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February 23

Sides of a pyramid

Given an arbitrary pyramid with four triangular sides, is there a relationship among the areas of the triangles? Bubba73 ^{You talkin' to me?} 02:27, 23 February 2022 (UTC)[reply]

Well, if you know the length of all six sides, Heron's formula will give you one, won't it? Bubba73 ^{You talkin' to me?} 03:00, 23 February 2022 (UTC)[reply]

Heron's formula gives us 4 equations in 10 variables (4 for the face areas, 6 for the edge lengths). To establish a relationship between the areas, we need to eliminate the 6 edge-length variables. Elimination is generally only possible if there are more equations than variables to be eliminated. Only in exceptional circumstances will this succeed otherwise. (For example, we can eliminate the two variables

a

and

b

from the pair of equations

~x=a{-}b,y=b{-}a~

to give

~x+y=0.

) I have no expectation that in this case the equations conspire to make this possible. What? There is no article on vanilla variable elimination? What happened to WHAAOE? --Lambiam 08:34, 23 February 2022 (UTC)[reply]

Thanks. In this case, it isn't just any 10 variables. Does the geometry impose restrictions to help eliminate variables? Bubba73 ^{You talkin' to me?} 01:00, 24 February 2022 (UTC)[reply]

No. The only geometric restrictions are the triangle inequalities plus the tetrahedron inequality stated by CiaPan below, which are not helpful in any way. I think that given any four areas satisfying the inequalities, there are many tetrahedra with these areas. --Lambiam 14:43, 24 February 2022 (UTC)[reply]

For a given A>0, there are infinitely many B's for which there is pyramid with side areas A, A, A, B. (The upper limit for B is 3A.) This implies there is no algebraic equality satisfied by the area. There is presumably an analog of the triangle inequality. --RDBury (talk) 13:45, 23 February 2022 (UTC)[reply]

A simplest, most obvious, is that the sum of areas of any three sides is greater than the fourth one... --CiaPan (talk) 21:00, 23 February 2022 (UTC)[reply]

It may be intuitively obvious, but I think a formal proof would be non-trivial. I have an idea of how it might be done though. Euclid's proof of Proposition 20 is non-trivial as well. --RDBury (talk) 21:53, 23 February 2022 (UTC)[reply]

We have a section of an article about this, added by Quantling a few months ago with no sources at all :/. --JBL (talk) 02:09, 24 February 2022 (UTC)[reply]

Yes, no side can have area greater than the sum of the other three sides' areas. Orient the tetrahedron so that the one side we want to bound is in the x-y plane. The fourth vertex will be at some point, which we orthogonally project to the x-y plane. The projected point sitting in the x-y plane, if connected by line segments to each of the three vertices of the base divides the base triangle into three parts. Each of these parts has an area and the sum of these areas is exactly the area of the base triangle. The original three sides that these parts came from have area at least as big as their projections into the x-y plane. Something like that, though the case that the fourth vertex has a projection outside of the base triangle also has to be handled. But surely some elementary textbook has a more rigorous proof of this. —

Q

uantling (talk | contribs) 02:29, 24 February 2022 (UTC)[reply]

If the apex projection falls outside the base, then the three sides' projections partially overlap outside the base. Then the sum of sides projections' areas is strictly greater than the area of a base and the inequality still holds. --CiaPan (talk) 11:17, 24 February 2022 (UTC)[reply]

But surely some elementary textbook has a more rigorous proof of this. Oh good lord. --JBL (talk) 12:36, 24 February 2022 (UTC)[reply]

We could prove the tetrahedral inequality using cross products at that the area of a triangle is one half the magnitude of the cross product of the vectors representing two of its sides. Suppose the tetrahedron has its vertices at $p_{0},p_{1},p_{2},p_{3}$ . Writing $A_{i}$ for the area of the triangle that is opposite the point $p_{i}$ and using the FOIL method to multiply things out, we have ${\begin{aligned}2A_{0}&=|(p_{2}-p_{1})\times (p_{3}-p_{1})|=|p_{2}\times p_{3}-p_{2}\times p_{1}-p_{1}\times p_{3}|\\2A_{1}&=|(p_{2}-p_{0})\times (p_{3}-p_{0})|=|p_{2}\times p_{3}-p_{2}\times p_{0}-p_{0}\times p_{3}|\\2A_{2}&=|(p_{3}-p_{0})\times (p_{1}-p_{0})|=|p_{3}\times p_{1}-p_{3}\times p_{0}-p_{0}\times p_{1}|\\2A_{3}&=|(p_{1}-p_{0})\times (p_{2}-p_{0})|=|p_{1}\times p_{2}-p_{1}\times p_{0}-p_{0}\times p_{2}|\,.\end{aligned}}$ We have that $|u|+|w|\geq |u+w|$ (the ordinary triangle inequality) and that $u\times w=-w\times u$ for any such vectors, and so ${\begin{aligned}2(A_{1}+A_{2}+A_{3})&=|p_{2}\times p_{3}-p_{2}\times p_{0}-p_{0}\times p_{3}|+|p_{3}\times p_{1}-p_{3}\times p_{0}-p_{0}\times p_{1}|+|p_{1}\times p_{2}-p_{1}\times p_{0}-p_{0}\times p_{2}|\\&\geq |p_{2}\times p_{3}-p_{2}\times p_{0}-p_{0}\times p_{3}|+|p_{3}\times p_{1}-p_{3}\times p_{0}-p_{0}\times p_{1}+p_{1}\times p_{2}-p_{1}\times p_{0}-p_{0}\times p_{2}|\\&\geq |p_{2}\times p_{3}-p_{2}\times p_{0}-p_{0}\times p_{3}+p_{3}\times p_{1}-p_{3}\times p_{0}-p_{0}\times p_{1}+p_{1}\times p_{2}-p_{1}\times p_{0}-p_{0}\times p_{2}|\\&=|p_{2}\times p_{3}-p_{2}\times p_{1}-p_{1}\times p_{3}|=2A_{0}\end{aligned}}$ Dividing by two gives the tetrahedral inequality, $A_{0}\leq A_{1}+A_{2}+A_{3}$ . Any side's area is not more than the sum of the other three sides' areas and we can combine all four such tetrahedral inequalities into the single expression $\max(A_{0},A_{1},A_{2},A_{3})\leq {\frac {1}{2}}\left(A_{0}+A_{1}+A_{2}+A_{3}\right)\,.$ — $Q$ uantling (talk | contribs) 20:49, 25 February 2022 (UTC)[reply]

