Problem of the week #1
Okay, so recent observations have led me to the conclusion that the general level of logical and mathematical ability amongst nomadlifers needs some encouragement. And so I am starting up a problem of the week section, where you mathematicians out there can show off your prowess, and the rest of you can be further educated in this wonderful field (especially Luli, who I know just loves it when my blog postings get educational). I might even think of a prize to bestow upon those who contribute the best answers.
And so to kick things off, a quick dabble into geometry. What is the side length (or area, if you prefer to answer that way) of the largest equilateral triangle that can fit entirely inside a unit square?
And for the more adventurous, once you get past that one, try the same question but in 3 dimensions. (i.e. what is the side length of the largest regular tetrahedron that can fit inside a unit cube?)
Creativity is encouraged.
And so to kick things off, a quick dabble into geometry. What is the side length (or area, if you prefer to answer that way) of the largest equilateral triangle that can fit entirely inside a unit square?
And for the more adventurous, once you get past that one, try the same question but in 3 dimensions. (i.e. what is the side length of the largest regular tetrahedron that can fit inside a unit cube?)
Creativity is encouraged.
16 Comments:
NERD !!! You are a bonofide nerd. I knew it before but now you have well and truely come out of the closest.
Jen x :)
Yep. And I know how to make popcorn too.
root two
Oh come on we can do better than offer random guesses. This is a post for people who are not lazy about details and enjoy meetings the rigours of logical proof. And the answer is not root two.
Giving it a fast look, it seems like the answer for question No 1 would be that the side lenght is 1.035 or 1/Cos(15�).
I'll try looking at the 3D one tonight see if I can come up with something.
Yippee, we have liftoff.
Indeed the side length for the 2d version is 1/cos(15�).
The 3d has a slightly larger side length, as I'm sure you've realized.
woops, i just re-read the question and realized my attention span made me miss a crucial part of the question. alas
and I have no clue what you guys are talking about.
Mathematics, Dody my friend. The unshaken Foundation of Sciences and the plentiful Fountain of Advantage to human affairs, to quote Isaac Barrow.
Embrace. Enjoy. And remember, the essence of mathematics is not to make simple things complicated, but to make complicated things simple.
Come on, Dody's just trying to pretend he's not nerdy enough to know. Everyone knows that he has this problem written on the back of his hand.
Mel,
I wish can fake false modesty in Math. My math ability is pathetic actually; I cannot think in Math.
i am working on a solution for the cube one...
Okay, well having posed the tetrahedron/cube problem before I actually knew the answer myself, I thought I better get around to solving it before I post the next problem. I hope it kept some other minds out there active for a little while too.
The maximum side length of a regular tetrahedron that can fit inside a unit cube is, to 14 decimal places, 1.06066017177982.
Or for those purists who prefer their answers in exact form, the side length is sqrt(8/(3x^2+2*sqrt(3)*x+1)) where x is the solution to the quartic equation:
?513*x^4+12*sqrt(3)*x^3-414*x^2-60*sqrt(3)*x+25=0? within the range of 0.9 to 1.
Alternatively, after solving the quartic for x, find alpha (between 0 and 90�) such that cos(alpha) = x, and beta (between 0 and 90�) such that cos(beta) = sqrt(3)/3, and then the side length can also be expressed as 1/sin(alpha + beta).
Problem of the week 2 shall involve no mathematics whatsoever. Pure logic is the order of the day here.
(and Jen, since I know you are wondering about it, sqrt stands for "square root" :P )
JERK ! I like POPCORN :)
hey dan, long time no see
Well the anonymous new entrant to the comments page, Leon, (who isn't actually anonymous to me at all) has just cut me down to size in the space of about 10 seconds after seeing this problem by googling the phrase "largest tetrahedron inside a cube" and then showing me this page. It seems that Mix, whether he knew it or not, was spot on with his initial answer. This solution is an orientation that I just completely overlooked...
Well, if nothing else, this exercise ended up giving me a learning experience, and so I say it was worthwhile. Cheers.
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