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I never thought to did these in 3D, I will think...

The picture shows the tilings

New tessellation. I hope you enjoy it!

Hi everyone;

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

It's one of my sequence

Thanks. By the way the formula was:

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

How about this sequence:

-2, -1, 8, 31, 74, 143, 244, 383, 566, 799...

Guess the pattern.

**berliner**- Replies: 7

We all know what are the tessellations. If you just look at a tiled floor you see a regular tiling, then there are aperiodic tiling, which don't have a pattern. Here there are two aperiodic tiling. In picture 1 there is an aperiodic tiling with two tiles. In picture 2 you will see where I got the tiles. In picture 3 there's another aperiodic tiling whit one only tile.

Hi julianthemath;

Nice sequence, look at A001844 in the OEIS. These are also centered square numbers. I'm sure that in "Add 1, then post a fact" I will include julianmath numbers

Then julianthemath squares can be obtained with 4n^4 + 1.

Hi Lahbennett84;

Welcome to the forum! I'm here from a short time but I understood that this forum is full of qualified people ready to help anyone (including me). I'm sure you'll enjoy this forum

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 = MR

Thesis: MA > MA and MA ⊥ r

If MA ⊥ r we must demonstrate that ∠MAR = 90° and ∠ARM' + ∠AMR = 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 MAR and PHR are similar, because ∠ARM'= ∠HR ̂P and PR : MR = HR : AR since that HR = 2AR and PR = 2MR.

Since that MAR and PHR are similar all their angles are congruent, so ∠HPR = ∠AMR and ∠PHR =∠MAR and since that ∠PHR = 90° then MA = 90°, then MA ⊥ 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 MA = 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 doesnt 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 = MA and MA > MA.

Hi everyone;

This is the last theorem of this topic.

Hypothesis: P∉r and PH ⊥ r

H, A, R ∈ r and HA = HR

PM = MH and PM = MR

Thesis: MA > MA and MA ⊥ r

If MA ⊥ r we must demonstrate that ∠MAR = 90° and ∠ARM + ∠AMR = 90°. For hypothesis PH ⊥ r, so ∠PHA = 90° and ∠HPR + ∠HRP = 90°. ∠AMR = ∠HRP because they have in common the line r and PR. Triangles MAR and PHR are similar, because ∠AMR = HR ̂P and PR : MR = HR : AR since that HR = 2AR and PR = 2MR.

Since that MAR and PHR are similar all their angles are congruent, so ∠HPR = ∠AMR and ∠PHR =∠MAR and since that ∠PHR = 90° then MA = 90°, then MA ⊥ 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 MA = 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 doesnt 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 = MA and MA > MA.

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.

Hi Bob;

Interesting problem, I found a solution, but it's a little bit complicated for grade 8 students. I'm trying to find an easier solution.

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

Hi Bob;

Those are two brilliant methods. I said that M' is the midpoint of HR because it's the orthogonal projection of M. Perhaps there's an axiom which say that The orthogonal projection of the midpoint of a segment is the midpoint of the orthogonal projection of the segment itself.

If it doesn't exist we can try to demonstrate it.

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;

It's impossible to demonstrate that the triangle is isosceles when PHR = 90. I never said that. I said that the triangle is isosceles only when PHR = 90.

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.

Hi Bob;

As I explained in the hypothesis PH is perpendicular to r, but don't worry. I'm trying to demonstrate that MRH 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.