Given is a right triangle ${\displaystyle \triangle ABC}$ with side length ${\displaystyle a_{0}=|AB|}$ and the ${\displaystyle \angle \beta }$ in vertex B. These two elements determine the triangle.

What is the general formula to determine the side length ${\displaystyle a_{1}}$ of any square ${\displaystyle \square DEFG}$ within the triangle provided that at least three vertices of the square touch the triangle?

Since at least three vertices of the square touch the triangle there must be a point ${\displaystyle D}$ where the square touches the line ${\displaystyle {\overline {AB}}}$. Similarily the vertices ${\displaystyle E}$ and ${\displaystyle G}$ touch the triangle sides ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {BC}}}$ respectively.

So for any inscribed ${\displaystyle \square DEFG}$ there is a right ${\displaystyle \triangle ADG}$ with two of the square's vertices and the vertex ${\displaystyle A}$ of the base triangle ${\displaystyle \triangle ABC}$.

The triangle ${\displaystyle \triangle ADG}$ has an angle ${\displaystyle \angle \delta }$ in the vertex ${\displaystyle D}$. Or put the other way: To any angle ${\displaystyle \delta }$ with ${\displaystyle 0<=\delta <={\frac {\pi }{2}}}$ there is a specific square ${\displaystyle \square DEFG}$ that satisfies the condition of being "three-touchy".
Within the ${\displaystyle \triangle ADG}$ the angle in vertex ${\displaystyle G}$ has the size ${\displaystyle \pi -{\frac {\pi }{2}}-\delta ={\frac {\pi }{2}}-\delta }$.

Concerning the vertices ${\displaystyle E}$ and ${\displaystyle F}$ the following has to be considered:
As long as ${\displaystyle 0<=\delta <=\beta }$ then the vertex ${\displaystyle E}$ touches the side ${\displaystyle {\overline {BC}}}$ of ${\displaystyle \triangle ABC}$ (see Figure 1).
Once ${\displaystyle \delta >\beta }$ then vertex ${\displaystyle E}$ dissolves from the side and it is vertex ${\displaystyle F}$ that touches ${\displaystyle {\overline {BC}}}$ of ${\displaystyle \triangle ABC}$.

Drawing a perpendicular from triangle side ${\displaystyle {\overline {AB}}}$ to vertex ${\displaystyle E}$ gives two further right triangles ${\displaystyle \triangle DHE}$ and ${\displaystyle \triangle BEH}$.
Within the ${\displaystyle \triangle DHE}$ the angle in vertex ${\displaystyle D}$ has the size ${\displaystyle \pi -{\frac {\pi }{2}}-\delta ={\frac {\pi }{2}}-\delta }$. So the angle in ${\displaystyle E}$ is math>\delta</math>.

Considering the above delivers the following equations:
(1) ${\displaystyle \quad a_{1}=|DG|=|FG|=|EF|=|DE|}$
(2) ${\displaystyle \quad a_{0}=|AB|=|AD|+|DH|+|BH|}$
(3) ${\displaystyle \quad {\frac {|AD|}{|DG|}}=cos{(\delta )}}$
(4) ${\displaystyle \quad {\frac {|EH|}{|DE|}}=cos{(\delta )}}$
(5) ${\displaystyle \quad {\frac {|DH|}{|DE|}}=sin{(\delta )}}$
(6) ${\displaystyle \quad {\frac {|EH|}{|BH|}}=tan{(\beta )}}$

Starting with equation (2) we have:
${\displaystyle \quad a_{0}=|AD|+|DH|+|BH|}$
${\displaystyle \quad \Leftrightarrow a_{0}=|DG|\cdot cos{(\delta )}+|DH|+|BH|\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+|DH|+|BH|\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+|DE|\cdot sin{(\delta )}+|BH|\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+|BH|\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {|EH|}{tan{(\beta )}}}\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {|DE|\cdot cos{(\delta )}}{tan{(\beta )}}}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {a_{1}\cdot cos{(\delta )}}{tan{(\beta )}}}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot \left(cos{(\delta )}+sin{(\delta )}+{\frac {cos{(\delta )}}{tan{(\beta )}}}\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}}{cos{(\delta )}+sin{(\delta )}+{\frac {cos{(\delta )}}{tan{(\beta )}}}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}}{sin{(\delta )}+cos{(\delta )}\cdot \left(1+{\frac {1}{tan{(\beta )}}}\right)}}\quad }$, rearranging

