Finding $a+b+c+d$, where $ab+c+d=15$, $bc+d+a=24$, $cd+a+b=42$, $da+b+c=13$

Question

Let $a,b,c,d \in \mathbb{R}$. Consider the following constraints:

\begin{cases} ab+c+d=15 \\ bc+d+a=24 \\ cd+a+b=42 \\da+b+c=13 \end{cases}

Calculate the value of $a+b+c+d$.

It is easy to use the Gröbner basis to get the value:

\begin{cases} 10849-4501 d+380d^2,-39409+2320c+3420d,-20+29b-9d,1801+2320 a-380 d\} \end{cases}

so the value of $a+b+c+d$ is $\frac{169}{10}$.

What I am curious about is how to use high schools mathematics to get an answer without too much complicated mathematical calculations ?

Answer 1

Playing around with the four LHSs, I tried to obtain polynomials with some symmetry. I first noted that $$\left((ab+c+d)+(cd+a+b)\right)-\left((bc+d+a)+(da+b+c)\right)=(b-d)(a-c)$$ then after replacing the $+$ sign with $\cdot$, we get $$(ab+c+d)\cdot(cd+a+b)-(bc+d+a)\cdot(da+b+c)=(b-d)(a-c)(a+b+d+c-1).$$ Putting all together, $$\begin{align} &\frac{(ab+c+d)(cd+a+b)-(bc+d+a)(da+b+c)}{(ab+c+d)+(cd+a+b)-(bc+d+a)-(da+b+c)}\\&\qquad= \frac{(b-d)(a-c)(a+b+d+c-1)}{(b-d)(a-c)}=a+b+c+d-1. \end{align}$$ Hence with the given values we find $$a+b+c+d=1+\frac{15\cdot 42-24\cdot 13}{15+42-24-13}=1+\frac{318}{20}=\frac{169}{10}.$$

Answer 2

Let $\,x=a+b+c+d\,$ then:

$$ab+c+d=ab\color{red}{-a-b+1-1+a+b}+c+d=(a-1)(b-1)-1+x\,$$

The system can then be written as:

$$ \begin{cases} (a-1)(b-1)=16-x \\ (b-1)(c-1)=25-x \\ (c-1)(d-1)=43-x \\ (d-1)(a-1)=14-x \end{cases} $$

It follows that:

$$ (a-1)(b-1)(c-1)(d-1) \;=\; (16-x)(43-x) \;=\; (25-x)(14-x) $$

The latter equality gives $\,2\left(169 - 10 x\right) = 0\,$.