Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

4. There are four rectangles with length a=18a=18. Their widths form a geometric sequence. The perimeter of the second rectangle is 60, and the third rectangle is a square. Determine the widths of the rectangles.

A number or a short expression. Spacing and $ signs are ignored.

Solution

4. Members of a geometric sequence b,bq,bq2,bq3b, b q, b q^{2}, b q^{3} ..... 1.5 points
Expression 2a+2bq=602 a+2 b q=60 ..... 0.5 points
a+bq=30bq=12a+b q=30 \Rightarrow b q=12 ..... 0.5 points
a=bq2bq2=18a=b q^{2} \Rightarrow b q^{2}=18 ..... 0.5 points
Solving the system ..... 1 point
12q=18q=3212 \cdot q=18 \Rightarrow q=\frac{3}{2} ..... 0.5 points
b=8b=8 ..... 0.5 points
Widths of the rectangles: 8,12,18,278,12,18,27 ..... 1 point

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.