If ${\displaystyle \delta =0\quad }$ then ${\displaystyle a_{1}={\frac {a_{0}}{\left(1+{\frac {1}{\tan \beta }}\right)}}}$
If ${\displaystyle \delta =\beta \quad }$ then ${\displaystyle a_{1}={\frac {a_{0}}{\sin \beta +\cos \beta \cdot \left(1+{\frac {1}{\tan \beta }}\right)}}={\frac {a_{0}}{\sin \beta +\left(\cos \beta +{\frac {\cos \beta }{\tan \beta }}\right)}}={\frac {a_{0}}{\sin \beta +\cos \beta +{\frac {\cos \beta }{\tan \beta }}}}}$
If ${\displaystyle \beta ={\frac {\pi }{4}}}$ (i.e. the triangle ${\displaystyle \triangle ABC}$ is a right isosceles triangle) then: ${\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +\cos \delta \cdot \left(1+{\frac {1}{1}}\right)}}={\frac {a_{0}}{\sin \delta +2\cdot \cos \delta }}}$
If ${\displaystyle \delta =0}$ and ${\displaystyle \beta ={\frac {\pi }{4}}\quad }$ then ${\displaystyle a_{1}={\frac {a_{0}}{2}}}$
If ${\displaystyle \delta =\beta ={\frac {\pi }{4}}\quad }$ then ${\displaystyle a_{1}={\frac {a_{0}}{{\frac {1}{\sqrt {2}}}+2\cdot {\frac {1}{\sqrt {2}}}}}={\frac {a_{0}}{3\cdot {\frac {1}{\sqrt {2}}}}}={\frac {{\sqrt {2}}\cdot a_{0}}{3}}}$

For a given ${\displaystyle \beta }$ the function ${\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +\cos \delta \cdot \left(1+{\frac {1}{\tan \beta }}\right)}}}$ can be transformed to ${\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +k\cdot \cos \delta }}\quad }$with ${\displaystyle \quad k=1+{\frac {1}{\tan \beta }}\quad }$.

The derivative is: ${\displaystyle \quad {\frac {da_{1}}{d\delta }}=a_{0}\cdot {\frac {-1}{\left(\sin \delta +k\cdot \cos \delta \right)^{2}}}\cdot \left(\cos \delta -k\cdot \sin \delta \right)=a_{0}\cdot {\frac {k\cdot \sin \delta -\cos \delta }{\left(\sin \delta +k\cdot \cos \delta \right)^{2}}}}$

For ${\displaystyle \delta =0\quad }$ the derivative is: ${\displaystyle \quad {\frac {da_{1}}{d\delta _{=0}}}=a_{0}\cdot {\frac {k\cdot 0-1}{\left(0+k\cdot 1\right)^{2}}}={\frac {-a_{0}}{k^{2}}}<0}$

For ${\displaystyle \delta =\beta \quad }$ the derivative is: ${\displaystyle \quad {\frac {da_{1}}{d\delta _{=\beta }}}=a_{0}\cdot {\frac {k\cdot \sin \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\left(1+{\frac {1}{\tan \beta }}\right)\cdot \sin \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\left(\sin \beta +{\frac {\sin \beta }{\tan \beta }}\right)-\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\sin \beta +\cos \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}>0}$

So in the interval ${\displaystyle 0<=\delta <=\beta }$ the derivative changes from negative to positive. This means there is a ${\displaystyle \delta =\phi }$ with ${\displaystyle {\frac {da_{1}}{d\delta _{=\phi }}}=0}$

To find ${\displaystyle \phi }$, we put
${\displaystyle {\frac {da_{1}}{d\delta _{=\phi }}}=0\quad \Rightarrow \quad a_{0}\cdot {\frac {k\cdot \sin \phi -\cos \phi }{\left(\sin \phi +k\cdot \cos \phi \right)^{2}}}=0\quad \Rightarrow \quad k\cdot \sin \phi -\cos \phi =0\quad \Rightarrow \quad k\cdot \sin \phi =\cos \phi \quad \Rightarrow \quad k\cdot {\frac {\sin \phi }{\cos \phi }}=1}$
${\displaystyle \quad \Rightarrow \quad \left(1+{\frac {1}{\tan \beta }}\right)\cdot \tan \phi =1\quad \Rightarrow \quad \left({\frac {\tan \beta +1}{\tan \beta }}\right)\cdot \tan \phi =1\quad \Rightarrow \quad \tan \phi ={\frac {\tan \beta }{\tan \beta +1}}}$

So, in the intervall ${\displaystyle \quad 0<=\delta <=\beta \quad }$ the smallest square with at least three vertices touching the surrounding right triangle occurs when ${\displaystyle \tan \delta ={\frac {\tan \beta }{\tan \beta +1}}}$

In the case of a right isosceles triangle with ${\displaystyle \beta ={\frac {\pi }{4}}}$ the smallest square occurs when ${\displaystyle \quad \tan \delta ={\frac {1}{1+1}}=0.5\quad \Rightarrow \quad \delta \approx 0.464\pi }$

Drawing a perpendicular from triangle side ${\displaystyle {\overline {AC}}}$ to vertex ${\displaystyle F}$ gives two further right triangles ${\displaystyle \triangle GFJ}$ and ${\displaystyle \triangle CJF}$.
Within the ${\displaystyle \triangle GFJ}$ the angle in vertex ${\displaystyle G}$ has the size ${\displaystyle \pi -{\frac {\pi }{2}}-({\frac {\pi }{2}}-\delta )=\delta }$.
The triangle ${\displaystyle \triangle CJF}$ is similar to the triangle ${\displaystyle \triangle ABC}$

