2022 AMC 12A Problems/Problem 3
Contents
Problem
Five rectangles, , , , , and , are arranged in a square as shown below. These rectangles have dimensions , , , , and , respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
Solution 1 (Area and Perimeter of Square)
The area of this square is equal to , and thus its side lengths are . The sum of the dimensions of the rectangles are . Thus, because the perimeter of the rectangle is , the rectangle on the inside must have a perimeter of . The only rectangle that works is .
~mathboy100
Solution 2 (Perimeter of Square)
Note that the perimeter of the square is the sum of four pairs of dimensions (eight values are added). Adding up all dimensions gives us . We know that as the square's side length is an integer, the perimeter must be divisible by . Testing out by subtracting all five pairs of dimensions from , only works since , which corresponds with .
~iluvme
Solution 3 (Observations)
Note that rectangle must be on the edge. Without loss of generality, let the top-left rectangle be as shown below: It is clear that so we can determine Rectangle
Continuing with a similar process, we can determine Rectangles and in this order. The answer is as shown below. ~MRENTHUSIASM
Solution 4 (Observations)
Let's label some points: By finding the dimensions of the middle rectangle, we need to find the dimensions of the other four rectangles. We have a rule: Let's make a list of all dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule: By applying the rule, we get , and .
By substitution, we get this list (This also tells us that the diagram is not drawn to scale.)
Notice how the only dimension not used in the list was and that corresponds with B so the answer is,
~ghfhgvghj10 & Education, the study of everything.
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.