Math Is Fun Forum

New tessellation. I hope you enjoy it!

Hi everyone;

I made a new tessellation. I hope you enjoy it!

Thanks. By the way the formula was:

a(n) = n^3-2n^2-1

Sorry Julianthemath, but in the OEIS there are 200000 sequences, I sent to them 40 sequences and they accepted just 10.

Some parts of math are fantastic, like fractals, but in my opinion also Euclidian geometry is awesome. To address these arguments we must learn some boring and important rules. Aniway, in this forum there are a lot of people ready to help.

Hi Bob;

This is the correct version, sorry, but I write an entire word document I realized that copying the text on the forum did not keep the symbols then I had to rewrite all and I made some mistakes.

Hypothesis: P∉r and PH ⊥ r
                  H, A, R ∈ r and HA = AR
                  PM = MH and PM’ = M’R
Thesis: MA > M’A and M’A ⊥ r
If M’A ⊥ r we must demonstrate that ∠M’AR = 90° and ∠ARM' + ∠AM’R = 90°. For hypothesis PH ⊥ r, so ∠PHA = 90° and ∠HPR + ∠HRP = 90°. ∠ARM' = ∠HRP because they have in common the line r and PR. Triangles M’AR and PHR are similar, because ∠ARM'= ∠HR ̂P and PR : M’R = HR : AR since that HR = 2AR and PR = 2M’R.
Since that M’AR and PHR are similar all their angles are congruent, so ∠HPR = ∠AM’R and ∠PHR =∠M’AR  and since that ∠PHR = 90° then M’A = 90°, then M’A ⊥ r.
In the first theorem of this topic we saw that distances from the midpoints of the oblique to the line are congruent. If we take PH as an oblique then M’A = MH, because M’ is the midpoint of PR and M is the midpoint of PH. We know that MA > MH, because M is point which doesn’t belong to r and MH is its distance to r and MA is an oblique and oblique carried out from the same point in a straight line, increasing the distance from the point to the line. So, MA > MH, MH = M’A and MA > M’A.

Hi Bob;

I used Erone's formula and I obtained this equation:

From this I obtained x=27.9374... and y: 32.3504...
It sounds like a strange result, but I can tell you that your solution is the best and more comprehensible and I'm a grade 8 student.

You're totally right, Bob. Well... I'll find something else

Hi Bob;

I've an alternative demonstration:
Tracing the segment MM' which is perpendicular to r and M' is the midpoint of HR HM' = M'R, so. The simmetry of axis MM' transforms M in M (because axis points are united), M' <-> M' and H <-> R because HM' = M'R and M' = M'. Since that extrems corresponds then segments, HM and MR, must correspond and since they correspond they are congruent.

hi Bob;

this is my demonstration but I dind't use the circle.

Hypothesis: P doesn't belong to r and H, R belong to r. M is the midpoint of PR.
Thesis: HM = MR

Begin to build the bisection s of ∠HMR, which is also the axis of simmetry of the angle. Since that s is the axis simmetry of ∠HMR:
Ss: HM <-> MR
Since that simmetry is an isometry and isometry manteins the distance between the points then HM = MR.
So, since that the triangle HMR has got two congruent line segments the triangle is isosceles.

Another theorem! I hope this one has not been demonstrated yet

Hypothesis: P, S ∉ r ∧ PH ⊥ r   
                  H, R  ∈ r  ∧ HS ∩ PR = {M}
Thesis: ∠SMR > ∠MRH
                                                                                                                                       
Let's indicate with π the straight angle. ∠SMR +∠HMR = π . ∠HMR belongs to the triangle MRH. So, since the sum of the interior angles of a triangle is 180 °, the sum of ∠MHR and ∠MRH plus ∠HMR give π. So, since supplementary angles of congruent angles are congruent, and that ∠HMR is congruent to itself then the sum of ∠MHR and ∠MRH is congruent to ∠SMR, then ∠SMR > ∠MRH. If HS is the median of PR then ∠SMR = 2∠MRH, because MRH is an isosceles triangle so the angles at the base are congruent.