Category: Calculus

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

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

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

Galvanizing Growth Python Simulation

import pandas as pd
from random import random
from math import floor, log
import matplotlib
from matplotlib import pyplot as plt
from numpy.random import normal
group = {}
growth = {}
countries = [('China', 137e7, 274e6), ('India', 127e7, 254e6), ('United States', 324e6, 648e5), ('Indonesia', 258e6, 516e5), ('Brazil', 206e6, 412e5), ('Pakistan', 202e6, 404e5), ('Nigeria', 186e6, 372e5), ('Bangladesh', 156e6, 312e5), ('Russia', 142e6, 284e5), ('Japan', 127e6, 254e5), ('Mexico', 123e6, 246e5), ('Philippines', 103e6, 206e5), ('Ethiopia', 102e6, 204e5), ('Vietnam', 953e5, 1906e4), ('Egypt', 947e5, 1894e4), ('Iran', 828e5, 1656e4), ('DR Congo', 813e5, 1626e4), ('Germany', 807e5, 1614e4), ('Turkey', 803e5, 1606e4), ('Thailand', 682e5, 1364e4), ('France', 668e5, 1336e4), ('United Kingdom', 644e5, 1288e4), ('Italy', 62e6, 124e5), ('Burma', 569e5, 1138e4), ('South Africa', 543e5, 1086e4), ('Tanzania', 525e5, 105e5), ('South Korea', 509e5, 1018e4), ('Spain', 486e5, 972e4), ('Colombia', 472e5, 944e4), ('Kenya', 468e5, 936e4), ('Ukraine', 442e5, 884e4), ('Argentina', 439e5, 878e4), ('Algeria', 403e5, 806e4), ('Poland', 385e5, 77e5), ('Uganda', 383e5, 766e4), ('Iraq', 381e5, 762e4), ('Sudan', 367e5, 734e4), ('Canada', 354e5, 708e4), ('Morocco', 337e5, 674e4), ('Afghanistan', 333e5, 666e4), ('Malaysia', 31e6, 62e5), ('Venezuela', 309e5, 618e4), ('Peru', 307e5, 614e4), ('Uzbekistan', 295e5, 59e5), ('Nepal', 29e6, 58e5), ('Saudi Arabia', 282e5, 564e4), ('Yemen', 274e5, 548e4), ('Ghana', 269e5, 538e4), ('Mozambique', 259e5, 518e4), ('North Korea', 251e5, 502e4)]
groupKeys = []
scale = 50
for i, j, k in countries:
  group[i] = [[0, j, 0, k]]
  growth[i] = [0]
  groupKeys.append(i)
group[groupKeys[0]] = [[13, countries[0][1]-13, 0, countries[0][2]]]
growth[groupKeys[0]] = [13]
e = 2.7183
def eDistribute(bottom, top, original):
  if(top == bottom):
    return top
  return top*e**((bottom-original)/(top-bottom))
def grow(sets, keys, past, index, spread=lambda a,b:1, expose=lambda a:1, expand=0.05, cut=5, spreadFactor=0.5, growthFactor=10):
  spreadSum = 0
  for i in keys:
    if(sets[i][index][0]+sets[i][index][1] == 0):
      adjExpand = 0
    else:
      adjExpand = expand * sets[i][index][1]/(sets[i][index][0]+sets[i][index][1])
    if(len(sets[i]) == index+1):
      sets[i].append([sets[i][index][0], sets[i][index][1], sets[i][index][2], sets[i][index][3]])
      past[i].append(0)
    for j in keys:
      if(i != j):
        currentSpread = spread(past[i], past[j])
        spreadSum -= currentSpread
        todaySpread = round(normal(1.5, 1)*eDistribute(0, spreadFactor, currentSpread)*sets[i][index][0]*expand)
        if(todaySpread > 0):
          if(len(sets[j]) == index+1):
            sets[j].append([sets[j][index][0]+todaySpread, sets[j][index][1]-todaySpread, sets[j][index][2], sets[j][index][3]])
            past[j].append(todaySpread)
          else:
            sets[j][index+1][0] += todaySpread
            sets[j][index+1][1] -= todaySpread
            if(sets[j][index+1][1] < 0):
              sets[j][index+1][0] += sets[j][index+1][1]
              sets[j][index+1][1] = 0
            past[j][index+1] += todaySpread
    currentSpread = expose(past[i])
    spreadSum -= currentSpread
    newGrowth = round(normal(1.5, 1)*eDistribute(0, growthFactor, currentSpread)*sets[i][index][0]*adjExpand)
    sets[i][index+1][0] += newGrowth
    sets[i][index+1][1] -= newGrowth
    past[i][index+1] += newGrowth
    if(sets[i][index+1][1] < 0):
      sets[i][index+1][0] += sets[i][index+1][1]
      past[i][index+1] -= sets[i][index+1][1]
      sets[i][index+1][1] = 0
    if(index >= cut):
      if(past[i][index-cut] > sets[i][index+1][0]):
        past[i][index-cut] = sets[i][index+1][0]
      sets[i][index+1][0] -= past[i][index-cut]
      sets[i][index+1][2] += past[i][index-cut]
  for one in sets:
    spreadSum += sets[one][index+1][0]
    if(sets[one][index+1][0] > sets[one][index+1][3]):
      spreadSum += 10*(sets[one][index+1][0]-sets[one][index+1][3])
  return spreadSum

