2005 AMC 10A Problem 19

After a $45^\circ$ turn, the middle square stands on one corner. Its diagonal, from the bottom tip straight up to $B$, has length $\sqrt{2}$ (the diagonal of a $1$ by $1$ square). So we only need the height of the bottom tip.

1. Where does it touch? The gap between the outer squares is $1$ inch wide. As the tilted square is lowered, its two lower edges slope outward at $45^\circ$. They hit the inner top corners of the neighboring squares. Those two corners are $1$ inch apart and $1$ inch above the base line.
2. Draw a small triangle. Join the two touching corners to each other and to the bottom tip. The two slanted sides lie along the tilted square's edges, which meet at a right angle at the bottom tip. So this is a right triangle with two equal legs, with the $1$ inch segment across the top as its longest side.
3. How far below the corners is the tip? In a $45$-$45$-$90$ triangle, the height from the right angle to the longest side is half that side. Half of $1$ is $\frac{1}{2}$. So the tip is $\frac{1}{2}$ inch below the corners, at height $1 - \frac{1}{2} = \frac{1}{2}$.
4. Go up to $B$. $B$ is one full diagonal above the tip:
$$\frac{1}{2} + \sqrt{2}.$$

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

A square turned $45^\circ$ is easiest to measure along its diagonals, and its edges make $45^\circ$ angles with the ground. Any flat line across it therefore forms a $45$-$45$-$90$ triangle with the lower tip. Find the exact touching points first, then add up the heights in steps: ground to corners, down to the tip, up to $B$.

Use coordinates. Put the middle of the gap at $x = 0$ and let the bottom tip be at height $h$. The lower edges go up one unit for every unit sideways, so the right edge reaches height $h + \frac{1}{2}$ at $x = \frac{1}{2}$. The neighbor's corner is at $(\frac{1}{2}, 1)$, so $h + \frac{1}{2} = 1$ and $h = \frac{1}{2}$. Then $B$ is at height $\frac{1}{2} + \sqrt{2}$.

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