Category: Math

Kayoed Kangaroo

This challenge considers an upset kangaroo making a series of discrete jumps in order to retaliate, and thus involves trigonometry.

Kangaroo chasing Sheep

Red Kangaroos in Australia compete with sheep for food. Sheep is grazing when it bumps into Kangaroo. The kangaroo wishes to learn the sheep some respect by either pummelling it, or pushing it into a nearby pond. Sheep wants to escape Kangaroo, but is slower than Kangaroo. However, Kangaroo is only capable of moving in jumps. Ergo, if Sheep stays far enough away to avoid pummelling, but too close for Kangaroo to jump, it is safe, and thus can avoid retribution indefinitely. Precisely formulated:

  • Kangaroo can jump up to 8 meters
  • Kangaroo cannot jump fewer than 5 meters
  • Kangaroo can jump at most fource per second
  • Kangaroo can pummel the sheep they are within 1 meter
  • Sheep can flee at 10 meters per second
  • Sheep can flee for 1 second before before Kangaroo pursues
  • There is a nearby pond, a perfect circle and 100 meters in radius
  • Sheep does not like swimming or being pummelled;
  • Kangaroo wishes to punish Sheep for perceived slight

Therefore, the challenge is thus: what is the maximum distance from the pond from which Kangaroo can punish Sheep?

Solution

Curt Cutting

This challenge is about cutting cards. And therefore about probability and combinatorics.

You have a deck of cards. For convenience, we’ll just use A-9 to have a nice round 40 cards. You cut some number of cards. At least one quarter and less than 3/4. That’s a neat 20 possible number of cards. To keep things simple, make them all have a probability of 5%. Now, from this cutting you cut a number of single cards, shuffling in between cuts of course. Formally, that’s random selection with replacement. At some point, you will cut a card that you have seen previously. At this point, you stop. This cuts short the data you receive about the deck of cards; you haven’t seen half of them. Of course, one can always estimate the remaining number of cards. So say you cut 10 cards before a repetition (11 cards total), and that for some reason you cannot visually approximate how many cards are in the deck. The challenge is to determine the expected number of cards in the deck.

The expected number of cards, to put it simply, is the “best” guess for how many cards are in the deck given your information. “Best” means minimizing the average difference between the actual number and your guess.

Solution

Cantankerous Circle

The average distance to the centre of a circle is two thirds the radius. For simplicity, say the radius is 1; the average is simply 2/3. How is this average defined? We will be exploring continuous extrapolations of averages using calculus, geometry, and more.

\int_a^b \operatorname f x \ \delta x \, \div (b-a)

We consider this to be the arithmetic mean value of f from a to b. This can be generalized to more than one dimension. But there are other types of average than the arithmetic mean. Consider the following other types of average, specifically applied to a circle:

\iint_R\ln{\operatorname f x}\ \delta A = \ln\text{GM}\cdot A \\[32pt] \iint_R{(\operatorname f x)}^2\ \delta A = \text{QM}^2\cdot A \\[32pt] \iint_R {1\over\operatorname f x}\ \delta A=\frac A{\text{HM}} \\[32pt] \operatorname f n < \text{median}\ \forall\ n \in R_1, \\[4pt] \operatorname f n > \text{median}\ \forall\ n \in R_2, \\[8pt] \iint_{R_1}\delta A = \iint_{R_2}\delta A,\quad R = R_1 \cup R_2

Since, at the origin, some of these means are discontinuous, ignore the boundary. The challenge is to determine the value of these averages for the distance from the centre.

SOLUTION

Macrocephalous Measurements

Here is a very interesting system of equations which is rather intriguing to solve.

Illustrating the mechanism measuring the capacity of the large container.

We begin with this setup, which is actually a mechanism to measure a volume of air. The capacity of the large container on the left is being measured. Say, for example, it is being used by Joanne the scientist. She pulls the handle of the pneumatic cylinder on the bottom to use the system. This pulls the two connected plungers in the middle, compressing the air in the container. Depending on the size of the container, pressure increases at a different rate. Once it is high enough, the horses are balanced and the plungers come to a stop. From how much they lowered, the volume is determined. A rather efficient mechanism, requiring only that the container be airtight and very little material. There is one small problem. How can the capacity possibly be calculated? For a start to this challenge, the supplementary air tank is half a litre, neglect air in the tubing, and let the amount pulled be 50 mL. Create a formula to find the capacity of the container given the volume by which the cylinder is displaced; the volume that was pulled out of it. Perhaps go further and find a general formula.

solution

Interlinear Interval

This challenge pertains to algebraic limits, and can be solved through calculus, clever arithmetic, graph theory, or other methods, though is only classified as the foremost.

n \in \mathbb{N} \\[4pt] \operatorname f : \mathbb{R}, \mathbb{Q} \to \mathbb{Q} \\[8pt] \operatorname f {(p, x)} = \begin{cases} p > x : x-\frac1{2n} \\[4pt] p < x : x+\frac1{2n} \\[4pt] p = x : x \end{cases}

