Question

Consider an equal-sided triangle connected to a square with side a. A straight line from
Q intersects the square at x and y. You have given x, find an equation for the intersection at y(x).

Answer 1

(This image is for reference and better understanding)

Draw a hypothetical line QCG such that it passes through middle and perpendicular to AD and EH.

$AC=\frac{a}{2};AB=x;BC=\frac{a}{2}-x$ ;

$EG=\frac{a}{2};EF=y;FG=\frac{a}{2}-y;$

In $\triangle QCA,QC=QA\ sin 60\degree=a\frac{\sqrt{3}}{2}$

$GC=HD=a$ ;$QG=a+\frac{\sqrt{3}}{2}a$

As $\triangle QBC$ and $\triangle QFG$ are similar,

So,$\frac{QC}{QG}=\frac{BC}{FG}$ ;

$\frac{\frac{\sqrt{3}}{2}a}{(\frac{\sqrt{3}}{2}+1)a}=\frac{\frac{a}{2}-x}{\frac{a}{2}-y}$ ;

$\frac{\frac{\sqrt{3}}{2}}{(\frac{\sqrt{3}}{2}+1)}=\frac{a-2x}{a-2y}$ ;

$\frac{{\sqrt{3}}}{(\sqrt{3}+2)}=\frac{a-2x}{a-2y}$ ;

$\sqrt{3}a-2\sqrt{3}y=(2+\sqrt{3})a-2(2+\sqrt{3})x;$

$y=\frac{2(2+\sqrt{3})x-2a}{2\sqrt{3}}$