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Schools Mathematics Grand Challenge

Week five's Puzzles

Problem 13:

The thirteenth problem was:

A conference centre contains five identical rooms. All of the rooms have square floors. Recently, the floors of the rooms have been painted, and a different pattern has been used for each room. Each room has a certain number of red circles painted on a white background in the patterns that you can see by clicking here.

• Room 1 has 9 red circles of equal area painted on a white background
• Room 2 has 16 red circles of equal area painted on a white background;
• Room 3 has 25 red circles of equal area painted on a white background;
• Room 4 has 36 red circles of equal area painted on a white background;
• Room 5 has 49 red circles of equal area painted on a white background;

Which room has the largest red-painted area on its floor? If one room has a greater red-painted area than the rest, give the number of that room (between 1 and 5 inclusive) as your answer. If all of the rooms have the same red-painted area, give 0 as your answer.
(You can ignore the black borders in the picture when figuring out the areas.)

The solution was:

The answer is that all rooms have the same red-painted area.

Suppose that there are horizontal and vertical lines which go through the points where the circles touch. These lines separate the floors into square tiles. Each square tile has exactly one red circle in it. Say that in a certain room the diameter of the circles is d. Then the area of one of the square tiles is d*d and the red painted area in one of the tiles is d*d*pi/4. So the total red painted area is pi/4 times the area of the room. This total red area doesn't depend on d and so is the same for each room.

Problem 14:

The fourteenth problem was:

For this problem, you should use the value 3.14 as an approximation for pi.

A communications company wanted to put a telephone cable around the equator of the earth. They had planned to place the cable along the surface but, due to a miscalculation, they ordered a cable that was 34.54m too short to reach all the way around. After much head-scratching, they decided that the best thing to do was to dig a trench of a fixed depth (same depth all the way around the equator), and place the cable at the bottom of the trench. After doing this, the cable was exactly the right length to fit once around the earth. Assuming that the equator is a circle, how deep in metres was the trench? Give your answer correct to one decimal place.

The solution was:

The answer is 5.5. Suppose that the radius of the earth is R and the depth of the trench is d. We must have that 2*pi*R-34.54=2*pi*(R-d). This simplifies to 34.54=2*pi*d. So d is 5.5.

Problem 15:

The fifteenth problem was:

Mary has a small pile of 14 cards, which contains 13 black cards and one red card. She deals the cards in the following way. First, she takes the card on top of the pack and places it on a table. The next card (the one on top of the remaining 13 cards) is then moved to the bottom of the pack. She repeats this process, placing a card on the table and then moving the next card to the bottom of the pack, until all of the 14 cards have been placed on the table. If the 12th card placed on the table is the red card, what was the position of the red card in the original pack?

Give your answer as a number between 1 and 14 (inclusive), where 1 means the top card, 2 the card second from the top, 3 the card third from the top and so on.

The solution was:

The answer is 8. If you actually wrote the numbers on the cards and went through the procedure, placing the cards on the table in a pile, you would end up with the cards in the order 12 4 8 14 10 6 2 13 11 9 7 5 3 1 where 12 is on top. So the 12th card placed is 8, so the red card originally was in position number 8.

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