Determine all real numbers $a, b, c, d$ for which $$ab + c + d = 3$$$$bc + d + a = 5$$$$cd + a + b = 2$$$$da + b + c = 6$$
Problem
Source: 2021 JBMO TST Bosnia and Herzegovina P1
Tags: algebra, system of equations
08.10.2022 00:08
nvm messed up
08.10.2022 00:22
08.10.2022 01:38
MathsSolver007 wrote:
reals, not integers
08.10.2022 02:51
4-1 $da+b+c-ab-c-d=3$ $da+b-ab-d=3$ $d(a-1)+b(1-a)=3$ $(d-b)(a-1)=3$ 2-3 $bc+d+a-cd-a-b=3$ $bc+d-cd-b=3$ $b(c-1)+d(1-c)=3$ $(b-d)(c-1)=3$ $(b-d)(c-1)=(d-b)(a-1)$ $c=2-a$ (1)+(2) $b+d=3$ $a+b-ab=2$ $(2-a)(3-b)+a+b=2$ $b(a-1) = 2(a-2)$ $b(a-1) =a-2$ $(a,b,c,d) = (2,0,0,3)$
08.10.2022 08:14
pinkpig wrote: MathsSolver007 wrote:
reals, not integers I did not assume that they are integers
08.10.2022 16:40
o oops i'm stupid and can't read properly
25.05.2023 11:31
My solution is the same as @MathsSolver007 so im not gonna post it again
20.02.2024 17:26
Take $k = a + b + c + d$ and $A = a-1, B = b-1, C = c-1, D = d-1$. One immediately has $$AB = 4-k, \quad BC = 6-k, \quad CD = 3-k, \quad DA = 7-k.$$Hence $ABCD = (4-k)(3-k) = (6-k)(7-k)$ which gives $k = 5$. This allows us to write $$B = -\frac1A, \quad C = -A,\quad D = \frac2A \text{ with constraint } k = A+B+C+D+4 = 5.$$Solving this yields the desired answer.