The function f id defined above. Consider evaluating f recursively, starting at 0.5, with x as the previous result and p a random number from 0 to 1. Eventually an integer would be reached, and the average time to reach one could be found given n. The challenge is to find this average for as many n as possible, then infinitely many, and perhaps all.

solution

Solution to Garbled Games

This challenge was to find the optimal strategy for player 5 of a certain game. This must be solved with probability density. Let the PDF before the guess of player n be fn, and that player’s best guess, x0. The derivative of the expected value must be zero, which can be simplified to:

x_0 \operatorname{f_n}x_0 = \int_0^{x_0}\operatorname{f_n}x \delta x

Solving for player 1, we get 3x0 = 2, and so x0 = 3/2. Using a the same process for players two through 5, we can find the answer. Shown is a table of the best guess, chance of victory, and expected value for players one two five, as well as the probability density function for the weight of the prize before each guess. The answer is thus a guess of exactly 0.1 kilograms, and you will win it exactly half the time.

\begin{array}{|r|rcl|} \hline \\ \text{Player} & \text{Guess} & \operatorname P {win} & \text{Value} \\[4pt] \hline \\ 1 & 3\over2 & 3\over4 & 9\over8 \\[4pt] 2 & 1 & 1\over2 & 1\over2 \\[4pt] 3 & 1\over3 & 1\over2 & 1\over6 \\[4pt] 4 & 5\over24 & 23\over48 & \approx.1 \\[4pt] 5 & 1\over10 & 1\over2 & 1\over20 \\[4pt] \hline \end{array}
  • Player 1
  • Player 2
  • Player 3
  • Player 4
  • Player 5

Garbled Games

This math challenge is a calculus challenge.

This challenge pertains to a game involving an infinite number of players, and 1 prize. The prize has a mass that is a random amount between 1 and 3 kilograms, just as likely to be less than 1.5 kg as more than 2.5. Each player in turn guesses a real number quantity of kilograms. The following is done for each player in turn: if the guess is less than the current mass of the prize: that mass is removed from the prize. Otherwise, that player receives nothing. You are player number 5. Assume you hear each of the previous players guess optimally. The challenge is to determine what is your best guess, and how much you will win on average.

solution

Scrupulist Square

This challenge is spans all the way from graph theory, geometry, and calculus—to algebra and combinatorics.

Example set of points for N = 15

Consider a square with diagonal 1. In it, N points are randomly arranged. A graph is created from these points, such that the probability any 2 points touch is equal to 1 minus their distance. Choose a point at random on this graph. The challenge is to calculate a probability regarding this point: what is the probability that a randomly selected node connected to this one is connected to more nodes? Each node has a loop, so if the node was connected to no other nodes the probability of this is 0. For N=1 or 2, this probability is 0. Try to calculate at least one of the following: chance for a specific N > 2, limit of chance as N approaches infinity, and most difficult, generalized probability in terms of N.

SOLUTION

Preposterous Primes

This scrupulous challenge is an algebraic one, and quite interesting.

Let a “prime factor connection” be a connection from two whole numbers A and B, such that:

  1. A and B share a common prime factor C
  2. A plus C equals B
  3. A divided by C is not prime

 For any whole number N, let the “prime connected set” of N be the set of all whole numbers that satisfy the following:

  1. For any number X in this set, N is part of its prime connected set
  2. All numbers prime connected to N are in this set
  3. This set is the minimum possible size given the first two requirements

 Let S(N) be the size of the prime connected set of N.

Does S(N) reach infinity eventually, or is it a fast growing but strictly finite function? If it reaches infinity, can you find the smallest value at which it is infinite? Try to prove it is so. If S(N) never reaches infinity, try to prove this. Good luck!

SOLUTION

Fascinating Frequencies BONUS

This challenge has calculus and algebra.

We begin this problem by defining a few variables and functions, see below.
Let a ‘functional triangle’ be a relationship between three functions shown, denoted by the two large circles.
The properties of such are shown.
Then, a ‘meta-functional triangle’ is denoted in the same manner be a functional triangle with the given additional property.
The challenge is to find a meta-functional triangle. Good luck!

\text{let }x,\,y,\,z\in\R,\;x\leq y\leq z\\[8pt] \text{if }\;\exists!\;\{x,\,y,\,z\}: \operatorname fa=\operatorname gb=\operatorname hc\;\; \forall\;\;\{a,\,b,\,c\}\equiv\{x,\,y,\,z\},\\[4pt] \text{then }\;f\bigcirc g\bigcirc h\\[8pt] \text{if }\ f\bigcirc g\bigcirc h\;\ \text{and }\ f’\circledcirc g’\circledcirc h’,\ \;\text{then}\;f\circledcirc g\circledcirc h
SOLUTION