There is a perfectly spherical apple with a radius 50mm. A worm has entered the apple, made a tunnel of length 99mm through it and left. Prove that we can slice the apple in two pieces through the center, so that one of them is untouched by the worm.
SOLUTION
Let the entering point is A, the leaving point is B and the center of the apple is C. Consider the plane P containing the points A, B and C and project the worm’s tunnel on it. Since 99 < 2×50, the convex hull of the tunnel’s projection will not contain the center C. Therefore we can find a line L through C, such that the tunnel’s projection is entirely in one of the semi-planes of P with respect to L. Now cut the apple with a slice orthogonal to P passing through the line L and you are done.
If you want to split a cubic piece of butter into 27 smaller cubes, you can easily do it using just 6 slices (imagine the Rubik’s Cube). However, after every slice you make you can also rearrange the pieces – stack them in different ways on top of each other so that the number of cuts possibly gets reduced. What is the minimum number of slices you need in order to accomplish the task?
SOLUTION
In order to separate the little cube in the center of the butter piece from the rest, you need 6 slices – that’s the number of sides it has. Therefore, you can’t accomplish the task with less than 6 cuts.
A hundred prisoners are locked up in a prison. The warden devises the following game: he writes 100 different numbers on the foreheads of the prisoners. Then, each of the prisoners inspects the numbers on the foreheads of the others and decides to put either a black or a white hat on his head. Once the prisoners put their hats on, the warden arranges them in a line according to the numbers on their foreheads, starting with the lowest one and ascending to the highest one.
If the hats in the resulting line alternate their colors, then the prisoners will be set free. If not, the prisoners will be executed.
Can the prisoners devise a strategy that will guarantee their freedom?
SOLUTION
Once the warden writes the numbers on the prisoners’ foreheads, they form a mental circle, arranging themselves alphabetically in it (or according to any other order they agree on). They include the warden in this mental circle and imagine he has infinity written on his forehead.
Then, each prisoner examines the sequence of 100 numbers written on the foreheads of the others, and computes the number of inversions, i.e. the pairs which are not in their natural order. The prisoners which count an even number of permutations put black hats on, and the prisoners that count an odd number of permutations put white hats on. The infinity symbol is treated as the largest number in the sequence.
For example, if a prisoner sees the sequence {2, -6, 15.5, ∞, -100, 10}, then he counts seven inversions, which are the pairs (2, -6), (2, -100), (-6, -100), (15.5, -100), (15.5, 10), (∞, -100), (∞, 10), and puts a white hat on.
Next, we prove that the devised strategy works. We consider two prisoners, P1 and P2, who are adjacent in the final line the warden forms. These two prisoners split the mental circle in two arcs: A and B.
The number of inversions P1 counts is:
I_1 = I(A)+I(B)+I(A,B)+I(A, P_1) + I(P_1, B),
where I(X) denotes the number of inversions in a sequence X, and I(X, Y) denotes the number of inversions in a pair of sequences (X, Y):
I(X) = \|x_i, x_j \in X : \quad i < j, \quad x_i > x_j\| \\
I(X, Y) = \|x \in X, y \in Y : \quad x > y\|
Similarly, the number of inversions P2 counts is:
I_2 = I(A) + I(B) + I(B, A) + I(B, P_2) + I(P_2, A)
We sum I_1 and I_2 to get:
\begin{align*}
I_1 + I_2 &= 2I(A) + 2I(B) + I(A, B) + I(B, A) \\
&+ I(A, P_1) + I(P_2, A) + I(P_1, B) + I(B, P_2) \\
&= 2(I(A) + I(B)) + \|A\|\|B\|+\|A\|+\|B\|
\end{align*}
Since \|A\| + \|B\| = 99 is an odd number, we see that I_1+I_2 is also odd. Therefore, one among P1 and P2 would have counted an even number of inversions, and one would have counted an odd number of inversions. Thus, their hats have alternating colors.
You are walking through the prairie when you find a madman wandering around talking to himself. The following is what you manage to hear of his speech:
“How? I – I’ll ask her. I owe her much, again. I’d a home on town, a tax as florid as out the coat, a virgin a year. Oh, yodel – aware you take all or I do. Never the road: I’ll land in the Anna-Marie land. Main can’s a sore gone; tennis is out t’car. Oh, line a canned turkey!”
Four grasshoppers start at the ends of a square in the plane. Every second one of them jumps over another one and lands on its other side at the same distance. Can the grasshoppers after finitely many jumps end up at the vertices of a bigger square?
SOLUTION
The answer is NO. In order to show this, assume they can and consider their reverse movement. Now the grasshoppers start at the vertices of some square, say with unit length sides, and end up at the vertices of a smaller square. Create a lattice in the plane using the starting unit square. It is easy to see that the grasshoppers at all times will land on vertices of this lattice. However, it is easy to see that every square with vertices coinciding with the lattice’s vertices has sides of length at least one. Therefore the assumption is wrong.
Napoleon has landed on a deserted planet with only two policemen on it. He is traveling around the planet, painting a red line as he goes. When Napoleon creates a loop with red paint, the smaller of the two encompassed areas is claimed by him. The policemen are trying to restrict the land Napoleon claims as much as possible. If they encounter him, they arrest him and take him away. Can you prove that the police have a strategy to stop Napoleon from claiming more than 25% of the planet’s surface?
We assume that Napoleon and the police are moving at the same speed, making decisions in real time, and fully aware of everyone’s locations.
SOLUTION
First, we choose an axis, so that Napoleon and the two policemen lie on a single parallel. Then, the strategy of the two policemen is to move with the same speed as Napoleon, keeping identical latitudes as his at all times, and squeezing him along the parallel between them.
In order to claim 25% of the planet’s surface, Napoleon must travel at least 90°+90°=180° in total along the magnitudes. Therefore, during this time the policemen would travel 180° along the magnitudes each and catch him.
A thousand people stand in a circle in order from 1 to 1000. Number 1 has a sword. He kills the next person (Number 2) and gives the sword to the next living person (Number 3). All people keep doing the same until only one person remains. Which number survives?
SOLUTION
First, we note that if the number of people is a power of 2, then the first person will survive every round. The greatest power of 2 that is less than 1000 is 512. Therefore, after 488 people die, there will be 512 remaining and the first one to kill the 489-th person will survive. This person has number 1+2×488=977.
Note: Magnus Carlsen spent almost 2 minutes solving this puzzle.
SOLUTION
The first move is 1. Qa8. If Black plays Qd4, then White responds with Re3#. Otherwise, White plays either Nc5# or Nd6#, so that Black cannot capture the Knight neither with the Rook or the Queen.
For this puzzle/game, you need to keep presenting various words to your friends and telling them whether they are “TOUCH” or they are “DON’T TOUCH”. Below, we have listed several words from each category.
Can you guess what determines whether a word is “TOUCH” or “DON’T TOUCH”?
SOLUTION
Words that make your lips touch when pronounced belong to the “TOUCH” category, while the others belong to the “DON’T TOUCH” category. The sounds “P”, “B”, “M”, and “W” that cause this are called “bilabial”.
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