a b =. Then the condition (a b)(a + b) = 7 tells us that a + b = 7. Hence, we can solve the system of equations to obtain a = 36 + a + b = 08 a b = a + b = 7 36 and b = q +. By AM-GM, this is geate than o equal to. The expession we ae tying to minimize is thus 08 =6p 6. Finally, we show that this is achievable. Equality holds in AM-GM pecisely when 08 = o when =6 p 6. Then a =4 p 6 and b = p 6. So this is achievable by finding x, y such that x + y =4 p 6 and xy = p 6. One such x, y is x = p p 6 and y = p p 6. Thus, the minimum possible value is indeed 6 p 6. 0. Conside a sequence defined ecusively by a n = + (a 0 + )(a + ) (a n + ). Find <a 0 < such that 05 X a n a n = a 0 +4 a 0 Answe: 3 05 n=0 Solution: Fist, notice that fo n>, a n =+(a 0 + )(a + ) (a n + )(a n + ) =+(a n )(a n + ) = a n and that a = a 0 +. Theefoe, fo n, a n =(a 0 + ) n. Next, notice that fo n>0 a n a n+ = (a 0 + )(a + ) (a n + ) a n = (a 0 + ) (a n + ) a n = (a n )(a n + ) = a n a n Theefoe, ou sum telescopes as follows 05 a 0 a 0 + X n= a n a n = a 05 0 a 0 + X n= = a 0 a 0 + a a n (a 0 + )(a + ) (a n + ) a 06 a n+
JHMT 05 Algeba Test Solutions 4 Febuay 05 Using the fact that a = a 0 + and a 06 =(a 0 + ) 05, we can simplify the above expession to a 0 a 0 + a 0 + (a 0 + ) 05 If we let x = a 0 +, then the condition in the poblem statement becomes x (x ) + x x (x )(x 3) + x 3x 3 (x )(x 3) 3 x 3 x 05 = x + (x ) x 05 =0 x 05 =0 x 05 =0 3(x 05 ) (x 3) = 0 3x 05 x =0 Since we want <a 0,0<x, so we can divide though by x to get 3x 05 =0 x =3 05 And hence a 0 = 3 05.