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, 2016

Geometry Difficulty 4.1 AIME Prove it Canada

Judy has square plates, each with side length 60 cm. A plate is Shanks-Decorated if identical shaded squares are drawn along the outer edges of the plate, as shown.

The diagram shows an example of a plate that is Shanks-Decorated with 12 shaded squares.

Judy’s first plate is Shanks-Decorated with 36 shaded squares. What is the side length of each shaded square?
When a second plate is Shanks-Decorated, an area of 1600 cm $^2$ is left unshaded in the centre of the plate. What is the side length of each shaded square?
A plate is Double-Shanks-Decorated if two layers of identical shaded squares are drawn along the outer edges of the plate, as shown. The diagram shows an example of a plate that is Double-Shanks-Decorated with 48 shaded squares.

A new plate is Double-Shanks-Decorated and an area of 2500 cm $^2$ is left unshaded in the centre of the plate. Determine the number of shaded squares.

Solution

A plate with 36 shaded squares has 10 shaded squares along each side of the plate, as shown.

[[IMAGE0]]

We can see this from the diagram, or by considering a plate with $s$ squares along each side.

In this case, we can count $s$ squares on the top edge and $s$ squares on the bottom edge, plus $s-2$ new squares on each of the left edge and the right edge. (The 2 corner squares on each of these edges are already counted.)

This means that there are $2s+2(s-2) = 4s-4$ squares along the edges.

Here, we want $4s - 4 = 36$ or $4s = 40$ , and so $s=10$ .

There are 10 squares along each side of the plate and the side length of the square plate is 60 cm, thus the side length of each of the shaded squares is $\frac{60}{10}=6$ cm.

Since the plate is a square, and there are an equal number of identical shaded squares along each edge of the plate, then the unshaded area in the centre of the plate is also a square.

The area of this unshaded square in the centre of the plate is 1600 cm $^2$ , and so each of its sides has length $\sqrt{1600}=40$ cm, as shown.

Consider the row of squares along the left edge of the plate.

Since the side length of the square plate is 60 cm and the side length of the inner square is 40 cm, then the sum of the side lengths of the two shaded corner squares is $60-40=20$ cm.

Therefore, each shaded corner square (and thus each shaded square) has side length $\frac{20}{2}=10$ cm.

[[IMAGE1]]
Using the same argument as in part (b), the area of the unshaded square in the centre of the plate is 2500 cm $^2$ , and so each of its sides has length $\sqrt{2500}=50$ cm, as shown.

The side length of the square plate is 60 cm and so the sum of the side lengths of 4 shaded squares (2 stacked vertically in the top two rows and 2 stacked vertically in the bottom two rows) is $60-50=10$ cm.

Therefore, each of these shaded squares (and thus each shaded square on the plate) has side length $\frac{10}{4}=\frac52$ cm.

[[IMAGE2]]

The side length of the square plate is 60 cm and each shaded square has length $\frac52$ cm, and so along an outside edge of the plate there are $60\div\frac52=60\times\frac25=12\times2=24$ shaded squares.

There are 2 rows that each contain 24 shaded squares along each of the top and bottom of the square, and 2 additional rows that each contain $24-4=20$ shaded squares along the left and right sides of the square.

That is, the total number of shaded squares on the plate is $4\times24+4\times20=96+80=176$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.