does an N-simplex of side x inside of an N-cube of side x always contain the center of the N-cube?

or to put it another way, for an N-cube of side 1, does a N-simplex of side 1 entirely inside the N-cube always including the point where all N coordinates are 0.5. If not, is it possible to have more than one N-simplexes inside an N-cube, and if yes, is there any way to calculate how many for each N?Naraht (talk) 07:07, 23 February 2022 (UTC)[reply]

The distance between a vertex of the unit hypercube and its centre equals

{\tfrac {1}{2}}{\sqrt {n}}.

If

n>4,

that distance exceeds

1.

It is always possible to position the regular unit

n

-simplex inside the

n

-cube so that it shares a vertex. Then no point of the simplex has a distance from that vertex exceeding

1,

so it does not contain the

n

-cube's centre. The diametrically opposite simplex is then disjoint. I suppose it may be possible to come up with an algorithm to determine how many simplices can be packed, but this is bound to be intricate, and proving the correctness of the algorithm will not be an elementary exercise. --Lambiam 10:08, 23 February 2022 (UTC)[reply]

In fact, by locking one vertex of the simplex to a hypercube vertex and keeping its centre on the line connecting that vertex to the centre of the hypercube, the

(n{-}1)

-simplex hyperface opposite the vertex is orthogonal to that line and its distance from the vertex is the height of the

n

-simplex. Already for

n=3,

we find

{\sqrt {\tfrac {n+1}{2n}}}<{\tfrac {1}{2}}{\sqrt {n}}.

--Lambiam 10:25, 23 February 2022 (UTC)[reply]

Thanx, I'd thought of locking one edge of the simplex to an edge of the cube, but didn't appreciate that this could be done with the corner. No clue on the algorithmNaraht (talk) 22:16, 23 February 2022 (UTC)[reply]

So, this is not the exact same problem, but a very closely related one is covered by this video by 3Blue1Brown. It deals a similar sort of question, dealing with inscribing a simplex inside of a sphere instead of a cube, but it DOES go into some problem-solving strategies which may be relevant to this problem as well. --Jayron 32 14:59, 25 February 2022 (UTC)[reply]

Ancient Greek anecdote

What's the name of the Greek mathematician in the story? The story goes like this. A student once asked, "What's the use of gemeotry?" The teacher then told his servant to give his student 3 tokens because geometry was useless at the time, so he gave the student the money to compensate for his learning. 137.26.237.163 (talk) 22:19, 23 February 2022 (UTC)[reply]

That story is told about Euclid.[1] CodeTalker (talk) 05:38, 24 February 2022 (UTC)[reply]

The anecdote comes from Stobaeus, who ascribes it to one Serenus, possibly Serenus of Antinoöpolis.^[2] The three obols of Stobaeus' anecdote were worth considerably more than the three pence of the retelling. This might be half the daily gain of a skilled worksman. --Lambiam 13:45, 24 February 2022 (UTC)[reply]

A version I heard in school when someone asked about what's the use of something in mathematics: "What's the use of a baby?" Similar questions might be asked of most of the things people learn in school: What's the use of knowing the plot of Hamlet? What's the use of knowing which decade WWI took place in? What's the use of knowing the life cycle of a frog? Sherlock Holmes decided that knowing the Earth revolved around the sun was useless and that he would try to remove that information from his brain to make room for something more relevant to his work. For some reason people tend to ask this type of question more often when it comes to mathematics. --RDBury (talk) 17:40, 24 February 2022 (UTC)[reply]

People ask this question of mathematics, because many people do, as their life's vocation, menial labor that often does not use many of the more esoteric aspects of mathematics. You know and I know that mathematics is as useful as any academic pursuit, useful in building mental capacity in general, even if one does not strictly use every mathematical concept on a daily basis, but for many people, the "how will this make me better at the specific tasks I need to do today" is as far as they see, and no, most of what is taught in the standard mathematical curriculum is not directly useful in that way. It's much more useful than that in making someone a better informed, more generally intelligent and wiser, and better thinking person, but many people aren't interested in that. --Jayron 32 19:06, 24 February 2022 (UTC)[reply]

Philip Greenspun did something like that in 1998, as a protest against high MIT tuition. From [3]:

I have decided to stop personally participating in the system of extracting money from MIT kids and their families. On Thursday, March 12, 1998, I guest-lectured an MIT class (on designing database-backed Web services). I calculated that the students were paying about $80 in tuition/lecture-hour. I withdrew a stack of $100 bills from my BankBoston account and I handed one out to each undergraduate in the course. I then proceeded to give my talk, telling the students that I was happy to teach them but I was not going to take their money.

2601:648:8202:350:0:0:0:C115 (talk) 20:41, 25 February 2022 (UTC)[reply]