savoy082

Pt En What if I told you that right now there is a place on Earth where there is no wind blowing to the sides? None at all! How can I know that? All we need is what is usually called the Hairy Ball Theorem . In less rigorous contexts, one can phrase the Hairy Ball Theorem as such: Theorem (Hairy Ball I): if you have a hairy ball, regardless of the way you comb its hair there will always be a spot where the hair points right up. In this particular image, the hair is pointing up both on the top and on the bottom. More formally, the Hairy Ball Theorem can be formulated like so: Theorem (Hairy Ball II): every continuous vector field over $S^2$ has at least a point where the tangential component is $0$. From this theorem it is actually quite easy to establish our interesting fact! If we think of the wind at the Earth's surface as a continuous vector field, the Hairy Ball Theorem says that there must be a point where the wind isn't blowing to the sides! Now all that is left...

This post's problem is a follow-up from this earlier post in which I talk a little bit about the problem of the water buckets and describe how I wrote a program to solve it. If you spot any mistakes here (or there) please let me know! Problem statement : Given $n$ buckets of capacities $c_1, \cdots, c_n$, prove that if the target value $t$ is not a multiple of $d = \gcd{(c_1,\cdots,c_n)}$ then the problem is unsolvable. Show that if all quantities in all buckets are multiples of $d$, then after one single move, that still holds. Hint 1 Solution : We will show that, regardless of the sequence of moves, all values in all buckets are - at all times - multiples of $d$. Let $w_i$ denote the value of bucket $i$. It is easy to see that in the beginning we have that each bucket has $w_k = 0 = \sum_{i=1}^n 0\cdot c_i$ and thus our property holds for the initial configuration. Now consider that for each bucket $k, w_k = \sum_{i=1}^n a_{k,i}c_i$ (i.e., all values are multiples of $d$). Reme...

    the scenery in which this proble m takes place Once again I bring you a problem alongside my proposed solution. If you find any mistakes or come up with a different solution, please let me know! Problem statement (and a lovely short story): The Shire is a lovely place where $N $ hobbits live in perfect harmony. Or at least they lived, until a hobbit decided to become an outside decorator and convinced some of its friends to paint their front doors with a very "fashionable" purple (all doors were yellow before that preposterous change). Overnight, the perfect balance and harmony in which the hobbits lived shattered, and hobbits whose doors were different colours couldn't stand one another. Worried, the great and wise Gandalf hurried to the Shire to try and settle this matter. This was what he decided to do: in alphabetical order, he would visit each hobbit. When visiting a hobbit $h $, he would change the colour of its door if and only if there were more hobbits mad...

This post contains a problem statement and my proposed solution. Please remember that I am only human and that I am (very) flawed. If you spot a mistake in my solution please  let me know. Problem statement : John and Mary have a bag full of integer numbers. In fact, the bag has $10^{10^{10}} $ numbers, each written on a plastic card, and the sum of all the $10^{10^{10}} $ integers in the bag is $0$. In turns, Mary and John are going to play with the bag, by doing the following: Picking two cards from the bag, let us say the cards have the numbers $a $ and $b $, and removing them from the bag; Inserting a new card in the bag with the number $a^3 + b^3$. Q: is there any initial number configuration and/or set of moves for which it is possible that, after $10^{10^{10}} - 1$ moves, the only card in the bag has the number $73$? The answer is "no". Can you show why? Hint 1 Look for an invariant of the game! That is, find a property of the game that does not change when Mary and...

This post's format will be a bit different from the usual. In this post, I will state a problem and then present my solution. More often than not (unfortunately) my solutions may have mistakes. If you find any, let me know! Problem statement : Let $[ABC] $ be an equilateral triangle of side length $1$. We now define a transformation of the triangle which moves one vertex and leaves the other two unchanged. To apply a transformation, start by picking the vertex you want to move. Assume it was $A $. Consider the line that goes through $A $ which is parallel to $[BC] $, and pick any point $A' $ in it. Your transformed triangle is $[A'BC] $. You can repeat this process as many times as you want. Q: Can the original triangle be transformed into an equilateral triangle of side length $2$? How/why? Try figuring this for yourself. The answer is "no". Can you show why? Hint 1 Look for an invariant of the transformation! That is, find a property of the triangle that does ...