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Marble Thirds (Marble Swirl Circle in the UK version) is a puzzle in Professor Layton and the Last Specter.

## Puzzle

US Version

The diameter of the circle below is divided into three equal lengths, and those lengths are used to make the marble-like design that splits the circle into three different-colored sections.

The ares of sections I and III are equal, but what about the area of section II? Is it less than, greater than, or equal to the area of I?

UK Version

The diameter of a circle is divided into three equal lengths and those lengths are used to make the marble-like design shown in the diagram.

The areas of section I and III are equal, but what about the area of section II? Is it less than, greater than or equal to the area of I? Choose from A, B and C.

## Hints

Click a Tab to reveal the Hint.

US Version

Bear in mind the following three sizes of semicircle:

• The biggest semicircles in the diagram.
• The medium-sized semicircles, with a length two-thirds the circle's diameter.
• The smallest semicircles, with a length one-third the circle's diameter.

UK Version

Bear in mind the following three sizes of semicircle:

• The biggest semicircles in the diagram.
• The medium-sized semicircles, with a length two-thirds of the circle's diameter.

US Version

Think of the area of the smallest semicircle as one.

The diameter of the medium semicircle is twice that of the small semicircle, so its area will be four times as large.

The big semicircle has a diameter three times as big as the small one, so its area will be nine times as large.

UK Version

For example, take the area of the smallest semicircle to be 1.
The diameter of the medium semicircle is twice that of the small semicircle, so the area will be four times as large. The big semicircle has a diameter three times as big as the small one, so the area will be nine times as large.

Use these ratios to help with your calculations.

US Version

To figure out the area of sections I or III, use the semicircles as follows:

Big - Medium + Small =

Replacing these with the areas we found in Hint 2 gives us:

9 - 4 + 1 = 6

This should help you figure out how to determine the area of section II.

UK Version

To work out the area of either I or III, use the three semicircles as follows:

Big - Medium + Small =

Replacing these with the areas we found in Hint 2 gives us:

9 - 4 + 1 = 6

So, can you see how it is possible to work out the area of II?

US Version

If you're followed along with all of the previous hints, the rest is easy.

The big semicircle has an area of nine. Since it's a semicircle, the area of the whole circle is twice that, or 18. From this, subtract the size of the areas of sections I and III, and you're left with the area of II.

All right then. What is it?

UK Version

If you've got this far, the rest is easy. Here's one way to do it:

The big semicircle has an area of 9. Multiplying that by two gives the area of the whole circle, 18.

Take away the areas of I and III, that we worked out in Hint 3, and you're left with the area of II.

## Solution

### Incorrect

US Version

There's a simple way to solve this puzzle. See if you figure it out.

UK Version

Try to think of a simple way of solving this.

### Correct

US Version

Precisely!

The areas of I and II are the same! The best way to think about it is to use the three sizes of semicircles as shown above. If the ratio of the diameters is 3:2:1, then the ratio of the areas is 9:4:1.

Using the ratios as a guide, we can compare the ares of I, II, and III and conclude that they are, in fact, equal.

UK Version

Marble-ous!

The areas of I and II are the same! The best way to think about it is to use the three sizes of semicircles as shown above. If the ratio of the diameters is 3:2:1, then the ratio of the areas is 9:4:1.

Using the ratios as a guide we can compare the areas of I, II and III, and conclude that they are, in fact, equal.

A big thanks to http://professorlayton4walkthrough.blogspot.com

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