impact = 0
impacts = {}
spreadFactor = 0.5
growthFactor = 1.25
meantime = 11
incperiod = 8
strictness = 100, 70
def adj(maxVal):
  def decorator(f):
    def wrapper(*args, **kwargs):
      raw = f(*args, **kwargs)
      if(raw < maxVal):
        return maxVal - raw
      return 0
    return wrapper
  return decorator
compensate = growthFactor*scale + spreadFactor*scale**2
def testAlgs(ban, close):
  i = 0
  impact = 0
  while i < 10:
    j = 0
    while i < 150:
      impacts[str(i)] = impact
      impact = floor(impact + compensate + grow(group, groupKeys, growth, i, spread=lambda a,b:ban(a, b, i-incperiod), expose = lambda a:close(a, i-incperiod), expand = 0.029, cut=meantime+incperiod, spreadFactor=spreadFactor, growthFactor=growthFactor))
      j += 1
  return impact / 10

Galvanizing Growth

This challenge on growth involves algebra, calculus, and some combinatorics.

Consider a pandemic outbreak. It begins with patient zero in a random country. For this challenge, only consider the countries with the 50 largest populations, rounded to four significant digits. Now, there are two ways the pandemic can grow. It can either expand within a country, or expand by travel to another country. There are then several methods to stop an outbreak: cutting travel, quarantining, and social distancing. The impact of a pandemic can be calculated by the net cost of these method, plus the cost of the sickness. Now, the more seriously we try to do any of these methods, the more costly and inefficient it will become. We express this with the first equation below for quarantine and social distancing, the second for travel restrictions. Now, let the expansion of a virus within a country be dictated by the third equation below, and the expansion between any two, the fourth. The cost of each case is then given by the fifth equation. The challenge is to find a general formula, given a segment of a countries net cases curve, to determine how much a country should enact social distancing, and how strictly it should quarantine victims. And another, complementary function, given curve segments for two countries, that will give the restriction on travel that the first should impose on the latter. The curves represent the data from an incubation period, x, ago, and this is randomly generated each pandemic be the sixth equation. They recover after a recovery time, y, given by the seventh. In the below equations, cn represents money spent on efforts in country n, in cost of cases in country n, pn is population, and r is random number 0 ≤ r < 1, different each time. anx gives the number of interactions between people, with an1 within a country, an2 between two. sn is the number of cases in country n, and gn is the number of new cases in that country from that country. gab is the number of new cases spread from b to a. Good luck! A python simulation of this has been published for you to test your strategy.

a_{n1}=125\ e^\frac{-c_n}{100}\\[8pt] a_{n2}=5\ e^\frac{-c}{5}\\[16pt] g_n=\left\lfloor a_{n1}s_n{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_n}{p_n}\right)\right\rfloor\\[16pt] g_{a_b}=\left\lfloor a_{b2}s_b{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_a}{p_a}\right)\right\rfloor\\[16pt] i_n=s_n+10\left(s_n-\frac{p_n}5+|s_n-\frac{p_n}5|\right)\\[16pt] x=5-3\sqrt2\operatorname{erfc^{-1}}{(2r)}\\[8pt] y=11-5\sqrt2\operatorname{erfc^{-1}}{(2r)}
SIMULATION
SOLUTION

Seriocomic Set

This challenge requires some knowledge of calculus and trigonometry, but mainly algebra.

We begin this challenge with a set of 4 numbers which has the following properties. Can you find them? If you can, good for you. But do you understand the use of this week’s adjective?

a,\,b,\,c,\,d\in\mathbb Z\\[8pt] a\leqslant b\leqslant c\leqslant d\\[8pt] a+b+3=a+d=\cos'{\pi}\\[8pt] \int \frac{d-c}a+1\;\;\delta a\bigg\lt\cos'{(a+b)\pi}\bigg\lt\int\frac{1-a}b-2\;\;\delta b\\[8pt] 0\lt|bc|+1\leqslant a+b+c+d\lt5\\[8pt] bc\notin\{a^2|a\in\mathbb N\}

Note: when calculating integrals, assume the ‘+c’ that you would add at the end is equal to the variable c.

SOLUTION

Salamandriform Sea

Geometry, calculus, and angle theorems are used in this challenge.

You are going on a vacation to an island in a sea. It’s a long journey. First, you have to drive to the sea. You park your car, take off your motorboat which you carried on top of it, and embark onto the sea. Then, you take your boat up to the island, park it, and walk up to your cottage. The sea and island are both ellipses, Whose foci are co-linear and point to your starting point. The sea and island both have their major axes twice the length of their minor axes, and you are a minor sea axis away from the closest point to the sea. Driving is 3 times faster than boating, which is thrice as fast as walking. The sea has only one island, and is 8 ninths water. What is the quickest route to your cottage?

Sophisticated Streetcar

Algebra, trigonometry, and calculus are used in today’s challenge.

You are taking a streetcar through a city. Your destination is in the middle of a large park, and you are late. You can jump of the streetcar at any time while it is on the side of the park. Your destination is 1 kilometre from the road. The streetcar moves at 15 m/s, and you move at 5 when you run, which you will to try to be on time. You calculate when you will need to jump off the streetcar to arrive as soon as possible, and jump of exactly then. For how long do you run?

Solution