Problem 17 · 2012 AMC 8
Medium
Geometry & Measurement
area-boundconstruction
A square with an integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
Show answer
Answer: B — Side 4.
Show hints
Hint 1 of 2
"Smallest possible" problems are squeezed from two directions. First squeeze from below: 10 pieces, each with integer side, so each has area at least 1 — what does that force about the big square's area?
Still stuck? Show hint 2 →
Hint 2 of 2
A lower bound only says "no smaller than". To prove the bound is actually reached, you must build an example at that size — here, an explicit way to cut a 4×4 into the right 10 squares.
Show solution
Approach: trap the answer: area lower bound, then a real construction
- Lower bound (rule out smaller): each of the 10 squares has integer side ≥ 1, so area ≥ 1, making the total area ≥ 10. A side-3 square has area only 9 < 10, so side ≥ 4.
- Upper bound (show 4 works): a bound is only believable if you can build it. Take the 4×4 square; fill the top 4×2 strip with two 2×2 squares, and the bottom 4×2 strip with 8 unit squares. That's 2 + 8 = 10 squares, exactly 8 of area 1. ✓
- Since side 3 is impossible and side 4 is achievable, the smallest is 4.
- The reusable habit: for any "smallest/largest that works" problem, prove a bound and exhibit an example hitting it — one half alone never settles the question.
Mark:
· log in to save