For what it's worth, here's the intended solution.
Substitute x = a² and y = b². We have
a² + b² = 6
(a² + a)(b²+b) = 33/4
Rearrange the second equation as:
ab(1+a)(1+b) = ab(1+a+b+ab) = 33/4
Now multiply both sides by 4:
2ab(2+2(a+b)+2ab) = 33
Notice that a² + b² = (a+b)² - 2ab = 6, so 2ab = (a+b)² - 6. Substitute it to find:
((a+b)²-6)((a+b)²+2(a+b)-4)=33
Now set u=a+b and expand:
u⁴ + 2u³ -10u² - 12u - 9 = 0
Use rational root theorem or whatever means to see that u=3 is a root. Then factor it
(u-3)(u³+5u²+5u+3) = 0
Notice that since u = a+b > 0, and the second factor is always positive for u>0, then u=3 is the only plausible root.
Then you have
a + b = 3
a²+b² = (a+b)² - 2ab = 6 => ab = 3/2
So, we find a and b by solving the quadratic z²-3z+3/2 to find a = (3+sqrt(3))/2 and b = (3-sqrt(3))/2 to find
x = (3+sqrt(3))²/4 and y= (3-sqrt(3))²/4
So the idea was to recast everything in an equation for the sum, which I chose the values to be very simple to solve.
Today's question has real world applications. Suppose you are a financial analyst working at an investment fund managing billions of dollars in assets, and you are not a corrupt individual (shoot, I just said it had real world applications, and now I make this unrealistic assumption). Someday, someone comes to you saying they have an insurance fund made up of shitcoins they created themselves. A possible example is if they tweeted the following:
Now, assume, just in theory, of course I'm not trying to imply anything here, that when they show you the daily value of this fund, instead of calculating it from the assets there, they just use the following Python code, with a Gaussian of mean 7500 and standard deviation 3000, which is summed to the value each day:
The question is: what is the probability that this code will increase the value of the fund each day it runs? Do you think this pattern is reasonable for a fund consisting of an asset whose price fluctuates like this?
Please, don't try to connect this problem to current events. This is a purely academic question :D
If you have two standard d12s, the probability distribution of the total value of rolling 2 of them is a triangular pmf. How many ways can you relabel the numbers on the d12s to get back the exact same distribution? (subject to the restriction that the values must be positive integers). Is there a generalizeable way find this do this for rolling n dice with k sides?
An example for d6s, is the standard dice both have faces {1, 2, 3, 4, 5, 6}, while the (unique) way of relabeling them such that the analogous situation is true is the first dice is labeled {1, 2, 2, 3, 3, 4} and the second is {1, 3, 4, 5, 6, 8}.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Apropos of nothing, anyone know an approximate closed form for: 2.405565034289246
This comes from a numeric solution, a, to minimizing the squared error between (1 + erf(x))/2 and 1/(1 + exp(-a x))
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
89/37 is not bad given the size of the denominator, around only 0.00016 off, but I suspect this isn't as close as you were looking for.
Of course, this came from continued fractions until hitting the relatively large denominator of 7; 89/37 = [2;2,2,7]
You may prefer the much closer 197sqrt(2) - 101 - 35pi - 24e with an error of 5.45*10^-9 and which was found by a brute-force computer search.
Current TAS:
[SNES] Jelly Boy
[NES] Street Fighter 2010
This problem was given to me via a Discord help chat, and I thought it was an interesting problem. The labels on the hypotenuses are referring to the entire hypotenuse. The student who gave this problem fell into a bit of a trap, and I was unable to convince him that the easy method was the intended one.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
I'm going to take a guess as to what OmnipotentEntity sees as the "easy method". I'm 90% sure it is this:
Let y and z be as shown in the image above based on the subdivisions of the bottom edge. Then:
(2x)2+y2=145 (A)
(5x)2+(y+z)2=689 (B)
(4x)2+(z+3)2=377 (C)
Now let's declare that x,y,z are natural numbers. (Just do this. Don't think about it.)
Equation (A) says that 145 = 29*5 is a sum of two squares. Checking shows that 145 = 82+92 = 122+12 (you can also use 29 = 32+22 and 5 = 22+12, and the Brahmagupta-Fibonacci identity to get the above sums of two squares). The first gives x=4, y=9, and then in (C) z=8, and (B) checks out. The second gives x=6, but then (C) has no solution for z, so reject that. Therefore x=4.
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If you are unsatisfied with the above argument, specifically the declaration that x,y,z are natural numbers, note that no other argument I've found is interesting. You are free to solve it in real numbers by squaring equation(s) a few times to get rid of radicals. WolframAlpha will happily tell you that the only solutions are x=±4.
I'd also guess this problem is intended to be solved by assuming that the sides of the triangles are natural numbers.
All three numbers 145, 377 and 689 can be represented as a sum of squared natural numbers. So this is the big hint to look for solutions in whole numbers.
But one thing I don't understand is: why include the number 3 in bottom-right part of the picture? Does having this number open the doors for an alternative way of solving the problem? Or did they include that number in order to confuse the reader?
Here's a pair of interesting problems I ran across recently.
