Archive for the ‘Puzzles’ Category.
I rarely post physics puzzles, but this one is too good to pass on.
Puzzle. A wireframe icosahedron is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the icosahedron?
While we are at it, another interesting question would be the following.
Puzzle. A wireframe cube is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the cube?
And this reminds me of a question I heard when I was preparing for an IMO many years ago.
Puzzle. A wireframe infinite square grid is assembled so that each of its edges has a resistance of 1. What is the total resistance between two neighboring vertices?
Here is the homework problem I gave to my PRIMES STEP students.
Puzzle. A man called his wife from the office to say that he would be home at around eight o’clock. He got in at two minutes past eight. His wife was extremely angry at this lateness. Why?
The expected answer is that she thought he would be home at 8 in the evening, while he arrived at 8 in the morning. However, my students had more ideas.
For example, one student extended the time frame.
- The man was one year late.
Another student found the words “got in” ambiguous.
- He didn’t get into his house two minutes past eight. He got into his car.
A student realized that the puzzle never directly stated why she got angry.
- The wife already got angry when he said he would be home around eight, as she needed him home earlier.
The students found alternative meanings to “called his wife from the office” and “minutes.”
- He had an office wife whom he called. But the wife at home was a different wife, and she was angry.
- “Two minutes past eight” could be a latitude.
I like including warm-up puzzles with every homework.
Puzzle. Farmer Giles has four sheep. One day, he notices that they are all standing the same distance away from each other. How can this be so?
The expected answer: The configuration is impossible in 2D. So, one of the sheep is on a hill or in a pit.
Some students thought big: The sheep could be placed at different locations around the Earth, forming a really big tetrahedron. In this case, we need to explain what it means for the farmer to “notice”, but this minor issue could be resolved in many ways.
Some of the students questioned the meaning of the word distance. They argued that if sheep are all touching each other, they are the same distance 0 from each other. One way this could happen is if the tails of the four sheep were entangled.
My PRIMES STEP program consists of two groups of ten students each: the senior group and the junior group. The senior group is usually stronger, and they were especially productive last academic year. We wrote four papers, which I described in the post EvenQuads at PRIMES STEP. The junior group wrote one paper related to the game SOS. The game was introduced in the following 1999 USAMO problem.
Problem. The game is played on a 1-by-2000 grid. Two players take turns writing an S or an O in an empty square. The first player who produces three consecutive squares that spell SOS wins. The game is a draw if all squares are filled without producing SOS. Prove that the second player has a winning strategy.
The solution is quite pretty, so I do not want to spoil it. If my readers want it, the solution for this grid, and, more generally, for any grid of size 1-by-n, is posted in many places.
My students studied generalizations of this game, and the results are posted at the arXiv: SOS. We tried different target strings and showed that:
- The SOO game is always a draw.
- The SSS game is always a draw.
- The SOSO game is always a draw.
Then, we tried a version where the winner needed to spell one of two target strings. We showed that:
- The SSSS-OOOO game is always a draw.
We tried several more elaborate variations, but I want to keep this post short.
I recently posted A Quadrilateral in a Rectangle puzzle.
Puzzle. A convex quadrilateral is inscribed in a rectangle with exactly one quadrilateral’s vertex on each side of the rectangle. Prove that the area of the rectangle is twice the area of the quadrilateral if and only if a diagonal of the quadrilateral is parallel to two parallel sides of the rectangle.
Now it is time for a solution where I use the sample rectangle pictured below. We draw lines parallel to the sides of the rectangle from every vertex of the quadrilateral. Now, we can find four pairs of congruent triangles where one triangle is inside the quadrilateral and the other is outside. In the picture below, the pairs are colored the same color. We see that green and red rectangles overlap, creating a brown rectangle. The fact that they overlap means that the quadrilateral’s area is less than half of the rectangle’s area.
It could go the other way, as the next picture shows. Here, the quadrilateral’s area is more than half of the rectangle’s area. In this case, we have an “underlap” as opposed to an overlap.
Therefore, the quadrilateral’s area is exactly half of the rectangle’s if and only if there is no overlap/underlap, implying that the thickness of the overlap/underlap rectangle is zero. This means that one of the diagonals of the quadrilateral has to be parallel to two sides of the rectangle.
I recently posted the following polyomino puzzle, and my readers are asking for a solution.
Puzzle. You are given a 5-by-7 rectangle with two corners cut out: A 1-by-1 tile is cut from the bottom left corner, and a 1-by-2 tile is cut out of the top right corner, as pictured. The task is to cut the resulting shape into two congruent polyominoes.
If a solution exists, there should be a transformation between the two congruent pieces. We can exclude a reflection and a central symmetry, as the union of the two shapes would have to be symmetric. We can exclude a translation: I leave it to the readers to explain why. What is left is a rotation or a glide. Let me remind you that a glide is composed of a reflection with respect to a line and a translation parallel to the line.
People often forget about glides, so this puzzle might be cool precisely because it is about glides. This is where we should start. We can safely assume that the top left corner belongs to piece A and, after the glide transformation, becomes the bottom right corner of piece B. It is also clear that the reflection line is parallel to the grid diagonals.
Then, we can start drawing the shapes. Piece A’s border starts from the top left corner and moves four squares down, then one square to the right, then one square down. Thus, piece B’s border starts at the bottom right corner and continues four squares to the left, then one square up, then one square to the left. In doing so, we reveal how the border of piece A continues. Thus, we can proceed in this manner to get the answer pictured below.
Here is another STEP homework question, which is a famous riddle.
Puzzle. Peter had ten cows. All but nine died. How many cows are left?
The wording is confusing on purpose. So, the students who are in a hurry subtract nine from ten and answer that only one cow is left. This answer is wrong. All but nine means that one cow died. So, the correct answer is nine.
One of my students decided that nine is the name of one of the cows, though it should have been capitalized. This means that all the cows except for Nine died, and only one cow, Nine, is left.
This student managed to find a legitimate explanation for the standard wrong answer.