I'm working through Hartshorne's Euclid and Beyond and have hit a wall with the following question, 14.5: "For each of he following problems, assume that you are working in the Cartesian plane PI over a field F of characteristic 0. Give necessary and sufficient conditions on the field F for the given configuration to exist. Assume that all lines that appear to be parallel are parallel, and apparent right angles are right angles."
He gives the solution as (13)^(1/2)
I've set up my axes as usual, and unit points at F=(0,1) vertically and B=(1,0) horizontally. I've been focused on using HC || IG, equating the slopes, etc. But it's unclear to me where he's getting his result.
Any suggestions on how to proceed with this?
I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$ and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $\frac{1+b}{a}=\frac{1}{1-a}=\frac{1-b}{b}$. Using $\frac{1}{1-a}=\frac{1-b}{b}$, we may solve for $a=\frac{1-2b}{1-b}$. Using $\frac{1+b}{a}=\frac{1-b}{b}$, we may solve for $a=\frac{b(b+1)}{1-b}$.
Putting these together, we have $\frac{b(b+1)}{1-b}=\frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=\frac{-3\pm\sqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=\frac{-3\pm\sqrt{13}}{2}$ is in the field we're working over.