Exercise 10.3.11

Proof. Here a solution is the vertical line $x=1$, corresponding to $m=\infty$, since the intersection of this line with $l_1$ and $l_2$ are the points $Q_1=(1,0),Q_2=(1,1)$ such that $Q_1{Q}_2=1$.

Any nonvertical line $l$ with slope $m$ through $P=(1,\frac{1}{2})$ has an equation

The intersection point of $l$ with $l_1:y=0$ and the intersection of $l$ with $l_2:y=x$ are the points

Therefore

Let $f(x)=4x^3+x^2-4x+1$, then

Therefore $f(x)$ has three real roots

This gives 4 solutions with $m=\infty$. (see figure.)

$f(x)$ has no rational root. Indeed , if $\alpha =\frac{p}{q},p,q\in \mathbb{Z},q>0,p\wedge q=1$ is a root of $f$, then $4p^3+p^{2}q-4p{q}^2+q^3=0$, so $p^3\mid 1,q^3\mid 2$, therefore $p=\pm 1,q=1$, so $\alpha =\pm 1$, but neither 1 nor $-1$ is a root of $f$. Since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, $f$ is irreducible over $\mathbb{Q}$ and the minimal polynomial of $m_1,m_2$ or $m_3$ over $\mathbb{Q}$ is $f$.

Let $L=\mathbb{Q}(m_1,m_2,m_3)$ the splitting field of $f$. The discriminant of $f$ is ${\mathrm{\Delta }}_f=316=2^2\times 79$, where $79$ is prime, so ${\mathrm{\Delta }}_f$ is not a square in $\mathbb{Q}$, so $\mathrm{Gal}(L:\mathbb{Q})\simeq S_3$, therefore $[L:\mathbb{Q}]=6$ is not a power of 2, so $m_1,m_2,m_3$ are not constructible numbers. Therefore the nonvertical lines are not constructible by straightedge and compass. □