Consider isosceles right triangle $ABC$ with $BC$ as the hypotenuse and $AB=AC=6$. $E$ is on $BC$ and $F$ is on $AB$ such that $AE+EF+FC$ is minimized. Compute $EF$.
My thought process:
I reflected triangle $ABC$ across $BC$ to get a square $ABA'C$. Then, I messed around with the placement of $F$ on $AB$, to see what results I could produce, with $E$ always being the midpoint. May someone help me on this?
On
Let $|AB|=|AC|=6=b$.
It is known that the minimal path would be the path of the light ray from the point $C$ to point $A$, reflected at points $F$ and $E$.
Consider points $C_1,A_1$ and $E_1$ as reflection of the points $C,A$ and $E$ with respect to $AB$.
Let $\angle A_1CC_1=\angle CFG=\angle GFE=\phi$, $\angle FEH=\angle HEA=\angle AE_1H=\angle HE_1C=\theta$.
Then
\begin{align} |EF|&=|E_1F|,\quad |EA|=|E_1A|=|E_1A_1| ,\\ |AE|+|EF|+|FC|&= |A_1E_1|+|E_1F|+|FC| \\ &=|CA_1| =\sqrt{(2b)^2+b^2}=\sqrt5\,b . \end{align}
\begin{align} \triangle BEF:\quad \angle BEF=90^\circ-\theta ,\quad \angle EFB=90^\circ-\phi \quad\Rightarrow\quad \theta=45^\circ-\phi , \end{align}
\begin{align} \triangle A_1CC_1:\quad \phi&=\arctan\tfrac12 =\arcsin\tfrac{\sqrt5}5 =\arccos\tfrac{2\sqrt5}5 ,\\ \theta&=45^\circ-\arcsin\tfrac{\sqrt5}5 =45^\circ-\arccos\tfrac{2\sqrt5}5 . \end{align}
\begin{align} \triangle BEF:\quad \frac{|EF|}{\sin45^\circ} &= \frac{|BF|}{\sin(90^\circ-\theta)} = \frac{\tfrac12|AB|}{\cos\theta} ,\\ |EF|&=\frac{b\sqrt2}{4\cos(45^\circ-\arctan\tfrac12)} \\ &= \tfrac{\sqrt2}4\, \frac{b}{ \cos 45^\circ \cos(\arccos\tfrac{2\sqrt5}5) + \sin 45^\circ \sin(\arcsin\tfrac{\sqrt5}5) } \\ &= \tfrac{\sqrt2}4\, \frac{b}{ \tfrac{\sqrt2}2 \tfrac{2\sqrt5}5 + \tfrac{\sqrt2}2 \tfrac{\sqrt5}5 }=b\,\tfrac{\sqrt5}6 , \end{align}
and since $b=6$, the answer is $|EF|=\sqrt5$.
Now, draw $\Delta BA'C'$ such that $A'$ is a mid-point of $CC'$.
Let $C'A\cap BC=\{E'\}$ and $C'A\cap BA'=\{F'\}$.
Thus, by the triangle inequality $$AE+FE+FC\geq AE'+E'F'+F'C'=AC'=\sqrt{6^2+12^2}=6\sqrt5.$$