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?

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

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

$2$

Answer: D

Concepts:transformation square (geometry)special right triangle

Difficulty rating: 1760

Solution:

When lowered, the rotated square's two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ The bottom vertex is centered between those corners, so each corner is horizontally $\tfrac12$ unit from it. A lower edge has slope $1$ in magnitude, so it rises $\tfrac12$ unit on the way to a corner. Therefore the bottom vertex is at height $1-\tfrac12=\tfrac12.$

Point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt2$ higher. Its height is therefore $\tfrac12+\sqrt2.$

Thus, the correct answer is D.

2005 AMC 10A Problem 19

Answer: D

Difficulty rating: 1760

Solution:

Problem 19 in Other Years