2008 AMC 10A Problem 2
A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is . The ratio of the rectangle's length to its width is . What percent of the rectangle's area is inside the square?
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Key insight
Only ratios are given, so give the square a side of 1. Then the rectangle is 2 wide and 4 long, and the square fills 1 out of 8.
Solution
1. The problem gives only ratios, not actual lengths, so you are free to pick an easy size. Let the square have side .
2. The rectangle's width is twice the square's side, so the width is .
3. The rectangle's length is twice its width, so the length is .
4. Now compare areas. The rectangle has area . The square has area .
5. So the square covers out of parts of the rectangle:
The answer is .
Why this works
A percent comparison stays the same if you blow the whole picture up or shrink it down. That is why you may choose the square's side to be . Follow the chain of ratios one link at a time (square side, then width, then length) and you get real numbers to work with.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Only ratios are given, so pick a convenient size: let the square have side .
The rectangle's width is twice the square's side, so the width is . The rectangle's length is twice its width, so the length is .
Rectangle area: . Square area: . The square occupies
of the rectangle.
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 6: Ratios and Percents
- Introduction to Geometry, Chapter 8: Quadrilaterals
- Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
- Introduction to Geometry, Chapter 9: Polygons (§9.3)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.