2005 AMC 10A Problem 19

After the $45^\circ$ turn, the middle square's diagonals are vertical and horizontal; $B$ is its top vertex and the diagonal from $B$ down to the bottom vertex has length $\sqrt{2}$.

The gap between the two outer squares is $1$ inch wide and $1$ inch tall. As the tilted square descends, its two lower edges (each sloping at $45^\circ$) first touch the outer squares at their inner top corners, which are $1$ inch apart at height $1$ above the base line.

Connect those two corners to the bottom vertex of the tilted square. This is an isosceles right triangle with hypotenuse $1$ (the horizontal segment between the corners) and right angle at the bottom vertex, so its height is half the hypotenuse, $\frac{1}{2}$. The bottom vertex therefore sits $\frac{1}{2}$ inch below the corners, at height $1 - \frac{1}{2} = \frac{1}{2}$.

Finally, $B$ is a full diagonal above the bottom vertex:
$$\frac{1}{2} + \sqrt{2}.$$

The answer is $\boxed{\textbf{(D)}\ \sqrt{2}+\frac{1}{2}}$.

A square rotated $45^\circ$ is best measured along its diagonals, and its edges make $45^\circ$ angles with the horizontal, so any horizontal chord across it is the hypotenuse of a $45$-$45$-$90$ triangle. Identify the exact contact points (the corners), then chain vertical distances: base to corners, corners down to the vertex, vertex up to $B$.

Coordinates: put the bottom vertex at $(0, y_0)$; the lower edges are $y = y_0 + |x|$. The outer squares' corners $(\pm\frac{1}{2}, 1)$ lie on these edges when $y_0 = \frac{1}{2}$, and $B = (0, y_0 + \sqrt{2})$.

• Assuming the tilted square's bottom vertex rests on the base line, giving B at height root 2, or forgetting to add the neighbors' height 1.
• Using the side length $1$ instead of the diagonal $\sqrt{2}$ for the vertical extent of the tilted square, giving $\frac{3}{2}$.