Problem:
Let the square $ABCD$ be on the side $l$ and the points $E$ and $F$ on the sides $BC$ and $CD$ respectively so that $\angle EAF= 45$. Find the ratio between the perimeter of the square and the perimeter of the triangle.
The drawing:
My idea:
First of all we can simply say that the perimeter of the square is $4l$
I noted $\angle DAF=x$ and $\angle EAB=y$.
As you can see in the drawing $\angle DFA=y+45$ and $\angle AEB=x+45$.(1)
We can simply show that $\angle PEA= \angle AFM= 45$(2)
From (1) and (2) we can say that $\angle PEC= y+45$ and $\angle MFC= x+45$
I thought of showing some congruences(I think that angles of trinagle are $45,75,30$ which means we can a right angles isoscel triangle and a $30,60,90$ triangle back to back). I also thought of taking the midpoint of $FE$ and applying the median theorem corresponding to the hypotenuse.
I don't know what to do forward! Hope one fo you can help me! Thank you!
Best Answer
Comment:
As can be seen in figure the altitude of triangle is always equal to $l$. If we suppose triangle is symmetric about diagonal AC of square , then we have:
$\angle EAB=\angle FAC=22.5^o$
$AE=AF=\frac l{\cos 22.5}$
$EF=2GE=2 AE \sin 22.5=2l \tan 22.5$
$P_{\triangle}=\frac{2l}{\cos 22.5}+2l\tan 22.5$
$\frac {P_\triangle}{P_\square}=\frac{\sin 22.5+1}{\cos 22.5}\approx 0.7494$
As can be seen the ratio is not constant, for example with triangle AHI we have:
$\frac {P_\triangle}{P_\square}=\frac{6.91+6.22+5.06=18.19}{4\times 6}\approx 0.7579$