Simple Math Problem DEBATE

Remember the pythagorean theorem, A2 + B2 = C2? The three letters correspond to the three sides of a right triangle. In a Pythagorean triangle, and all three sides are whole numbers. Let's extend this idea to three dimensions. In three dimensions, there are four numbers. (A,B,C,G) The first three are the dimensions of a box, and G is the diagonal running from one of the top corners to the opposite bottom corner.


Just as there are some triangles where all three sides are whole numbers, there are also some boxes where the three sides and the spatial diagonal (A, B, C, and G) are whole numbers. But there are also three more diagonals on the three surfaces (D, E, and F) and that raises an interesting question: can there be a box where all seven of these lengths are integers?


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According to my calculations, A2+G= Nobody cares.

 

A2+G would actually = CG+5C

 

Hey how old are you even?

 

no.

 

A cube with dimensions of 0 would

 

Or a rectangular prism with dimensions of 1n x 2n x 2n with a corner to corner length of 3n

 

nvm, re-read the question lol

 

I understood everything untill the inreger part because I just finished NCEA lvl 1 and we did algebra. But its the holidays and I completely forgot some of the small stuff. what is an integer again?

 

A whole number (no fractions, decimals, or roots) ← some courses teach that integers and whole numbers have different definitions but you know what I mean

 

Yeah... I'm more interested in the math that actually means something in the real world.

 

trust me math is in the real world. all the idiots in my class would say exactly that

 

A whole number is just the positive values, an integer includes the negative values.

 

Where in New Zealand are you? I live in Auckland

 

ItsColinGuys Ok, so 'a', 'b' and 'c' must be integers. I pictured the cross-section of a cuboid and the lengths (as you called 'g') is √(a^2+√((b^2)+(c^2))). As you said, numbers like 3, 4 and 5 fit nicely into the Pythagorous theorem. However, since we are talking about 3-d objects, not only should √(a^2+√((b^2)+(c^2))) equal an integer but √(c^2+√((b^2)+(a^2))) must also equal that as well. I have made a graph: (b^2)=(a^2)+(x^2) and y=(a^2)+(x^2) to found out where they intersect. a is the first length of a triangle and b is the hypotenuse of the triangle. You can see that when b=5 and a=3, x must equal 4. Or when b=13 and a=12, x must equal 5. This is the way to find out perfect triangles where all the sides are integers. However, you cannot keep change the hypotenuse with opposite side or the adjacent side and have the new hypotenuse equal an integer. Like in the example, 3^2+4^2=5^2. You can't expect 5^2+3^2 to equal an integer. Therefore, the last part of √(a^2+√((b^2)+(c^2))) -> √((b^2)+(c^2))) cannot be an integer. So √a^2+... must equal an integer. You can't square an integer, add it to an irrational number, then take the square of that number and expect that number to be an integer.

In conclusion: The closest you can get to solving this problem is to have a, b, and c as integers and have two 'top corner to bottom corners' be integers. You can't get closer than that. Hopefully you can understand all this...

 

Rangiora - its near Christchurch somewhere. tbh NCEA lvl 1 was pritty easy.

 

I just finished NCEA Level 2 and I did both Stats and Calc. So I had to do both subjects in the same 3 hours because they both classify as Math . It was fun though

 

im doing level 2 calc this year and im not looking forard to it. also physics.

 

I did that in Level 2 too. It's really not too hard, do all the work in class then you have no homework . Feel free to ask me for maths help anytime, I love it.

Talking about maths and physics, I did make a projectile motion graph. Let's say you kick a soccer ball at an angle of 30 degrees at 5m/s. It can track it's motion until it lands. Here it is here, https://www.desmos.com/calculator/9bea8udoov

 

To me at least, I found the whole relativity / space side of physics to be the most interesting by far.