Considering the above delivers the following equations:
(1) ${\displaystyle \quad a_{1}=|DG|=|FG|}$
(2) ${\displaystyle \quad |AC|=|AG|+|GJ|+|CJ|=a_{0}\cdot \tan \beta }$
(3) ${\displaystyle \quad {\frac {|AG|}{|DG|}}={\frac {|AG|}{a_{1}}}=\sin \delta \quad \Rightarrow |AG|=a_{1}\cdot \sin \delta }$
(4) ${\displaystyle \quad {\frac {|GJ|}{|FG|}}={\frac {|GJ|}{a_{1}}}=\cos \delta \quad \Rightarrow |GJ|=a_{1}\cdot \cos \delta }$
(5) ${\displaystyle \quad {\frac {|CJ|}{|FJ|}}={\frac {|AC|}{|AB|}}=\tan \beta \quad \Rightarrow |CJ|=|FJ|\cdot \tan \beta =a_{1}\cdot \sin \delta \cdot \tan \beta }$

Applying equations (3), (4) and (5) to equation (2) leads to
${\displaystyle a_{1}\cdot \sin \delta +a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta \cdot \tan \beta =a_{0}\cdot \tan \beta }$
${\displaystyle \quad \Leftrightarrow a_{1}\cdot (\sin \delta +\cos \delta +\sin \delta \cdot \tan \beta )=a_{0}\cdot \tan \beta }$
${\displaystyle \quad \Leftrightarrow a_{1}\cdot (\cos \delta +\sin \delta \cdot (1+\tan \beta ))=a_{0}\cdot \tan \beta }$
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}\cdot \tan \beta }{\cos \delta +\sin \delta \cdot (1+\tan \beta )}}}$

If ${\displaystyle \delta ={\frac {\pi }{2}}\quad }$ then ${\displaystyle a_{1}={\frac {a_{0}\cdot \tan \beta }{0+1\cdot (1+\tan \beta )}}={\frac {a_{0}\cdot \tan \beta }{1+\tan \beta }}={\frac {a_{0}}{{\frac {1}{\tan \beta }}+1}}={\frac {a_{0}}{1+{\frac {1}{\tan \beta }}}}}$

If ${\displaystyle \delta =\beta \quad }$ then ${\displaystyle \quad a_{1}={\frac {a_{0}\cdot \tan \beta }{\cos \beta +\sin \beta \cdot (1+\tan \beta )}}={\frac {a_{0}}{{\frac {\cos \beta }{\tan \beta }}+{\frac {\sin \beta }{\tan \beta }}\cdot (1+\tan \beta )}}={\frac {a_{0}}{{\frac {\cos ^{2}\beta }{\sin \beta }}+\cos \beta \cdot (1+\tan \beta )}}}$

If ${\displaystyle \beta ={\frac {\pi }{4}}\quad }$ (i.e. the triangle ${\displaystyle \triangle ABC}$ is a right isosceles triangle) then ${\displaystyle \quad a_{1}={\frac {a_{0}\cdot 1}{\cos \delta +\sin \delta \cdot (1+1)}}={\frac {a_{0}}{\cos \delta +2\cdot \sin \delta }}}$

If ${\displaystyle \delta ={\frac {\pi }{2}}}$ and ${\displaystyle \beta ={\frac {\pi }{4}}\quad }$ then ${\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}\cdot 1}{0+1\cdot (1+1)}}={\frac {a_{0}}{2}}}$

If ${\displaystyle \delta =\beta ={\frac {\pi }{4}}\quad }$ then ${\displaystyle a_{1}={\frac {a_{0}\cdot 1}{{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}\cdot (1+1)}}={\frac {\sqrt {2}}{3}}\cdot a_{0}}$

Putting ${\displaystyle A}$ on ${\displaystyle 0+0i}$, will put the other vertices in both figures to:
${\displaystyle Pos(B)=a_{0}+0i}$

${\displaystyle Pos(C)=0+(a_{0}\cdot \tan \beta )i}$

${\displaystyle Pos(D)=(a_{1}\cdot \cos \delta )+0i}$

${\displaystyle Pos(E)=(|AD|+|AH|)+(|EH|)i=(a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta )+(|EH|)i=(a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta )+(a_{1}\cdot \cos \delta )i}$

${\displaystyle Pos(F)=(|FJ|)+(|AG|+|GJ|)i=(a_{1}\cdot \sin \delta )+(|AG|+|GJ|)i=(a_{1}\cdot \sin \delta )+(a_{1}\cdot \sin \delta +a_{1}\cdot \cos \delta )i}$

${\displaystyle Pos(G)=0+(a_{1}\cdot \sin \delta )i}$

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