Author: zachary

Complex Coins

You will need to use algebra and combinatorics to solve this challenge.

A normal coin has 2 flat sides and one thin round edge. When tossed, the coin always lands on a flat side. A brand new pencil has 2 small flat sides and 1 long round side. When tossed, the pencil always lands on the round side. Somewhere in the middle is the 3 sided coin, where all 3 sides are equally likely to be rolled. You happen to have 5 such 3 sided coins*. You call the 3 sides heads, tails, and round, or H, T, and R for short. You and your friend John a game with these coins. For every H, you get a point. For every T, John gets a point. But for every R, you both get a point. First to 5 points wins. You start playing, and roll 2 tails and a round. What is the probability that you win?

*Despite various lengthy efforts, nobody has been able to produce a perfect 3 sided coin as described in this challenge. So far. Some have come close, see “How thick is a three-sided coin?” from Standupmaths on Youtube, available below.

Solution

Solution to Vile Voting

We wanted to find the standings and satisfaction of Brutus and Craw’s vote. Brutus will vote Red, Orange, Yellow, Purple, Blue, Green. Then Craw moves his extremes to the middle to vote Blue, Purple, Green, Red, Yellow, Orange. Now Brutus moves his extremes to the middle to get Orange, Yellow, Red, Green, Purple, Blue. It will never end! Serves them right.

Table \space of \space votes\\[8pt] \begin{matrix} &\#1&\#2&\#3&\#4&\#5&\#6\\ B_1&R&O&Y&P&B&G\\ C_1&B&G&P&R&Y&O\\ B_2&Y&R&O&G&P&B\\ C_1&G&P&B&O&R&Y\\ \end{matrix}

Vile Voting

Algebra and combinatorics are needed to solve this challenge.

Brutus and Craw are voting over their preference of colours. Their satisfaction is measured by how close their preferences are to the ones agreed on; they do not care about each other. The vote is determined by each player prioritizing the colours of the rainbow. Brutus and Craw do not know about indigo. Their first pick gets 5 points, their second 4, and so on such that their last gets 0.The colour with the most is first, the second most, second, and so on. Brutus and craw receive “points” of satisfaction, based on how close the vote is to their personal preferences. For each preference the number of points starts at one less than the previous, with 5 for 1st. It goes down by 2 if it’s below it’s position, and then decreases by one until it stops at 0. Except for fifth position, which cannot drop 2 because it started at only 1. In case of a tie, the position that comes first is picked randomly. First, Brutus states his preferences: red, orange, yellow, purple, blue, green. Then Craw would states his: purple, blue, green, yellow, orange, red. Except that Craw is a meanie and adjusts his to make it unfair. But Brutus then adjusts his, and they go back and forth until neither can improve their score. The challenge is to figure out what the votes, standings, and satisfaction will be for Brutus and Craw will be when they finish, if they ever do.

Satisfaction\\[16pt] \begin{matrix} & 1st & 2nd & 3rd & 4th & 5th & 6th\\ \#1 & 5 & 3 & 2 & 1 & 0 & 0\\ \#2 & 4 & 4 & 2 & 1 & 0 & 0\\ \#3 & 3 & 3 & 3 & 1 & 0 & 0\\ \#4 & 2 & 2 & 2 & 2 & 0 & 0\\ \#5 & 1 & 1 & 1 & 1 & 1 & 0\\ \#6 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{matrix}
Solution

Solution to Fathomless Fraction

We wanted to figure out the value of the fathomless fraction. We can start by saying the value of this fraction equals x1. x1 = 1 + 1 / x2, and so on. We can define this a recursive sequence, where xn = n + 1 / xn+1. But this doesn’t help if we don’t know any of the later terms. It approaches infinity, of course. xn>n, but xn<n+1. Maybe we could figure out the value of a term based on the previous one? xn-1 = n-1 + 1 / xn. Subtract n-1 from both sides. xn-1 -n +1 = 1 / xn. Take the reciprocal of both sides to find that xn = 1 / xn-1-n+1. We can make an alternate version of the fathomless fraction, call it fathomless equivalent fraction, or FEF for short. We soon realize that both lead to the exact same fraction. But x1 = x-1+1, and so xn=x-n+1, and we can make another fraction. But it returns to x1! It’s like with √x+1, it cannot be simplified. The answer is simply ≈1.433.

x_1=1+\frac1{2+\frac1{3+\frac1{4+…}}}\\[16pt] x_n=n+\frac1{x_{n+1}}\\[16pt] x_n=\frac1{x_{n-1}-n+1}\\[16pt] x_1=\frac1{\frac1{1+x_{-1}}}\\[16pt] x_n=x_{-n}+n\\[16pt] x_1=1+\frac1{2-2+\frac1{x_1-1}}=x_1\\[16pt] x_1 \approx 1.433

Solution to Baffling Bungee

So, we’re trying to find Bob’s final height after his bungee jump. The spring constant of C5H8, normal rubber, is ≈88. The equation for the force is 1/2 kx2, where k is the spring constant and x is the portion of the natural length. When Bob begins his jump, the rubber is inactive. It’s length is zero. Thus the force of it’s pull is ≈88*02. But gravity pulls down with a force of ≈9.807 m/s2. After this we cannot continue. The simplest way is to code. Simulate the bungee jump in many microscopic steps. Calculate the pull of the elastic at each point. This will show where the lowest point of the jump is. The stable point is easily calculated. 44x2≈9.807. x2≈9.807 / 44 ≈ 0.2229. x ≈ 0.4721. The bottom of the jump is a whopping 101.7 metres down, making the answer ≈102.3 metres above the water.

