USA Junior Math Olympiad Problem | AIME 2023 | Algebra

[@devondevon4366]

6 and -10 for x, y and z Easy, another approach xy + 4z =yz + 4x since both = 60 4z-yz = 4x- xy z ( 4-y)= x (4-y) z=x zx + 4y =yz + 4x 4y-yz = 4x-zx y (4-z) = x(4-z) y= x Hence x=y=z x^2 + 4x= 60 substituting x=y=z into zx + 4y =60 x^2 + 4x -60 =0 (x-6)(x+10) =0 x = 6 and -10 hence y=z= 6 and -10

[@guyhoghton399]

_xy + 4z = 60_ ... ① _yz + 4x = 60_ ... ② _zx + 4y = 60_ ... ③ Subtract ② from ① to get _(x - z)(y - 4) = 0_ ... ④ and since you get the same equations ① - ③ by permuting x, y, z cyclically we can permute them in ④ to get _(y - x)(z - 4) = 0_ ... ⑤ _(z - y)(x - 4) = 0_ ... ⑥ *Case ⑴* _x = y = z_ Then ① becomes _x² + 4x - 60 = 0_ ⇒ _(x - 6)(x + 10) = 0_ ⇒ _x = y = z = 6_ or _x = y = z = -10_ *Case ⑵* At least one of _x, y, z_ is different from the other two. Let that be _x_ . _y ≠ x ≠ z_ ∴ from ⑤: _z = 4_ and from ④: _y = 4_ ∴ ① becomes _4x + 16 = 60_ ⇒ _4x = 44 ⇒ x = 11_ Again we can permute _x, y, z_ cyclically to get solutions where _y = 11_ and again where _z = 11_ . Hence the solutions are: (6, 6, 6), (-10, -10, -10), (11, 4, 4), (4, 11, 4), (4, 4, 11).

[@kaushikbasu9707]

4and 4 and 11 respectively

[@devondevon4366]

x, y, and z = 6 and -10

[@honestadministrator]

x y + 4 z = y z + 4 x = z x + 4 y = 60 (x - z)( y - 4) = (y - x)(z -4) =0 (z - y)( x -4) = 0 case I: y - 4 = 0 implies x z = 44 at x - 4 = 0 , z = 11 Hereby (x, y, z) € {(4, 4, 11),(4, 11, 4),(11, 4, 4)} case II : x = y = z , x^2 + 4 x - 60 = 0 (x + 10 )( x - 6) = 0 Hereby (x, y, z) € {(6, 6, 6),(-10, -10, -10)}