2005 AMC 10A Problems/Problem 19
Contents
Problem
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, 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 from the line on which the bases of the original squares were placed?
Solution
Consider the rotated middle square shown in the figure. It will drop until length is 1 inch. Then, because is a triangle, , and . We know that , so the distance from to the line is
.
Note
(Refer to Diagram Above)
After deducing that , we can observe that the length from to the baseline is . This can be obtained by subtracting from the side length of the square(s), which is .
Adding these up, we see that our answer is .
- sdk652
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.