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Oct 11

Diagonal in a Rectangle

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A 1000 × 1004 rectangle is split into 1 × 1 squares. How many of these squares does the main diagonal of the large rectangle pass through?

Notice that the number of small squares the main diagonal passes through is equal to the number of horizontal and vertical lines it intersects. Indeed, every time the diagonal goes through the interior of one square to the interior of another, it must intersect one of these lines.

There are 1000 + 1004 = 2004 lines which are intersected by the main diagonal. However, on four occasions (which is the greatest common divisor of 1000 and 1004), the main diagonal intersects one horizontal and one vertical line at the same time, which results in double-counting., so we must subtract 4 from the answer.

Therefore, the answer is 1000 + 1004 – 4 = 2000.

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Sep 25

Rhombuses

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A regular hexagon is split into small equilateral triangles and then the triangles are paired arbitrarily into rhombuses. Show that this results into three types of rhombuses based on orientation, with equal number of rhombuses from each type.

Color the rhombuses based on their type and imagine the diagram represents a structure of small cubes arranged in a larger cube. If you look at the large cube from three different angles, you will see exactly the three types of rhombuses on the diagram.

Alternatively, the problem can be proven more rigorously by considering the three sets of non-intersecting broken lines connecting the pairs of opposite sides of the hexagon, as shown on the image below. The type of each rhombus is determined by the types of the broken lines passing through it. Therefore, there are n² rhombuses of each type, where n is the length of the hexagon’s sides.

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Sep 13

Eight Queens Puzzle

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Rearrange the eight queens so that no two of them attack each other. For an extra challenge, make sure that no three of them lie on a straight line.

The original puzzle has 10 unique solutions, up to rotation and symmetry. With the additional restriction imposed, there is only one solution.

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Sep 11

Batting Averages

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David Justice and Derek Jeter were professional baseball players. In 1997 they had the following conversation:

David: Did you know that in both 1995 and 1996 I had better batting averages than you?

Derek: No way, my batting average over the last two years was definitely higher than yours!

It turned out that both of them were right. How is it possible?

This is the so called Simpson’s paradox. The reason it occurred is that during 1996 both players had high averages and Derek Jeter had many more hits than David Justice. In 1996 both players had low averages and David Justice had many more hits than Derek Jeter. You can see their official statistics below.

Player199519961995-1996
Derek Jeter12/48 (.250)183/582 (.314)
195/630 (.310)
David Justice104/411 (.253)45/140 (.321)149/551 (.270)
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Aug 27

String of Words

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There is a common 9-letter word in the English language, such that if you keep removing its letters one by one, the resulting 8 words are still valid. What is this word?

Remark: The removed letters do not need to be from the beginning or the end of the word.

The word is STARTLING -> STARTING -> STARING -> STRING -> STING -> SING -> SIN -> IN -> I.