Challenges cover a variety of topics
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!
We’re trying to solve a rather insane system of equations. Due to the size of this problem, It’s publishing was delayed. Also, the solution is not typed, that would take a very long time, but shown is a scan of my rough work.
The initial equation system problem was:
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)}The solution is:
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 / 10This 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)}Begin with the equation above and the assumption that any leftover constants after integrating always equal c. We can simplify to the 7 simpler expressions below. Consider only up to before the statement regarding their sum, that the sum is positive. There are at this point four possible tuples (a, b, c, d). They are -2, -1, 0, and 2; -3, 0, 1, and 3; -4, 1, 2, and 4; and -5, 2, 3, and 5. Now, the first and last of these have sums that are out of bounds. The first of the remaining options is eliminated by the final clue. This leaves us with an answer of (a, b, c, d) = (-4, 1, 2, 4).
a,\ b,\ c,\ d\in\Z\\[8pt] a\leqslant b\leqslant c\leqslant d\\[8pt] a+b=-3\\[8pt] 2c-d-b\lt3\\[8pt] 2b-c-a\gt0\\[8pt] 0\lt a+b+c+d\lt5\\[8pt] \sqrt{bc}\notin\Z\\[16pt] \Rightarrow\therefore(a,b,c,d)=(-4,1,2,4)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.
To begin, the total of the perimeters of rectangles X Y and Z is 46 cm. This is equal to 2(A+B+C+2D+E). So A+B+C+2D+E is 23 centimetres. If we can find the value of D, we can solve the problem. Knowing that rectangles X and Y have equal perimeters and X and Z have equal areas, we can form a system of equations below. There are only 2 possible values of D, 3 cm and 13 cm. 13 cm obviously doesn’t work. D=3, and so A+B+C+D+E = 20 cm.
A+B=C+D\\AB=DE\\A+B+C+2D+E=23\\AB+CD+DE=39\\[8pt] \\D(C+2E)=39\\C,D,E\in\Z\\\Rightarrow D=3This challenges covers geometry and algebra.
We have 3 rectangles, X Y and Z. Rectangle X has sides with lengths A and B. Rectangle Y, C and D, and rectangle Z, D and E. Rectangle X has the same perimeter as rectangle Y and the same area as rectangle Z. The total perimeter of the three rectangles is 46 centimetres, and the total area, 39 square centimetres. All sides have integral centimetre lengths. Find the sum A+B+C+D+E.
In this challenge, the objective is to find the distance between 2 rods pulling elastics given that those elastics have equal heights. Now, because the elastics have different nominal lenghts and spring constants, they will only have equal lengths at a specific distance. This is because they must both be pulling with equal force. The force exerted by the two elastics is given by the first 2 equations below. They are equal, so we can solve them like a system of equations. We get that both are stretched to a length of 9 cm. The elastics are folded over twice, so their total length is thus only 4.5 centimetres. The combined force of the 2 elastics is 24 cmkg/s2. The weight resting upon the elastics will exert a downward force of 1 g * ≈9.807 m/s2 ≈ 9.807 mg/s2. In order for the elasitics to support it’s weight, the elastics must have an elasics must have this upwards force component. Let the angle below horizontal of the elastics be theta. θ ≈ asin(≈9.807/240), and the vertical distance between the rods is given by 4.5 * cos(asin(≈0.04086)). Using trigonometric identities, we simplify to the equation below. We get a distance of ≈8.992 centimetres.
\vec F_A=-{2kg\over s^2}(D-3cm)\\[8pt] \vec F_B=-{3kg\over s^2}(D-5cm)\\[16pt] \vec F_A=\vec F_B\\[8pt] {Dkg\over s^2}={9cmkg\over s^2}\\[8pt] D=9cm\\[32pt]F_A+F_B=24{cmkg\over s^2}\\[8pt] F_G\approx9.807{mg\over s^2}\\[16pt] x\approx4.5\cdot2\cdot\cos{\Bigg(\arcsin{\bigg({\approx9.807\over240}\bigg)}\Bigg)}\\[8pt] \approx\sqrt{1-\approx{0.04086}^2}\\[8pt] \approx8.992cmThis challenge is about elastics.
We begin with 2 elastics. Elastic A has nominal length 3 centimetres and spring constant 2 kilograms per square second. This notation seems unfamiliar but remember that we multiply by the displacement, in metres, and get a force, in mkg/s2. Elastic B has nominal length 5 centimetres and spring constant 3 kg/s2. The long elastic it threaded through the small one. The 2 ends of the long elastic are folded together. 2 identical rods are threaded through them. They are then pulled apart. The elastics are held in place at equal heights on the rods so they cannot move up or down. They are at equal heights and can rotate frictionlessly. The rods are then fastened to the surface of the earth some distance apart. The rods do not sway, bend, or in any way stop being vertical. A weight is placed halfway between the 2 rods, resting on the elastics. It weighs 1 gram. The weight does not cause any friction on the elastics. Elastics A and B are stretched to equal lengths. How far apart are the rods?