4. Lucia, after drawing the square $A B C D$ with a unit side, draws a circle centered at $C$ with a radius equal to the side of the square. She then indicates with $X$ the intersection between the diagonal $A C$ and the circle, and with $Y$ the intersection of the line $D X$ with the side $A B$. What is the length of the segment $A Y$?
(A) $\frac{\sqrt{2}+1}{6}$
(B) $\sqrt{2}-1$
(C) $\frac{\sqrt{2}}{3}$
(D) $\frac{3+\sqrt{2}}{8}$
(E) $\frac{\sqrt{2}}{2}$
4. The answer is $\mathbf{( B )}$. Draw the line $r$ perpendicular to side $A D$ and passing through $X$. Let $H$ be the intersection of $r$ and $A D$, and let $K$ be the intersection of $C B$ and $r$. Let $J$ be the projection of $X$ onto side $C D$. First, we show that triangles $D H X$ and $D A Y$ are similar. Indeed, $\angle X D H = \angle Y D A$ since they are common to both triangles, and $\angle D H X = \angle D A Y = 90^{\circ}$, the first being the angle formed by the two perpendicular lines $H X$ and $A D$, and the second being the angle of a square; the first similarity criterion for triangles leads to the conclusion we sought. From this, we obtain

$$
\frac{H X}{D H} = \frac{A Y}{A D} = \frac{A Y}{A B}.
$$

Now notice that $C K D J$ forms a square: indeed, it has three right angles $90^{\circ} = \angle C J X = \angle X K C = \angle K C J$, the first two because they are angles formed by two perpendicular lines and the last one because it is an angle of a square. Therefore, we have $K X = C K = C J$. Let $\ell$ be the side length of the square. Since $C X = C B = \ell$, the radius of the circle, we have that $C X = \sqrt{2} \cdot C K$. $K C D H$ is a rectangle because it has all right angles, so we can obtain the relation $C K = D H$, and by the difference of segments $H X = H K - X K = D C - J C$. Combining what we have found, we get that $D H = C K = \frac{\ell}{\sqrt{2}}$ and that $H X = D C - C J = \ell - \frac{\ell}{\sqrt{2}} = \left(\frac{2-\sqrt{2}}{2}\right) \ell$. We can now conclude, since

$$
\frac{A Y}{A B} = \left(\frac{2-\sqrt{2}}{2}\right) \frac{1}{\sqrt{2}} = \sqrt{2} - 1
$$[^0]