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The Bicycle Problem

If you pull straight back on a pedal of a bicycle when it is at its lowest position, will the bicycle move forward or backward?

The surprising answer is that (usually) the bicycle will move backward.

When a bicycle moves forward, the trajectory its pedal traces with respect to the ground is called a trochoid. Depending on the selected gear of the bicycle, that trochoid could be:

  1. Curtate trochoid (for almost all gears of most bicycles)
  2. Prolate trochoid (if the gear is very low and the bicycle moves slowly)
  3. Common trochoid, a.k.a. cycloid (if the wheels of the bicycle and the pedal spin at identical speeds, practically never happens)
curtate trochoid
prolate trochoid
common trochoid (cycloid)

Since we are fixed with respect to the ground, by pulling the pedal backward, we are causing it to move leftward along the trochoid and therefore the bicycle will be moving backward. We note that despite that, the pedal will be moving forward with respect to the bicycle (but not with respect to the ground).

You can see a visual explanation of this puzzle in the video below.

Leave No Squares

How many matchsticks do you need to remove so that no squares of any size remain?

Nine matchsticks are enough, as seen from the solution below.

To see that eight matchsticks are not enough, notice that removing an inner matchstick reduces the number of 1×1 squares at most by 2. Since there are 16 such small squares, in order to get rid of them all, we need to remove only inner matchsticks. However, in this case, the large 4×4 square will remain.

Conflicting Words

What unique feature do the following words share?

FRIEND, FEAST, THERE, THOROUGH, FLIGHT, WONDERFUL, RESIGN, ENDURING, PEST, COVERT

Each of these words contains its antonym as a sub-word:

FRIEND – FIEND, FEAST – FAST, THERE – HERE, THOROUGH – ROUGH, FLIGHT – FIGHT, WONDERFUL – WOEFUL, RESIGN – REIGN, ENDURING – ENDING, PEST – PET, COVERT – OVERT

Circular Racetrack

Suppose you are on a one-way circular racetrack. There are 100 gas cans randomly placed on different locations of the track and the total amount of gas in the cans is enough for your car to complete an entire circle. Assume that your car has no gas initially, but you can place it at any location on the track, then pick up the gas cans along the way and use them to fill the tank. Show that you can always choose a starting position so that you can complete an entire circle.

Imagine you put your car at any location, but instead with an empty tank, you start with enough gas to complete the circle. Then, simply track the amount of gas you have and locate the point on the track where it is the lowest. If you choose that location as a starting point, you will be able to complete the track.

12 Balls, 1 Defective

You have 12 balls, 11 of which have the same weight. The remaining one is defective and either heavier or lighter than the rest. You can use a balance scale to compare weights in order to find which is the defective ball and whether it is heavier or lighter. How many measurement do you need so that will be surely able to do it?

It is easy to see that if we have more than 9 balls, we need at least 3 measurements. We will prove that 3 measurements are enough for 12 balls.

We place 4 balls on each side of the scale. Let balls 1, 2, 3, 4 be on the right side, and balls 5, 6, 7, 8 on the left side.

CASE 1. The scale does not tip to any side. For the second measurement we place on the left side balls 1, 2, 3, 9 and on the right side balls 4, 5, 10, 11.

If the scale again does not tip to any side, then the defective ball is number 12 and we can check whether it is heavier or lighter with our last measurement.

If the scale tips to the left side, then either the defective ball is number 9 and is heavier, or it is number 10/11 and is lighter. We measure up balls 10 and 11 against each other and if one of them is lighter than the other, then it is the defective one. If they have the same weight, then ball 9 is the defective one.

If the scale tips to the right side, the procedure is similar.

CASE 2. Let the scale tip to the left side during the first measurement. This means that either one of the balls 1, 2, 3, 4 is defective and it is heavier, or one of the balls 5, 6, 7, 8 is defective and it is lighter. Clearly, balls 9, 10, 11, 12 are all genuine. Next we place balls 1, 2, 5, 6 on one side and balls 3, 7, 9, 10 on the other side.

If the scale tips to the left, then either one of the balls 1, 2 is defective and it is heavier, or ball 8 is defective and lighter. We just measure up balls 1 and 2 against each other and find out which among the three is the defective one.

If the scale tips to the right, the procedure is similar.

If the scale does not tip to any side, then either the defective ball is 4 and it is heavier, or the defective ball is 8 and it is lighter. We just measure up balls 1 and 4 against each other and easily find the defective ball.

Not Twins

A teacher enters the classroom and sees on the first row two students sitting next to each other, looking completely identical. She asks them if they are twins, but the students simultaneously reply that they are not. After checking in the records, the teacher furthermore discovers that the two children have the same mother and father. What is the explanation?

The students are three of a triplet.

Chessboard Madness

You have unlimited number of knights, bishops, rooks and kings. What is the biggest number of pieces (any combination) you can place on a chessboard, so that no piece is attacked by another one?

If we put 32 knights on all black squares, then no two pieces will attack each other. Now let’s see that if we have more than 32 pieces, then there will be two which attack each other. Split the chessboard in 8 rectangular sectors of size 2×4. It is not hard to see that if we have more than 4 pieces in the same 2×4 sector, then 2 of them will attack each other. Therefore we can place at most 4 × 8 = 32 pieces on the chessboard.

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Shark Attack

A man stands in the center of a circular field which is encompassed by a narrow ring of water. In the water there is a shark which is swimming four times as fast as the man is running. Can the man escape the field and get past the water to safety?

Yes, he can. Let the radius of the field is R and its center I. First the man should start running along a circle with center I and radius R/4. His angular speed will be bigger than the angular speed of the shark, so he can keep running until gets opposite to it with respect to I. Then he should dash away (in a straight line) towards the water. Since he will need to cover approximately 3R/4 distance and the shark will have to cover approximately 3.14R distance, the man will have enough time to escape.