How find the maximum possible length of OC, where ABCD is a square, and AD is the chord of the circle? Given a circle $o(O(0,0), r=1)$. How to find the maximum possible length of $OC$, where $ABCD$ is a square, and $AD$ is the chord of the circle?
I have no idea how to do this, can this be proved with simple geometry?
 A: This might not exclusively be using simple geometry, but here is a solution.
Without loss of generality, we can assume that the point $D$ is located at $(1,0)$, and that the point $A$ is located with a nonnegative $y$ coordinate.
We can construct a function which gives the $y$ coordinate of the $A$ point given the $x$ coordinate, it will simply be a function following a half-circle.
The equation for the circle is: $y^2+x^2=1$, and to construct a function outlining this circle, we simply rearrange it to $y=\pm\sqrt{1-x^2}$. Since we want the upper half of the circle, we want the function $y=\sqrt{1-x^2}$.

We now have 2 points, one fixed at $(1,0)$ and one located somewhere on the upper half of the circle perimeter.
We assume that the square is pointing away from center of the circle. And to find $C$ point of the square, we can do the following:
For the line $AD$ the change in $x$ is $1-x$ and the change in $y$ is $\sqrt{1-x^2}$
Since the line $DC$ is ortogonal to $AD$, the change in $x$ and $y$ have ben exchanged, so the coordinate of $C$ is
$$C=\left(1+\sqrt{1-x^2},1-x\right)$$
For finding the distance from $(0,0)$ we use the pythagorean theorem and get
$$
\begin{align}
|OC|&=\sqrt{\left(1+\sqrt{1-x^2}\right)^2+(1-x)^2}\\
&=\sqrt{2\sqrt{1-x^2}-2x+3}
\end{align}
$$

The maximum distance is simply the maximum of this function. Since the function is exclusively positive in the domain the $x$ coordinate of the maximum does not change by taking the square
$$2\sqrt{1-x^2}-2x+3$$
For finding the maximum, take the derivative and put it equal to zero, since that gives the point where the curve turns around.
$$\frac d{dx}2\sqrt{1-x^2}-2x+3 = \frac{-2x}{\sqrt{1-x^2}}-2$$
$$
\begin{align}
\frac{-2x}{\sqrt{1-x^2}}-2&=0\\
\frac{-2x}{\sqrt{1-x^2}}&=2\\
-2x&=2\sqrt{1-x^2}\\
4x^2&=4(1-x^2)\\
x^2&=1-x^2\\
2x^2&=1\\
x&=\pm\frac1{\sqrt{2}}
\end{align}
$$
We introduced a extra solution when we took the square, check both solutions and find that the $x$ coordinate is
$$x=-\frac1{\sqrt2}$$
And plug that into the formula and find the distance is
$$
\begin{align}
&\sqrt{2\sqrt{1-\frac12}+2\frac1{\sqrt2}+3}\\
=&1+\sqrt2
\end{align}
$$
Therefore the solution is $1+\sqrt2$
A: 
With the diagram as labeled, we see that
$$\begin{align}
|\overline{OC}|^2 &= \cos^2\theta + ( \sin\theta+2\cos\theta )^2 \\
&= \cos^2\theta + \sin^2\theta + 4 \cos\theta\sin\theta + 4 \cos^2\theta \\
&= 3+2 \sin 2\theta + 2 \cos 2\theta  \\
&= 3+2\sqrt{2}\left( \sin 2\theta \cos45^\circ + \cos 2\theta \sin45^\circ \right) \\
&= 3+2\sqrt{2}\sin(2\theta+45^\circ)
\end{align}$$
This value is clearly maximized when $\sin(2\theta+45^\circ) = 1$ (which happens for $\theta = 22.5^\circ$), so that
$$|\overline{OC}|^2 = 3 + 2\sqrt{2} = ( 1 + \sqrt{2} )^2 \quad\to\quad |\overline{OC}| = 1 + \sqrt{2}$$
