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April 2

Closure of the union of sets

Is this true: $cl\left(\cup _{i}A_{i}\right)=cl\left(\cup _{i}cl\left(A_{i}\right)\right)$ ? $cl\left(\cup _{i}A_{i}\right)\subset cl\left(\cup _{i}cl\left(A_{i}\right)\right)$ is trivially obvious, so all there is to show now is $cl\left(\cup _{i}cl\left(A_{i}\right)\right)\subset cl\left(\cup _{i}A_{i}\right)$ . I believe the answer is "yes". If $x$ is an interior point of some $A_{i}$ then it is clearly in that set. If $x$ is a boundary point of some $A_{i}$ then it will be in $cl\left(\cup _{i}A_{i}\right)$ (although not necessarily as a boundary point). Widener (talk) 11:31, 2 April 2012 (UTC)[reply]

cl(A_i) is the minimal closed set containing A_i. Since

cl\left(\cup _{i}A_{i}\right)

is also a closed set containing A_i, it must contain cl(A_i). Therefore

\cup _{i}cl\left(A_{i}\right)\subset cl\left(\cup _{i}A_{i}\right)

. Since

cl\left(\cup _{i}A_{i}\right)

is closed, this implies

cl\left(\cup _{i}cl\left(A_{i}\right)\right)\subset cl\left(\cup _{i}A_{i}\right)

. Rckrone (talk) 17:31, 2 April 2012 (UTC)[reply]

Morley triangle

Is there a formula for the side of the Morley triangle, preferably solely in terms of the sides of the reference triangle?

Is there a formula for the area of the Morley triangle, preferably solely in terms of the area of the reference triangle? Duoduoduo (talk) 15:41, 2 April 2012 (UTC)[reply]

Naively, isn't the second question impossible? An affirmative answer is tantamount to saying that all triangles of a given area have the same Morley triangle, and vice versa. Aren't there two different-area triangles with a same-area Morley triangle, or two same-area triangles with different-area Morleys triangles? (apologies if I've missed something obvious, but a positive answer to the the second question seems to imply a much stronger claim than it would for the first. I'd also be interested to see proofs contraindicating my claims...) SemanticMantis (talk) 02:10, 3 April 2012 (UTC)[reply]

According to the Mathworld article First Morley Triangle,

It has side lengths

a^{'}=b^{'}=c^{'}=8R\sin(A/3)\sin(B/3)\sin(C/3)

where R is the circumradius of the reference triangle and A, B, C are the angles of the reference triangle. By extension, the area could be expressed using Heron's formula in terms of the sides. From this, can it be concluded that both questions are impossible? Duoduoduo (talk) 14:27, 3 April 2012 (UTC)[reply]

I'm not sure I follow your logic. Can you describe the reasoning a little more? Actually, I think this formula might lead to an answer to your first question. We can find R in terms of known sides of the reference triangle, and also find all angles in terms of the sides, right? SemanticMantis (talk) 21:56, 3 April 2012 (UTC)[reply]

You're right. I think what I had in mind when I wrote that was that there must be no algebraic formula involving only roots of real numbers -- algebraically trisecting an arbitrary angle requires solving a cubic with three real roots, which in general can be done algebraically only by taking cube roots of non-real numbers. Duoduoduo (talk) 15:06, 4 April 2012 (UTC)[reply]

Ok, I think we've learned 1. That the second question cannot have an affirmative answer. 2. That there is an answer to the first question as-written, but it is a not closed form, "by radicals" type formula. Anyone disagree? (not looking to pick a fight, I just haven't worked this all out rigorously enough to thoroughly trust my reasoning) SemanticMantis (talk) 15:16, 4 April 2012 (UTC)[reply]