PE=\frac{kx^2}2\\[16pt] 44x^2=9.807\\ x=0.4721\\[16pt] h_f=\Delta x+\approx0.5279m\\ \quad h_f=h_i-\approx0.4721m\because h_i=\Delta x+1m\\ \Delta x=101.7m\\ \Rightarrow h_i=101.7m+1m\\ \therefore h_f=101.7m+0.5279m=102.3m

Baffling Bungee

This week’s science challenge requires the principals of physics and chemistry to solve.

Diagram of Baffling Bungee Jump.

Bob the bungee jumper goes bungee jumping over water on a cord made out of C5H8. The thickness of the rope is 3 centimetres. It’s natural length, the length it will have if not acted on by a force, is 1 metre. Bob starts the jump with a velocity vector with a magnitude of 1 metre per second and an angle of 45°. The lowest point of the jump is 1 metre above the water. Afterwards, the cord pulls Bob up to a stable position higher above the water. The challenge is to calculate the elasticity of the rope to figure out exactly how high.

Solution

Solution to Recursive Renewals

We want to find the optimal strategy for renewing a book at the Leibniz Library. First, let’s look at the maximum number of days you can keep a book before renewing it. 8 at first, then 7, 7, 6, and then 4. And that’s it, you cannot renew the book again. The fine equation will start above 0 when you renew the book a fifth time. But the hold equation will max out at 100% after only 10 days. The best is clearly 16 days. But it’s very likely it will be put on hold and you’ll be stuck with 14, or worse, 8. After 2 days, the hold equation ≈22%, and the loss would be 1 day if you don’t know. But if you don’t renew, you will probably gain a day. So don’t renew then. But on day 3, you can lose 2 days, and the hold equation now ≈24%. It’s less than 33%, so the day gained would still be worth it. On day 4, the hold equation is now at ≈29%, more than 25%, so you renew. We start again on day 5, with ≈40%, it’s less than 50%, so we don’t renew. But on day 6, we have ≈51%, far more than 33%. Although we lose the possibility of an extra day, it’s quite unlikely, and not nearly enough to make it worth not renewing. So we ought to renew. Not until day 8 can we gain another day. By then the hold equation is already ≈79%. It doesn’t matter if it has been renewed yet, if it hasn’t yet we do. By the time we can gain another day, the hold equation will have reached 100%. So our answer is to renew on days 4, 6, and 8.

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}

Solution to Spinning Stick

Diagram of the Spinning Stick.

We wanted to find the velocity of the ball when it rolls off of the spinning stick. The velocity of a point stick will be 2π metres per minute times it’s distance from the origin, in whatever direction it’s traveling in. This is 90° clockwise of the direction of the stick. This vector makes a circle with radius 2π metres, and it completes a rotation at the same speed. Knowing this, we find the centripetal force the stick experiences to be 4π2 metres per minute squared multiplied by the distance, another 90° clockwise of the stick. This makes centripetal force towards the origin, as you probably already know. If to spin in a circle the ball must accelerate, then it wont. The stick does not hold it in, and so it will quickly fly away. Since acceleration is proportional to distance, velocity increases acceleration. If you try solving Captivating Circles 5, you will see that such relationships relate to e. The rotation doesn’t interfere at all, though you’d think it would. When you have they are the same size and the proportions of change equal, their sum increases exponentially. Every unit of time when one would double the other if the other wasn’t returning the favour, they both multiply by e. They multiply at the same rate even if they aren’t the same size. The difference will decrease by a factor of e in the same duration of time. If one increases the other faster by a factor of n, the other will approach being larger by a factor of √n. The distance from the correct ratio will also decrease by a factor of e in the same duration of time as before. Using this knowledge, we can calculate the velocity, using the duration of time shown before as the unit for time, and take it’s derivative to get distance. Solve this equation to equal 2, and we find the time when the ball leaves the stick. Put time back into the equation for velocity, and we have our answer. 12.3134 metres per minute.

\frac d{dx}\frac{e^t-e^{-t}}{2\sqrt{2\pi}}=\frac{e^t+e^{-t}}{2\sqrt{2\pi}}\\[16pt] t=log(2\sqrt{2\pi}+\sqrt{8π-1})\approx2.295\\[16pt] \overrightarrow v\approx\frac{e^{2.295}-e^{-2.295}}{2\sqrt{2\pi}}\approx12.313

Recursive Renewals

You will need to do algebra, combinatorics, calculus, and trigonometry, and also use multiple matrices, to solve today’s challenge.

When you take a book out at the Leibniz Library, you must return it before the fine equation, f(x), rises above zero. If not, your fine is determined by the fine equation. The variable r represents the number of times you have renewed the book, beggining at zero. The variable d counts days and increases by 1 every day, beggining at zero. The variable t counts days and increases by 1 each day, beggining at 0. Each time you renew the book, the variable r in the fine equation, which begins at zero, increases by 1, and the variable t is reset to 0. You may not renew the book if someone else has it on hold or the fine equation is above 0. The probability that someone puts your book on hold on any given day is determined by the hold equation, h(x). How can you maximize the number of days you have with your book without paying anything?

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}
Solution