Question

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated $45^\circ,$ as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{2} + \dfrac{1}{2}$

$2$

Solution:

When lowered, the rotated square's two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ Working out the geometry, the square's bottom vertex settles at height $\frac{1}{2}.$ The point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt{2}$ higher, so its height is $\frac{1}{2} + \sqrt{2}.$

Thus, the correct answer is D.

2005 AMC 10A Problem 19

Solution:

Problem 19 in Other Years