Math mini-contest problem for Day 7 on D.Buzz for April 2021 ๐Ÿ˜Ž

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Math problem for Day 7 on D.Buzz for April 2021 ๐Ÿ˜Ž

Holovision must store "The Source" - which can potentially provide limitless energy - into a tesseract whose enclosed hyperspace is 8 metersโด. What is the surface area of the tesseract that can contain The Source?



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My answer is "No Solution".

I have no way of even knowing how to solve this problem and the sci-fi just throws me off. A Tesseract, hyperspace and etc? Whatever I go with the simplest thing that comes to my mind.

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I'm sorry that the problem is not interesting for you. ๐Ÿ˜ฐ From my perspective, it is still in geometry that may be asked in high school math contests. ๐Ÿ˜

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Ok wild ass guess. If I want to find the edge length of a square I would square, a cube I would cube, so let's assume this holds and we can ^(1/4) . The edge length of the terrsecat is (8)^(1/4) or 1.68. The surface area in real space is 6x1.68x1.68 = 16.97 in real space.

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you're close! only the number of faces has to be scaled up: while it's 6 faces in 3D, it has 24 faces in 4D. Pretty hard to image...

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A = 24 a^2, where edge length a = fourth-root of the volume = 8 ^(1/4) = 1.682 => The surface area is 67.882.

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Answer for Day 7 Math problem

67.88 square inches ๐ŸŽฏ

Solution

Let's get started with some facts:

  1. A tesseract is the 4-dimensional analog of a cube.
  2. The enclosed hyperspace of a tesseract is equal to its side length raised to the 4th power.
  3. A square has 4 "one-dimensional surfaces" which are line segments of equal length and has a "surface area" of 4 * (side length).
  4. A cube has 6 two-dimensional surfaces which are all squares and has a total surface area of 6 * (side length)ยฒ.
  5. A tesseract has 8 "three-dimensional surfaces" which are all cubes. An analogy is by drawing a cube, then drawing another cube inside it, then connecting their respective vertices. From the 2 cubes, we have 12 surfaces, then the surfaces formed by the connecting lines form another 12 surfaces. Therefore, we have 24 two-dimensional surfaces.
  • It is also possible to perform the analogy by drawing two cubes separate from each other, then connect their respective vertices. It will show the same result, but it will be significantly harder to visualize than if one of the cubes is drawn inside/outside of the other.

To get the side length of the tesseract, we just get the fourth root of the enclosed hyperspace of the tesseract.

side length = โดโˆš(8 inchesโด) โ‰ˆ 1.6818 inches

We can now directly solve for the surface area of the tesseract given its side length.

surface area of tesseract โ‰ˆ 24 * (1.6818 inches)ยฒ โ‰ˆ 67.8823 inchesยฒ

Winner: @minus-pi ๐Ÿ…

The award of 1 HIVE has been sent to @minus-pi's Hive account. ๐Ÿ’ฐ

  • I'm sorry that @jfang003 did not like the "science fiction" thing. ๐Ÿ˜ฐ I just wanted my Math mini-contest problems for D.Buzz to be as Hive-themed as possible. ๐Ÿ™ Though, an online search will reveal that the formula for the surface area (2D) of a tesseract is 24 * (side length)ยฒ, or that there are 6 cubes comprising the 3D surface of a tesseract. ๐Ÿ˜
  • @failingforwards's answer was close, if only they realized that there are 24 two-dimensional surfaces on a tesseract instead of 6 surfaces! Thanks to @minus-pi for pointing that out! ๐Ÿ˜… By the way, I would like to point out that based on the official contest rules, every participant should have posted or commented at least once on the D.Buzz community to be qualified to win. ๐Ÿ™‚

Mentions: @holovision, @ahmadmanga (@ahmadmangazap), @eturnerx (@eturnerx-dbuzz), @dkmathstats, @paultactico2, and @appukuttan66 ๐Ÿค“
Special mentions: @dbuzz, @chrisrice, @jancharlest, and @mehmetfix ๐Ÿคฏ

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