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Here are some math puzzles I made! They are, in my opinion, interesting yet difficult.

1) There are n lights in a room. They all start on. Each minute, a light is picked at random, and if it is on, it is turned off, and if it is off, it is turned on. This process continues until the whole room is dark, in which case a monster comes out and eats you. How many cycles must pass before the chance that you have been eaten exceeds 50% if n = 4? I was contemplating making this way harder by adding in more states that the lights could be in and having a greater n, but I decided to make it simple. The reason for this, and the ultimate inspiration for this problem, comes from this program I made: https://www.khanacademy.org/cs/dots/4659495756365824
Each of the tiles are a 'light' and there are 6 states. The program (theoretically) stops when there are no tiles with a state of 0.

2) If f, g, and h are all functions such that f'(x)=-h(x), g'(x)=-f(x), and h'(x)=-f(x), and f(0)=1, g(0)=0, and h(0)=0, what is f(2)? This was also inspired by a program I made which "graphs" f, g, and h: https://www.khanacademy.org/cs/cyclesys ... 0208182272

3) Two points are picked within a circle and a line segment is drawn between them. What is the probability that the length of this segment is greater than the length of the circle's radius?

4) A triangle, XYZ, is drawn with vertices at (0,0), (a,0), and (b,c). A circle, C, is drawn such that C circumscribes XYZ. What is the area, in terms of a, b, and c, of the triangle defined by the tangents to C at XYZ's vertices?

5) Usually, the distance traveled on a certain path is found purely by the changes in x and y along the path. Imagine that instead, the distance is the area under the path when it is raised to coincide with a certain function, f, of x and y. If f(x,y)=2^(√(x^2+y^2)), what is the shortest path between (1,1) and (2,0)?

6) Lines are drawn continuously from the parametric function x1=sin(6t), y1=sin(7t) to the parametric function x2=sin(8t), y2=sin(9t). What is the length of a full circuit around the envelope of these lines? This was also inspired by a program I made: https://www.khanacademy.org/cs/linefill ... 2073726976

7) What is the probability in terms of n that the sum of n d6's (n six-sided dice) will be n+5?

8) Two planets of mass 1 are located on opposite sides of a sphere of radius length 1, and the force of gravity between them is determined by a 2D gravity law (the force is the inverse distance). This means that the acceleration of planet A due to the gravity of planet B is the inverse distance to planet B units/sec^2. Initially, the first planet is traveling at 1 unit/sec and the second one is at rest. How many seconds does it take them to collide if gravity does not travel further than the circumference of one of the sphere's great circles (e.g.-it doesn't go around the sphere for more than one rotation)? What if the force of gravity continues on indefinitely? (Something to keep in mind is that gravity goes both ways around the sphere).
P.S.: This got me wondering- Why doesn't something's size or how well something's angled to 'catch' gravitons affect how many gravitons it 'catches'/how much gravity it feels?

9) Here's one that's not exactly mathy-pondering, but just a mean equation I made: Solve for x:
x  ^  floor(  (x-7)! / (x-1)  )   =   x * sin( pi*x )  -  x^2 * cos( pi*x )

These do sound interesting, I'll need to give them a shot when I can.

According to Wolfram|Alpha, the answers to #9 are:
x ≈ -3.4170344530530901101473201680273743900547278318290051667318273
x ≈ 2.42575585187886002704321456582478539823626869915741463453487636
x ≈ -2.42575585187886002704321456582478539823626869915741463453487636
x ≈ -1.39852377777532072817480550071052471277687799255522637343581504
x ≈ -0.812179385267707066171077370612593789272093172239982978724541838
x ≈ 1.08470376300726793341777188918186908772341310150383709318645447
x ≈ 3.40121555233023750873673801837378648604706237902046064183146884
x ≈ 4.44528833611703135851517399210623465739415477342088774622319007

I could get more precise values, but chose not to.

NealCruco wrote:According to Wolfram|Alpha, the answers to #9 are:
x ≈ -3.4170344530530901101473201680273743900547278318290051667318273
x ≈ 2.42575585187886002704321456582478539823626869915741463453487636
x ≈ -2.42575585187886002704321456582478539823626869915741463453487636
x ≈ -1.39852377777532072817480550071052471277687799255522637343581504
x ≈ -0.812179385267707066171077370612593789272093172239982978724541838
x ≈ 1.08470376300726793341777188918186908772341310150383709318645447
x ≈ 3.40121555233023750873673801837378648604706237902046064183146884
x ≈ 4.44528833611703135851517399210623465739415477342088774622319007

I could get more precise values, but chose not to.

Um... There's actually an integer value that satisfies the equation... So, yeah...

Well, then Wolfram|Alpha got it wrong. *shrug* I didn't do any computation myself, so the error isn't my fault.

exfret wrote:Um... There's actually an integer value that satisfies the equation... So, yeah...

Oooh, I was about to say "ZERO!"... but then I realized that would mean there's a negative factorial in the equation, which is a no-no.

The way I create mean equations is this:

1) Choose a simple integer, like, for example, -5.
2) Do some operations on that number, like (-5)^2-(-5)/3
3) Add in some parts that make the calculations easier for you and the solving harder for nobody because you're the only person who'll ever appreciate the equation you made, like multiplying that (-5)/3 part by some operation on -5 that results with three. For example, you could add in an absolute value, put a factorial on the end, and divide by 40, which would give you (-5)^2-(-5)/3*abs(-5)!/40, which you can simplify to (-5)^2-(-5)*(abs(-5))!/120
4) Find the value of this expression. In this case it would be 25-(-5)/3*3=25+5=30
5) Replace all instances of the simple integer you chose with 'x' or a similar variable. In this case, this would result in x^2-x*|x|!/120. Now we just need to make an expression involving x's that has the same value, in this case 30.
6) So, we could do (31^32)^(2^(-5))-1, which could also be written as ((36+(-5))^32)^(2^(-5))-1 to make it even more confusing
7) Now replace -5's in this expression with x's. This would get us ((36+x)^32)^(2^x)-1
8) Set both expressions equal. This results in the horribly nasty equation x^2-x*|x|!/120=((36+x)^32)^(2^x)-1. Usually, I would also go on to refine the equation and make it more elegant while trying to retain some of its difficulty, but this is just for the sake of example, so it'll do.
9) Check to make sure the value you picked is correct. I'm not going to show the work, but this equation works.

I like how creating an equation is one of those things you can easily make near impossible to reverse engineer while expending little effort on your part.

Okay, in that case, I'm going to have to turn this up... to 11!!

wut-wuuuut!

testtubegames wrote:Okay, in that case, I'm going to have to turn this up... to 11!!

wut-wuuuut!

Oh wait, I "must have typed it wrong." The REAL equation is
x ^ floor( (x-7)! / (x-1) ) = x * sin( pi*x ) - x^2 * cos( pi*x ) + 1

exfret wrote:Oh wait, I "must have typed it wrong." The REAL equation is
x ^ floor( (x-7)! / (x-1) ) = x * sin( pi*x ) - x^2 * cos( pi*x ) + 1

Grrr