Let x = logb a. Use the logarithm property logb a = 1/loga b to rewrite the equation as x + 1/x = sqrt(293). Multiply everything by x and move all terms to the same side to get the quadratic equation x2 - x sqrt(293) + 1 = 0. The roots of the quadratic are x = (sqrt(293) +- sqrt(289))/2. Note that 289 = 172, which means that sqrt(293) - sqrt(289) < 1. Since a > b > 1, however, x must be greater than 1, so we can eliminate the difference as a value of x. Therefore, x = (sqrt(293) + 17)/2.
Use another logarithm change-of-base property to rewrite logab b = (log b)/(log ab) = (log b)/(log a + log b) in some non-a and non-b base. Then 1/logab b = (log a + log b)/(log b) = 1 + logb a = 1 + x. Likewise, 1/logab a turns out to be 1 + loga b = 1 + 1/x. Their difference is x - 1/x. Substitute x = (sqrt(293) + 17)/2 from above and simplify things to get the solution 17, which is choice B.
Instead of solving explicitly for x, it's easier to square x + 1/x to get x^2 + 1/x^2 + 2 = 293. Notice that the goal formula reduces to x - 1/x, and if you square that you get x^2 + 1/x^2 - 2, which is 293 - 4 = 289 from our above formula. Then just take the square root of 289 = 17.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Instead of solving explicitly for x, it's easier to square x + 1/x to get x^2 + 1/x^2 + 2 = 293. Notice that the goal formula reduces to x - 1/x, and if you square that you get x^2 + 1/x^2 - 2, which is 293 - 4 = 289 from our above formula. Then just take the square root of 289 = 17.
I did eventually get to answer=17 through some algebraic manipulation of my own, but this method works as well.
Although I was thinking of why OmnipotentEntity had the multiple choice in the first place.
I mean, I was very tempted to go with the following "solution":
With some algebraic manipulation similar to before, logba + logab = sqrt(293) and the goal is to find logba - logab. But a>b>1, so logab is clearly between 0 and 1. That means the answer is between sqrt(293)-2 and sqrt(293), and the only number in the multiple choice list that fits is 17. Therefore the answer is 17.
For this second one, find all solutions in the Reals, prove you have all of them.
Nice that this is a Diophantine problem in disguise.
Since |cos(anything)|≤1 and |1/sin(anything)|≥1, the equality can only possibly be true if cos(88*pi2/x) = ±1 and sin(3x) = ±1.
sin(3x) = ±1 when 3x = pi/2 + pi*n, or in other words, x = (1+2n)*pi/6. Note that sin(3x) = +1 when n is even, and -1 when n is odd.
cos(88*pi2/x) = ±1 when 88*pi2/x = pi*m, or in other words, x = 88*pi/m. Note that cos(88*pi2/x) = +1 when m is even, and -1 when m is odd.
So x is a solution if and only if x is both of the form 88*pi/m and of the form (1+2n)*pi/6, where m and n are either both even or both odd.
So we have:
88*pi/m = (1+2n)*pi/6
88*6 = m*(1+2n)
m*(1+2n) = 11*3*24
Since 1+2n is always odd, it follows that m is a multiple of 24, and so m and n are both even. Then 1+2n must be a divisor of 33 (positive or negative) that is 1 mod 4, and now we can list the possibilities for 1+2n:
1+2n = 33: n=16 and m=16, x=11*pi/2
1+2n = 11: no, not 1 mod 4
1+2n = 3: no, not 1 mod 4
1+2n = 1: n=0 and m=88*6, x=pi/6
1+2n = -1: no, not 1 mod 4
1+2n = -3: n=-2 and m=-88*2, x=-pi/2
1+2n = -11: n=-6 and m=-48, x=-11*pi/6
1+2n = -33: no, not 1 mod 4
So there are exactly four solutions for x: x=11*pi/2, x=pi/6, x=-pi/2, x=-11*pi/6.
Now, assume, just in theory, of course I'm not trying to imply anything here, that when they show you the daily value of this fund, instead of calculating it from the assets there, they just use the following Python code, with a Gaussian of mean 7500 and standard deviation 3000, which is summed to the value each day:
The question is: what is the probability that this code will increase the value of the fund each day it runs? Do you think this pattern is reasonable for a fund consisting of an asset whose price fluctuates like this?
Please, don't try to connect this problem to current events. This is a purely academic question :D
Let y and z be as shown in the image above based on the subdivisions of the bottom edge. Then:
(2x)2+y2=145 (A)
(5x)2+(y+z)2=689 (B)
(4x)2+(z+3)2=377 (C)
Now let's declare that x,y,z are natural numbers. (Just do this. Don't think about it.)
Equation (A) says that 145 = 29*5 is a sum of two squares. Checking shows that 145 = 82+92 = 122+12 (you can also use 29 = 32+22 and 5 = 22+12, and the Brahmagupta-Fibonacci identity to get the above sums of two squares). The first gives x=4, y=9, and then in (C) z=8, and (B) checks out. The second gives x=6, but then (C) has no solution for z, so reject that. Therefore x=4.
----
If you are unsatisfied with the above argument, specifically the declaration that x,y,z are natural numbers, note that no other argument I've found is interesting. You are free to solve it in real numbers by squaring equation(s) a few times to get rid of radicals. WolframAlpha will happily tell you that the only solutions are x=±4.
For this second one, find all solutions in the Reals, prove you have all of them.
