This problem can be solved graphically, but can it be solved mathematically? 


I can solve it graphically, but not mathematically. Graphically, I found X and Y to be equal, X = Y = 1.33
 A: Let $ t = \angle ABC $
then
$ x = (y + 2) \sin t  $
$ x = 1 \cdot \sin t + y \cos 45^\circ = \sin t + y \cos 45^\circ $
$ \dfrac{1}{x} = \dfrac{(y+1)}{(y + 2) \cos t } $
The last equation becomes
$ x(y + 1) = (y + 2) \cos t $
Use the first equation,
$  (y+1) (y+2) \sin t = (y+2) \cos t $
So
$\tan t = \dfrac{1}{y+1} $
It follows that
$ \cos t = \dfrac{y+1}{\sqrt{y^2 + 2 y + 2 }} $
$ \sin t = \dfrac{1}{\sqrt{y^2 + 2 y + 2 }} $
Substitute into the second equation,
$\dfrac{y + 2}{\sqrt{y^2 + 2 y + 2}} = \dfrac{1}{\sqrt{y^2+ 2 y + 2}} + \dfrac{y}{\sqrt{2}} $
And this simplifies to
$ y + 2 = 1 + \dfrac{y}{\sqrt{2}} \sqrt{y^2 + 2 y + 2} $
And further to
$ \sqrt{2} (y + 1) = y \sqrt{ y^2 + 2 y + 2} $
So that
$ 2 (y^2 + 2 y + 1) = y^2 (y^2 + 2 y + 2 ) $
And finally,
$ y^4 + 2 y^3 - 4 y - 2 = 0 $
whose solution is (from wolframalpha.com)
$ y = \dfrac{1}{2} (-1 + \sqrt{2} \sqrt[4]{3} + \sqrt{3} ) = 1.29663026289$
It follows that
$ t = \tan^{-1} \bigg( \dfrac{1}{y+1} \bigg) = 23.5292985676^\circ $
And
$ x = (y + 2) \sin t = 1.31607401295 $
A: Draw a vertical line through point $D$. Call the intersection with interval $AB$ point $E$. Also draw a horizontal line through point $D$. Call the intersection with interval $AC$ point $F$. This way the triangle $ABC$ is divided into a square with sides $AE = ED = DF = AF = L$ and with diagonal $y = AD = \sqrt{2} L$, plus the triangles $EBD$ and $FDC$. Furthermore let $\phi$ be the angle at point $B$.
Now it is given that $BD = AD + 1 = \sqrt{2}L+1$. From triangle $EBD$ we derive $L = BD \sin(\phi)$. From triangle $FDC$ we derive $L = \cos(\phi)$. Three equations to solve, with three unknowns.
Eliminating $BD$ we get $L = (\sqrt{2}L+1)\sin(\phi)$. Write this as $L - \sin(\phi) = \sqrt{2}L\sin(\phi)$. Squaring both sides and substituting $L = \cos(\phi)$ we get
$$2\sin^2(\phi)\cos^2(\phi) + 2\sin(\phi)\cos(\phi) -1 =0$$
Using $\sin(2\phi) = 2\sin(\phi)\cos(\phi)$ and solving the resulting quadratic equation leads to:
$$\sin(2\phi) = -1 + \sqrt{3}$$
Therefore the exact solution is:
$$\phi = 0.5 \arcsin(-1 + \sqrt{3})$$
$$x = \cos(\phi) + \sin(\phi) = 1.316074...$$
$$y = \sqrt{2}\cos(\phi) = 1.296630... $$
A: Sage gave up on this, but
Using the figure from Piquito, and the set of equations below,
Geogebra CAS solved the following:
$$\begin{align}
X^2+w^2&=(y+2)^2\quad &\Rightarrow Pythagoras\\
X&=(y+2)\cdot\sin(t)\quad &\Rightarrow \text{sine definition}\\
-2yX\cos\left(\frac{\pi}{4}\right)+y^2+X^2 &=1^2 \quad &\Rightarrow\ \text{law of cosines}\\
-2\cdot y\cdot w\cdot \cos\left(\frac{\pi}{4}\right)+y^2+w^2 &=(y+1)^2 \quad &\Rightarrow \text{law of cosines}\\
\end{align}
$$
No analytic solution was found.
$$
\begin{align}
X &= 1.316074012952\\
y &= 1.296630262902\\
w &= 3.022535406347\\
t &=0.410663730686793 = 23.52929856790132 degrees
\end{align}
$$
A: Re-thinking the figure for greater symmetry ...

Let the angle bisector at right $\angle A$ meet hypotenuse $\overline{BC}$ at $D$, and define $m:=|BD|$, $n:=|CD|$ (with $m\geq n > 0$). The problem imposes the condition $|AD|=m-n$.



*

*The Angle Bisector Theorem tells us that, for some $\lambda>0$, we can write $$|AB|=\lambda m \qquad |AC|=\lambda n \tag1$$

*The Pythagorean Theorem then tells us
$$(m+n)^2 = (\lambda m)^2+(\lambda n)^2 \quad\to\quad
2 m n = (\lambda^2-1) (m^2 + n^2) \tag2$$

*Finally, Stewart's Theorem tells us
$$\begin{align}
(\lambda m)^2 n+(\lambda n)^2m &=(m+n)\left((m-n)^2+mn\right)\\[4pt]
\to\qquad(\lambda^2+1) m n &= m^2+n^2\tag3
\end{align}$$
From here, combining equations via $(\lambda^2-1)(3)-(2)$ yields
$$(\lambda^4-3)mn = 0 \qquad\to\qquad \lambda = \sqrt[4]{3} \tag{$\star$}$$

For the problem at hand: $n=1$, so that
$$x:=\lambda n = \sqrt[4]{3} = 1.3160\ldots$$ and (solving quadratic $(2)$, and using the fact that $m\geq n$ to establish that the "$\pm$" should be "$+$")
$$m =\frac12 \left(1 + \sqrt{3} \pm \sqrt[4]3\sqrt{2}\right) \quad\to\quad y := m-n =\frac12 \left(-1 + \sqrt{3} + \sqrt[4]3\sqrt{2}\right) = 1.2966\ldots$$
