Special Rectangle

Let the semicircle be of radius $1$ and its center be the origin of the $xy$ -plane, $BC=AD=h$ and $P(x,y)$ . Let $AX=a$ and $BY=b$ . Then

$\begin{cases} a = 1 + x + \dfrac {y(1-x)}{y+h} = 1 + \dfrac {y+hx}{y+h} \\ b = 1 - x + \dfrac {y(1+x)}{y+h} = 1 + \dfrac {y-hx}{y+h} \end{cases}$

$\implies$ When $\begin{cases} x = - 1, & y = 0 & \implies a = 0,\ b= 2, & a^2 + b^2 = 4 \\ x = 0, & y = 1 & \implies a = b = \dfrac {h+2}{h+1}, & a^2 + b^2 = 2\left(\dfrac {h+2}{h+1}\right)^2 \\ x = 1, & y = 0 & \implies a = 2,\ b= 0, & a^2 + b^2 = 4 \end{cases}$

For $a^2+b^2$ to be independent of $(x,y)$ , then $2\left(\dfrac {h+2}{h+1}\right)^2 = 4$ $\implies h = \sqrt 2$ . Putting it in $a$ and $b$ .

$\begin{cases} a = \dfrac {2y+\sqrt 2x + \sqrt 2}{y + \sqrt 2} \\ b = \dfrac {2y-\sqrt 2x + \sqrt 2}{y + \sqrt 2} \end{cases}$

Then

$\begin{aligned} a^2 + b^2 & = \left(\frac {2y+\sqrt 2x + \sqrt 2}{y + \sqrt 2}\right)^2 + \left(\frac {2y-\sqrt 2x + \sqrt 2}{y + \sqrt 2}\right)^2 \\ & = \frac {2(2y+\sqrt 2)^2 + 4x^2 }{(y + \sqrt 2)^2} \\ & = \frac {8y^2 + 8\sqrt 2 y + 4 + 4x^2}{(y+\sqrt 2)^2} & \small \blue{\text{Note that }x^2 + y^2 = 1} \\ & = \frac {4y^2 + 8\sqrt 2 y + 8}{y^2 + 2\sqrt 2 y + 2} \\ & = 4 \end{aligned}$

That is $a^2+b^2$ is independent of $(x,y)$ when $h = \sqrt 2$ and $\dfrac {AD}{AB} = \dfrac h2 = \dfrac {\sqrt 2}2 \approx \boxed{0.707}$ .

Let $|\overline {AB}|=a, |\overline {AD}|=b$ . Consider two positions of the point $P$ , one at $A$ and the other at the midpoint of the semicircular arc. In the first case, points $A, P, X, Y$ are identical, so that $|\overline {AX}|^2+|\overline {BY}|^2=a^2$ . In the second case, $|\overline {AX}|=|\overline {BY}|=\dfrac{a^2+ab}{a+2b}\implies |\overline {AX}|^2+|\overline {BY}|^2=2a^2\left (\dfrac{a+b}{a+2b}\right )^2$ . So $a^2=2a^2\left (\dfrac{a+b}{a+2b}\right )^2\implies a^2=2b^2\implies \dfrac{b}{a}=\dfrac{1}{\sqrt 2}\approx \boxed {0.7071067812}$ .

The problem looks slightly daunting at first, but when really thinking about it, one can find out the constant value of $AX^2+BY^2$ using a simple case.

This simple case is such:

When P is at point B, or A, assuming that the value of $AX^2+BY^2$ is constant for all points of P, the value of $AX^2+BY^2$ = $AB^2+0+0$ = $AB^2$

We have just deduced that the value of $AX^2+BY^2$ which will remain constant will be $AB^2$ .

Using algebra, we can apply this to the example of P being at the top of the semicircle. If P is at the top, then we can visualise the divisions created by the triangle of $AB$ like this:

The triangle's points $Y$ and $X$ divide $AB$ 's segments $AY$ and $BX$ equally, and this can be written algebraically as $a$ . Hence the total length of $AB$ is $2a+x$

Therefore, combining the fact that the constant value of $AX^2+BY^2 = AB^2$ and the constant value for this case where P is at the top is $2(a+x)^2$ , we can write the equation as follows:

$2(a+x)^2 = (2a+x)^2$

This simplifies into $2a^2+4ax+2x^2 = 4a^2+4ax+x^2$

Simplifying, we get: $2a^2 = x^2$

Therefore $x = \sqrt{2} a$

Therefore, using similar triangles for case 2 where P is at the top of the semicircle, we can find AD in terms of a by equating the ratios of height and base:

$\Large \frac{\frac{1}{2} AB}{x} = \frac{\frac{1}{2} AB+AD}{AB}$

Which is:

$\Large \frac{\frac{1}{2} (2a+\sqrt{2} a)}{\sqrt{2} a} = \frac{\frac{1}{2} (2a+\sqrt{2} a)+AD}{2a+\sqrt{2} a}$

Isolating AD, we get $AD = (1+ \sqrt{2})a$

writing the ratio and cancelling the $a$ s, we get $\Large \frac{1+ \sqrt{2}}{2+ \sqrt{2}}$

Which is equal to $\boxed{0.7071067812}$

I used the two particular cases : X=A and X=centre of circle. And then I used Thalès twice.

We consider two position of P.First one is when P coincides with A.That gives us, $AX^2+BY^2=AB^2$ .Then we consider when P is the midpoint of the arc AB.In that case we get two equations, $(1-\frac{1}{\sqrt{2}})*cot\theta=k$ and $cot\theta=2k+1$ where $k=\frac{AD}{ab}$ and $\theta=angleADY$ . Solving these two eqautions gives us the value of k.