<s>[INST] Simple method to solve a geometry question for junior high school student Rencently, my sister asked me a geometry question that came from her mock examination, please see the following graph.

Here,


*

*$\angle DOE=45°$

*the length of $DE$ is constant, and $DE=1$. Namely, $OD,OE$ are changeable.

*$\triangle DEF$ is equilateral triangle.



Q: What is the maximum length of $OF$?


My solution
Denote $OD,OE,\angle ODE$ as $x,y,\theta$, respectively.
Via sine theorem
$$
  \begin{cases}
    ED^{2} = OE^{2} + OD^{2} - 2OE \times OD\cos \angle EOD \\[6pt]
    \cos \theta = \dfrac{EO^{2} + ED^{2} - OD^{2}}{2 EO \times ED}
  \end{cases}
$$
$$
\begin{align}
  1^{2} &= x^{2} + y^{2} - 2xy\cos 45^{\circ} \\
  &= x^{2} + y^{2} - \sqrt{2} xy
\end{align}
$$
$$
  \implies
  \begin{cases}
    \color{red}{xy} = \dfrac{x^{2} + y^{2} - 1}{\sqrt{2}} \color{red}{\leq}
    \dfrac{x^{2} + y^{2}}{2} \implies x^{2} + y^{2} \color{red}{\leq}
    2 + \sqrt{2} \\[6pt]
    \cos \theta = \dfrac{x^{2} + 1 - y^{2}}{2x}
  \end{cases}
$$
Via cosine theorem
$$
  \frac{y}{\sin \theta} = \frac{DE}{\sin \angle EOD} = \frac{1}{\sin
    45^{\circ}} \implies \sin \theta = \frac{y}{\sqrt{2}}
$$
$$\begin{align}
  OF^{2} &= EO^{2} + EF^{2} - 2EO \times EF\cos \angle OEF \\
  &= x^{2} + 1^{2} - 2x\cos(\theta + 60^{\circ}) \\
  &= x^{2} + 1 - 2x(\cos \theta \cos 60^{\circ} - \sin \theta \sin
    60^{\circ}) \\
  &= x^{2} + 1 - 2x\left(\frac{x^{2} + 1 - y^{2}}{2x} \frac{1}{2} -
    \frac{y}{\sqrt{2}} \frac{\sqrt{3}}{2}\right) \\
  &= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3} xy}{\sqrt{2}} \\
  &= \frac{x^{2} + y^{2} + 1}{2} + \frac{\sqrt{3}}{\sqrt{2}}
    \frac{x^{2} + y^{2} - 1}{\sqrt{2}} \\
  &= \frac{(\sqrt{3} + 1)(x^{2} + y^{2})}{2} + \frac{1 - \sqrt{3}}{2}
  \\
  &\color{red}{\leq} \frac{(\sqrt{3} + 1)(2 + \sqrt{2})}{2} + \frac{1 -
    \sqrt{3}}{2} = \frac{1}{2}(3 + \sqrt{3} + \sqrt{2} + \sqrt{6})
\end{align}
$$
However, for junior high school student, she doesn't learn the following formulae:


*

*sine theorem

*cosine therem

*$\cos(x+y)=\cos x \cos y-\sin x \sin y$

*fundamental inequality $x y\leq \frac{x^2+y^2}{2}$


Question


*

*Is there other simple/elegant method to solve this geometry question?



Update
Thanks for MXYMXY's hint

Here, the line $O'F$ pass the center of the circle. Namely, $O'D=OF$
In Rt $\triangle O'OF$, the inequality $O'F>OF$ holds.
 [/INST] Practical approach.Take a cardboard in the shape of an equilateral triangle of side 1. Slide two vertices along arms of the $ 45 ^0$ lines,note the other vertex is farthest when the setup is symmetrical, lying along bisector of $45^0 $ corner.
Only new trig ratio to be known is $ \tan 67.5^0$ is $ \sqrt2 +1 $
So the required length is: 
$$ \frac12 \; (\tan 60^0 + \tan 67.5^0 ) =\frac {\sqrt3 + \sqrt2 +1}{2}. $$
</s>