# Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$

This is a question from Math Olympiad.

If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.

I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.

• Could you provide some background about methods you’ve tried that have failed and about your mathematical skillset? Commented Feb 9, 2018 at 13:44
• Wouldn't this be solving multiple equations with multiple unknowns? Commented Feb 9, 2018 at 15:02
• If all else failed, I would personally try finding a Groebner basis of the corresponding ideal, and then if that doesn't directly give $x,y,z$, hopefully at least dividing $xy+yz+zx$ by the ideal would give a constant remainder. (Keeping my fingers crossed that the problem would also be solvable for $x,y,z \in \mathbb{C}$ and doesn't on the contrary require using some inequalities.) Commented Feb 9, 2018 at 20:16
• I added an answer with the Gröbner basis idea (before seeing your comment). It does involve using the inequality at some point, but it remains clean, fortunately ! Commented Feb 9, 2018 at 20:38
• My gut feeling (also because it is a quiz for young adults) is that there is some clever way to exploit the symmetry of the equations without going the whole 9 yards. Commented Feb 12, 2018 at 11:06

We can obtain $yz+zx+xy=2$ simply by finding the values of $x$, $y$, and $z$. We are given $$y^2+yz+z^2=1,\qquad(1)$$$$z^2+zx+x^2=4,\qquad(2)$$$$x^2+xy+y^2=3,\,\qquad(3)$$with $x,y,z>0$. Subtracting eqn $3$ from eqn $2$, and noticing the factor $z-y$ on the LHS, gives $$(z-y)(x+y+z)=1.$$Similarly, from eqn $2$ minus eqn $1$, we get$$(x-y)(x+y+z)=3.$$The ratio of the latter two equations yields $x-y=3(z-y)$, or$$x=3z-2y.$$ Now substituting for $x$ from the above equation into eqn $3$ yields$$y^2-3yz+3z^2=1.$$By subtracting this from eqn $1$, we find $2yz-z^2=0$, or$$z=2y$$(since $z\neq0$). Substitution for $z$ into eqn $1$ now gives $7y^2=1$. It follows that $y=1/\surd7$, because $y>0$, and then $z=2/\surd7$ and $x=4/\surd7$, giving$$yz+zx+xy=2.$$

• Nice and straightforward, no magic involved. Gets my vote! Commented Feb 10, 2018 at 3:26
• I like this solution especially because I hate geometry. Commented Feb 10, 2018 at 6:05
• Precisely my solution, +1. Commented Feb 11, 2018 at 20:39

To my surprise, this problem can be solved using geometry.

Identify the Euclidean plane with complex plane $\mathbb{C}$. and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity.

Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$.
The sides of the triangle equal to

\begin{align} a^2 = BC^2 &= | y\omega - z\omega^2|^2 = y^2 + yz+ z^2 = 1\\ b^2 = AC^2 &= | x - z\omega^2|^2 = x^2 + xz + z^2 = 4\\ c^2 = AB^2 &= | x - y\omega|^2 = x^2+xy+y^2 = 3 \end{align} Since $a^2 + c^2 = b^2$, this is a right angled triangle with area $\mathcal{A} = \frac12 ac =\frac{\sqrt{3}}{2}$.

An alternate way to compute the area is cut the triangle into 3 pieces along the line $OA$, $OB$ and $OC$. This gives us

$$\mathcal{A}= \frac12 (xy + yz + xz)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4} (xy+yz+xz)$$ Combine these two results, we find: $$xy+yz+xz = 2$$

Update

For a pure algebraic answer, one can substitute above expression of $a^2,b^2,c^2$ into Heron's formula for area of triangle, $$\mathcal{A} = \frac14\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$ simplify and obtain following algebraic identity

$$\begin{array}{rl} 3(xy+yz+zx)^2 =& \phantom{+0}((x^2+xz+y^2) + (y^2+yz+z^2) + (z^2+xz+x^2))^2\\ &-2((x^2+xz+y^2)^2 + (y^2+yz+z^2)^2 + (z^2+xz+x^2)^2) \end{array}$$ Using given values of $a, b, c$, we find $$3(xy+yz+xz)^2 = (3+4+1)^2 - 2(3^2+4^2+1^2) = 64 -2(26) = 12 = 3(2)^2$$ This leads to the same conclusion $xy+yz+xz = 2$ as before.

• It is precisely from a geometric construction that I get the below inequality!
– user284001
Commented Feb 9, 2018 at 12:27
• It also implies that $x^2 + y^2 + z^3$ is equal to $3$, right? Commented Feb 9, 2018 at 14:07
• @EricDuminil yes, if you mean $x^2+y^2+z^2 = 3$. Commented Feb 9, 2018 at 14:17
• @achillehui: Yes, that was a typo. Commented Feb 9, 2018 at 14:27

I would like to mention an other method, more algorithmic but also much more general to this problem.

First compute a Gröbner basis (say, using the lexicographic order) of the three polynomials $$x^2 + xy + y^2 - 3 \\ y^2 + yz + z^2 - 1 \\ x^2 + xz + z^2 - 4$$ One obtains $\left\{7 z^3-4 z,2 y z-z^2,2 y^2+3 z^2-2,x+2 y-3 z\right\}$.

We want to evaluate $xy+yz+zx$, which is presumably a constant, under the given conditions. This means that $xy+yz+zx-a$ belongs to the ideal generated by the three above polynomials, for some constant $a$ to be determined. Let's evaluate the remainder of the Euclidean division of $xy+yz+zx-a$ by the Gröbner basis given above: one finds $$-2-a+7z^2 \, .$$ This should be zero, if our relation is to be a consequence of the original relations. So we deduce $a=7z^2-2$.

