## Magic Liquid

You buy a bottle with a letter from the merchant, the merchant tells you that when you drink the liquid in the bottle it grants you eternal life, he supposedly deciphered this from the letter.
After you get home you decide to study the letter if it really says what the merchant told you, can you figure out if the bottle really grants eternal life?

You come to a fork in the road.
To the left is an empty well made from stone.
On the right is a pirate’s buried treasure.
Ahead you only see a tall straight tree.
The night is dark with only a dying moon in the sky.

Source: Puzzling StackExchange

The objects described in the last paragraph have the following shapes:
fork in the road = T
empty well = O
buried treasure = X
straight tree = I
dying moon = C
The 5 letters form the word “TOXIC”, which suggests you shouldn’t drink from the bottle.

## Move a Card

Can you move just one card so that you get a correct identity?

Yes, you can, as shown below.

## Slicing Butter

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 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?

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.

## Chessboard Infection

On a standard 8×8 chessboard there are 7 infected cells. Every minute each cell which has at least 2 infected neighbors gets infected as well. Is it possible for the entire chessboard to get infected eventually?

The total perimeter of the infected regions never increases. If there are 7 infected cells initially, their total perimeter is at most 28. The perimeter of an 8×8 square is 36. Therefore it is impossible to infect the entire chessboard.

## Worm in an Apple

There is a perfectly spherical apple with 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.

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.

## Pirate’s Treasure

Five pirates steal a treasure which contains 100 gold coins. The rules for splitting the treasure among the pirates are the following:

1. The oldest pirate proposes how to split the money.
2. Everybody votes, including the proposer.
3. If there are more than 50% negative votes, the proposer gets thrown in the water and the procedure repeats.

Given that the pirates are very smart and bloodthirsty (if they can kill another without losing money, they will do it), how should the oldest pirate suggest to split the money among the five of the in order to maximize his profit?

Solve the problem backward. Let the pirates be called A, B, C, D, E, where A is older than B, B is older than C, C is older than D and D is older than E.
If there are only two pirates left – D and E, then the D will keep all the treasure for himself.
If there are three pirates left – C, D, and E, C can propose to give just 1 coin to E and keep the rest for himself. Pirate E will agree because otherwise, he will get nothing.
If there are four pirates left – B, C, D, and E, then B can propose to give just 1 coin to D and keep the rest for himself. Pirate D will agree because otherwise, he will get nothing.
Now if there are five pirates – A, B, C, D and E, A should give coins to at least two other pirates, because otherwise at least three of them will vote negative. Clearly, B will always vote negative, unless he gets offered 100 coins and D will also vote negative, unless he gets 2 coins or more. Pirate A can offer to give one 1 coin to C, 1 coin to E and keep the rest for himself and this is the only optimal proposal – 98:0:1:0:1.

## A Maze Puzzle for the Day

Here’s a little maze puzzle I originally built a couple of years ago, that seems apropos to reprise now:

Can you make it from the A in the top left of this grid to the Z in the bottom right, always going either up one letter (for instance, A to B or G to H) or down one letter (for instance, N to M)? The alphabet wraps around, so you can go from Z up to A or A down to Z too. Try as hard as you can (and remember that you can always work backward if you get stuck forwards), and see where you get!

Remark: Solving the maze is not the same thing as solving the puzzle. Read those instructions carefully!

Source: Puzzling StackExchange

Notice this puzzle is published on April 1st. Actually, it doesn’t have a standard solution. If you connect every two consecutive letters which appear next to each other in the grid, you will get two disconnected components, one of which contains the START and the other contains the END. The first component has 5 dead-ends – at letters A, P, R, I, L, and the second component has 5 dead-ends – at letters F, O, O, L, S. These two spell out “April Fools”, which is the real solution of the maze.

## The Madman’s Speech

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!”

What is the man really talking about?

Source: Puzzling StackExchange

The man is actually reciting the states in America, even though you can’t hear him well:

“How? – I” = Hawaii
“I owe her” = Iowa
“much again” = Michigan
“I’d a ho…” = Idaho
“…me on town, a” = Montana
“tax as” = Texas
“florid a…” = Florida
“…s out the coat, a” = South Dakota
“virgin a year.” = Virginia
“Oh yo…” = Ohio
“…del – aware” = Delaware
“You ta…” = Utah
“…ke all or i do” = Colorado
“Never the” = Nevada
“road: I’ll land” = Rhode Island
“in the Anna-…” = Indiana
“…Marie land” = Maryland
“Main” = Maine
“can’s a s…” = Kansas
“…ore gone;” = Oregon
“tennis i…” = Tennessee
“…s out t’car. Oh line a” = South Carolina
“canned Turkey” = Kentucky

## Grasshoppers

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?

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.