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General Puzzles, Problems & Riddles! (View 1st Post BEFORE You Answer or Ask!)

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  • _raoul_raoul Posts: 280 Pool Pro
    Here's an interesting one to pass the time.

    A point inside an equilateral triangle has distances 3,4, and 5 from the three vertices of the triangle. What is the area of the triangle?
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited December 2016
    Nanny Ogg wrote: »
    When you google it you are supposed to get this.....

    x4npa4yq0rn7.png

    Ah! Could have been more clear with the equation next time! :D
    _raoul wrote: »
    Here's an interesting one to pass the time.

    A point inside an equilateral triangle has distances 3,4, and 5 from the three vertices of the triangle. What is the area of the triangle?

    I know the answer and solution for this problem, but I am going to leave this for other users to work out.

    Sets of Possible Methods:
    1. Equilateral triangle is cyclic polygon in the circle.
    2. Forming extra right triangles with the given side is another way to approach this problem. I recommend this geometric strategy instead of standard approach (as shown below).
    3. Law of cosine is also doable. However, you need to consider 4 parameters for system of 4 equations.
    Post edited by Michael Huang on
  • Michael HuangMichael Huang Posts: 4,114 mod
    Since the problems aren't solved, I am going to remove them.

    I have moved this thread to Forum Games category...
  • _raoul_raoul Posts: 280 Pool Pro
    edited January 2017
    Nanny Ogg wrote: »
    When you google it you are supposed to get this.....

    x4npa4yq0rn7.png

    Ah! Could have been more clear with the equation next time! :D
    _raoul wrote: »
    Here's an interesting one to pass the time.

    A point inside an equilateral triangle has distances 3,4, and 5 from the three vertices of the triangle. What is the area of the triangle?
    I know the answer and solution for this problem, but I am going to leave this for other users to work out.

    Sets of Possible Methods:
    1. Equilateral triangle is cyclic polygon in the circle.
    2. Forming extra right triangles with the given side is another way to approach this problem. I recommend this geometric strategy instead of standard approach (as shown below).
    3. Law of cosine is also doable. However, you need to consider 4 parameters for system of 4 equations.

    Here's how us guys who are better at programming than geometry solve this kind of thing.

    If a side of the equilateral triangle is x, you can use Heron's formula four times so triangle (3,4,x)+(3,5,x) + (4,5,x) = (x,x,x), which gives a horrible unsolvable equation. Solve it numerically and plug into Wolfram to give the closed-form solution x=sqrt(25+12*sqrt(3)).
  • omgsunomgsun Posts: 2,239 Pool Champion
    edited March 2017
    _raoul wrote: »
    Nanny Ogg wrote: »
    When you google it you are supposed to get this.....

    x4npa4yq0rn7.png

    Ah! Could have been more clear with the equation next time! :D
    _raoul wrote: »
    Here's an interesting one to pass the time.

    A point inside an equilateral triangle has distances 3,4, and 5 from the three vertices of the triangle. What is the area of the triangle?
    I know the answer and solution for this problem, but I am going to leave this for other users to work out.

    Sets of Possible Methods:
    1. Equilateral triangle is cyclic polygon in the circle.
    2. Forming extra right triangles with the given side is another way to approach this problem. I recommend this geometric strategy instead of standard approach (as shown below).
    3. Law of cosine is also doable. However, you need to consider 4 parameters for system of 4 equations.

    Here's how us guys who are better at programming than geometry solve this kind of thing.