What is the value of $z^2$? Well, the first term of the Gröbner basis tells us that $z(7z^2-4)=0$, so $z^2 = 0 , 4/7$, and therefore $a= \pm 2$.

To fix the sign, we need the last piece of information, namely that $x,y,z \geq 0$. Clearly, if $z=0$ then the last term of the basis tells us that $x+2y=0$, which is forbidden (since $x=y=z=0$ is not a solution). Therefore, $z^2 = 4/7$ and $a=2$.

Suppose $x,y,z > 0$ satisfy the system $$\begin{cases} x^2 + xy + y^2 = 3&\;\;\;(\text{eq}1)\\[4pt] y^2 + yz + z^2 = 1&\;\;\;(\text{eq}2)\\[4pt] z^2 + zx + x^2 = 4&\;\;\;(\text{eq}3)\\ \end{cases}$$ From $(\text{eq}2)$, since $y,z > 0$, we get $y,z < 1$, hence from $(\text{eq}1)$, we must have $x > 1$.

Let $a = x + y + z$, and let $b = xy + yz + zx$.

The goal is to find the value of $b$.

Since $x,y,z > 0$, and $x > 1$, we get $a > 1$.

Identically, we have $x^2 + y^2 + z^2 = a^2 -2b$.

From the sum $(\text{eq}1)+ (\text{eq}2) +(\text{eq}3)$, we get $$2(x^2 + y^2 + z^2) + (xy + yz + zx) = 8$$ hence $$2a^2-3b=8\qquad(\text{eq}4)$$

Subtracting $(\text{eq}2)$ from $(\text{eq}1)$, we get \begin{align*} &(x^2 +xy + y^2) - (y^2 + yz + z^2) = 2\\[4pt] \implies\;&(x^2 - z^2) + (xy - yz) = 2\\[4pt] \implies\;&(x-z)(x+y+z) = 2\\[4pt] \implies\;&x-z =\frac{2}{a}\\[4pt] \implies\;&z^2-2zx+x^2 = \frac{4}{a^2}\qquad(\text{eq}5)\\[4pt] \end{align*} Subtracting $(\text{eq}2)$ from $(\text{eq}3)$, we get \begin{align*} &(z^2 +zx + x^2) - (y^2 + yz + z^2) = 3\\[4pt] \implies\;&(x^2 - y^2) + (zx - yz) = 3\\[4pt] \implies\;&(x-y)(x+y+z) = 3\\[4pt] \implies\;&x-y =\frac{3}{a}\\[4pt] \implies\;&x^2-2xy+y^2 = \frac{9}{a^2}\qquad(\text{eq}6)\\[4pt] \end{align*} Subtracting $(\text{eq}1)$ from $(\text{eq}3)$, we get \begin{align*} &(z^2 +zx + x^2) - (x^2 + xy + y^2) = 1\\[4pt] \implies\;&(z^2 - y^2) + (zx - xy) = 1\\[4pt] \implies\;&(z-y)(x+y+z) = 1\\[4pt] \implies\;&z-y =\frac{1}{a}\\[4pt] \implies\;&y^2-2yz+z^2 = \frac{1}{a^2}\qquad(\text{eq}7)\\[4pt] \end{align*}

From the sum $(\text{eq}5)+ (\text{eq}6) +(\text{eq}7)$, we get $$2(x^2 + y^2 + z^2) - 2(xy + yz + zx) = \frac{14}{a^2}$$ hence $$a^2-3b=\frac{7}{a^2}\qquad(\text{eq}8)$$ Subtracting $(\text{eq}8)$ from $(\text{eq}4)$, we get \begin{align*} &(2a^2-3b) - (a^2-3b) = 8-\frac{7}{a^2}\\[4pt] \implies\;&a^2 = 8-\frac{7}{a^2}\\[4pt] \implies\;&a^4 - 8a^2 + 7=0\\[4pt] \implies\;&(a^2-1)(a^2-7)=0\\[4pt] \implies\;&a^2 = 7\qquad\text{[since $a > 1$]}\\[4pt] \end{align*} Then from $(\text{eq}4)$, we have $$b = \frac{2a^2-8}{3}$$ hence $$b = \frac{2(7)-8}{3} = 2$$

The given equations $(1),(2),(3)$ can be expressed as: $$\begin{cases} x^3-y^3=3(x-y) \\ y^3-z^3=y-z \\ z^3-x^3=4(z-x)\end{cases} \stackrel{+}{\Rightarrow} x=3z-2y \ \ (4)$$ Plug $(4)$ in $(1)$ and consider it with $(2)$: $$\begin{cases} 3z^2-3zy+y^2=1 \\ y^2+yz+z^2=1\end{cases} \Rightarrow 2z^2-4zy=0 \Rightarrow z=2y \ \ (5)$$ Put $(5)$ to $(4)$ to find $x=4y$. Now you can put $z=2y$ and $x=4y$ to $(1)$ to find $y=\frac{1}{\sqrt{7}}$. Hence: $$xy+yz+zx=\frac{4}{7}+\frac{2}{7}+\frac{8}{7}=2.$$

Possible Hint:

For $x, y, z \in \mathbb{R}^{+}$ $$(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$$

• For a proof of this inequality, see here or the linked questions. Commented Sep 2, 2020 at 11:47