Maths Olympiad Prep

Track / Stage 5 / 208 of 400 #808 of 1964

Problem 808

AIME late
Geometry Difficulty 5.6 Find the answer

5-6. On a rectangular table of size xx cm ×80\times 80 cm, identical sheets of paper of size 5 cm ×8\times 8 cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right corner. What is the length xx in centimeters?

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A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 77.

Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
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Let's call such a rectangle the current one. Notice that for each new current rectangle, both the width and the height are 1 cm larger than the previous one. Initially, when there was only one sheet of paper, the width of the large rectangle was 8 cm, and at the end, it was 80 cm. Thus, a total of (808):1=72(80-8): 1=72 sheets of paper were added. The height of the current rectangle also increased by 72172 \cdot 1 cm, initially it was 5 cm, so x=5+72=77x=5+72=77.

Solution II. As in the first solution, let's look at the length and width of the current rectangles. Again, notice that for each new current rectangle, both the length and the width are 1 cm larger than the previous one. However, we will draw a different conclusion: specifically, the difference between the width and the height of the current rectangle is always the same! (Such a value that does not change during a certain process is called an invariant.) Since initially the width was 3 cm greater than the height, i.e., 85=38-5=3 cm, at the end it should also be 3 cm greater, so the answer is x=803=77x=80-3=77 cm.

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