    If a side of the equilateral triangle is x, you can use Heron's formula four times so triangle (3,4,x)+(3,5,x) + (4,5,x) = (x,x,x), which gives a horrible unsolvable equation. Solve it numerically and plug into Wolfram to give the closed-form solution x=sqrt(25+12*sqrt(3)).

    heron's formula is the answer to life, no questions asked
    1. for math
    2. most people don't know it idk why
  • Abdul-lattīfAbdul-lattīf Posts: 96 Pool Player
    Solve:

    9x-7i>3(3x-7u).
  • _raoul_raoul Posts: 280 Pool Pro
    P25: 9,003,674?
  • _raoul_raoul Posts: 280 Pool Pro
    P25: 9,003,674?
  • _raoul_raoul Posts: 280 Pool Pro
    P25: 9,003,674?
  • _raoul_raoul Posts: 280 Pool Pro
    Ah, okay how about 9,003,000 for P25? Octagons instead of triangles.
  • _raoul_raoul Posts: 280 Pool Pro
    Ah, okay how about 9,003,000 for P25? Hexagons instead of triangles.
  • _raoul_raoul Posts: 280 Pool Pro
    Ah, okay how about 9,003,000 for P25? Hexagons instead of triangles.
  • _raoul_raoul Posts: 280 Pool Pro

    You are right about the number of solutions. However, the calculation seems to be incorrect. Keep trying though!

    Hint: Start off by playing around with exponents, which is the good start to solve for x. As you mentioned, you want to obtain of the Lambert-W form.

    Good thing you did a lot of research on the method! I don't think anyone sees that coming! :smiley:

    I double-checked the math on my solution and it looks okay, so to verify I solved it numerically for the principal branch of n=0, giving

    x= 0.000619235135215395 - i * 9.64910373153867e-007

    Plugging it back into the left and right sides of your equation, they are both equal.

    left = pow(1621.0*x, 2016.0) = -2015.9999999954 - i * 0.412858624584096
    right = pow(2016.0, left+2017.0) = -2016.00002831113 - i * 0.412858634433167
  • _raoul_raoul Posts: 280 Pool Pro

    You are right about the number of solutions. However, the calculation seems to be incorrect. Keep trying though!

    Hint: Start off by playing around with exponents, which is the good start to solve for x. As you mentioned, you want to obtain of the Lambert-W form.

    Good thing you did a lot of research on the method! I don't think anyone sees that coming! :smiley:

    I double-checked the math on my solution and it looks okay, so to verify I solved it numerically for the principal branch of n=0, giving

    x= 0.000619235135215395 - i * 9.64910373153867e-007

    Plugging it back into the left and right sides of your equation, they are both equal.

    left = pow(1621.0*x, 2016.0) = -2015.9999999954 - i * 0.412858624584096
    right = pow(2016.0, left+2017.0) = -2016.00002831113 - i * 0.412858634433167
  • _raoul_raoul Posts: 280 Pool Pro
    Check the algebra very carefully. The exponent 2016 does not apply for 1621 in the exponent of the right-hand side.

    Oh, I see. Then it is just the same thing except a=2016^2017/1621^2015 and z=1621*x^2016, so

    x= [(Wn(-2016^2017/1621^2015*log(2016)))/(-1621*log(2016))]^(1/2016)
  • _raoul_raoul Posts: 280 Pool Pro
    Check the algebra very carefully. The exponent 2016 does not apply for 1621 in the exponent of the right-hand side.

    Oh, I see. Then it is just the same thing except a=2016^2017/1621^2015 and z=1621*x^2016, so

    x= [(Wn(-2016^2017/1621^2015*log(2016)))/(-1621*log(2016))]^(1/2016)
  • _raoul_raoul Posts: 280 Pool Pro
    I doubt if any of us here have seen anything like that path finding problem before. If you want anyone here to take a stab at it, you'll need to point us straight at the relevant math or a very similar problem.
  • _raoul_raoul Posts: 280 Pool Pro
    I doubt if any of us here have seen anything like that path finding problem before. If you want anyone here to take a stab at it, you'll need to point us straight at the relevant math or a very similar problem.
  • _raoul_raoul Posts: 280 Pool Pro
    I doubt if any of us here have seen anything like that path finding problem before. If you want anyone here to take a stab at it, you'll need to point us straight at the relevant math or a very similar problem.
  • _raoul_raoul Posts: 280 Pool Pro
    I doubt if any of us here have seen anything like that path finding problem before. If you want anyone here to take a stab at it, you'll need to point us straight at the relevant math or a very similar problem.
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