Given $a,b,c \in \mathbb{R}$, and $x,y,z \in \mathbb{R}$, with $x,y,z$ not all zero. Find $a^2+b^2+c^2+2abc$

Question

This question is to be solved in about $3$ minutes, without a calculator.

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Let $a,b,c$ be any real numbers. Suppose that $x,y,z$ are real numbers, not all simultaneously zero, such that

$$x=cy+bz$$

$$y=cx+az$$

$$z=bx+ay$$

What is the value of $a^2+b^2+c^2+2abc$?

$\text{(A) }0\space\space\space\space\space\text{(B) }\frac{1}{2}\space\space\space\space\space\text{(C) }\frac{\sqrt{3}}{3}\space\space\space\space\space\text{(D) }1\space\space\space\space\space\text{(E) }5$

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My attempt (requires a long time, therefore not a feasible way to be used in the exam):

Start with rewriting the $3$ equations as

$$\text{something}=0$$

$$\text{something}=0$$

$$\text{something}=0$$

These are linear homogenous equations, and will have a non-trivial solution if the determinant of the coefficient matrix of $(\text{something})$'s is equal to zero.

To evaluate this determinant, easily, we carryout a couple of raw operations, ...

Opening the determinant and simplifying, we get $a^2+b^2+c^2+2abc=1$, which is option $\text{(D)}$

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So I am not totally stuck on solving this problem, but I am not sure if there is a clever trick to solve it under $3$ minutes.

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Your help would be appreciated. THANKS!

Answer 1

This is just a trick, not a complete algebraic answer .

Let $a=0$ and $x\neq 0$ . We get $x=c^2x+b^2x=(b^2+c^2)x$, which gives $b^2+c^2=1$ . Hence, the answer is $1$ .

Answer 2

This was a way to test your attention in class. You have a linear homeneous system $-x+cy+bz=cx-y+az=bx+ay-z=0$ which has a non trivial solution (= non zero) if and only if the determinant of the system (namely $-1+2abc+a^2+b^2+c^2$) is zero.