# Mixed volume of three squares

I have 3 following polynomials (for a coefficient I will just put K becuase they are not important here, but of course I don't mean they are all equal coefficients):

$$K \cdot b c + K \cdot b + K\cdot c + K$$ $$K \cdot ac + K \cdot a + K \cdot a + K$$ $$K \cdot ab + K \cdot a + K \cdot b + K$$

So, all three of them are bilinear polynomials with exponents: $(1,1), (1,0), (0,1), (0,0)$ I guess they all share same Newton polytope which is the square with the vertices same as their exponents.

My literature says: "These three equations have two complex solutions. Indeed, the mixed volume of the three Newton squares equals $2$."

Here is my question: I don't understand how the mixed volume of these three squares equals $2$. I thought it's $1$ because of the following: $$Vol(t_1P_1+t_2P_2+t_3P_3)=(t_1+t_2+t_3)^3$$ where $P_1,P_2,P_3$ are all Newton polytopes of mine polynomials - squares with vertices$(1,1),(1,0),(0,1),(0,0)$. Mixed volume is by definition the coefficient of $t_1 \cdots \cdot t_n$ in the volume divided by $n!$. So, in my opinion it would be $6$ divided by $3!$ which is $1$. Where am I making a mistake?

Maybe, if you are reading this and you also don't see a mistake, can you give me an idea of some other method with which I can count the number of solutions of this polynomial system ( but considering that I don't know the coefficients)?

Thank you for your answers.

## 1 Answer

Your three squares lie in 3-d, so they are not $(1,1),(1,0),(0,1),(0,0)$ but rather $$(0,1,1),(0,1,0),(0,0,1),(0,0,0),$$ $$(1,0,1),(1,0,0),(0,0,1),(0,0,0),$$ $$(1,1,0),(1,0,0),(0,1,0),(0,0,0).$$

Thus, $$Vol(t_1P_1+t_2P_2+t_3P_3)=(t_1+t_2)*(t_1+t_3)*(t_2+t_3).$$ The coefficient of this in $t_1t_2t_3$ is 2.

The "divided by $n!$" is not part of the definition of mixed volume. You sometimes see it because in lattice point theory volume is often considered normalized to the volume of a unimodular simplex. That is, the $1/n!$ is there simply to get back to Euclidean volumes.