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Hi!

## #2 Re: This is Cool » Berliner's aperiodic tiling » 2013-04-24 07:36:53

I never thought to did these in 3D, I will think...
The picture shows the tilings

## #3 Re: This is Cool » Berliner's aperiodic tiling » 2013-04-24 04:07:19

New tessellation. I hope you enjoy it!

## #4 Re: This is Cool » Berliner's aperiodic tiling » 2013-04-16 16:57:30

Thanks! I would like to make these more bigger, but I have a problem to post the picture beacuse of its size.

## #5 Re: This is Cool » Berliner's aperiodic tiling » 2013-04-16 06:41:30

Hi everyone;

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

## #6 Re: This is Cool » Julianthemath's number 3 » 2013-04-14 21:33:04

It's one of my sequence

## #7 Re: This is Cool » Julianthemath's number 3 » 2013-04-14 20:30:47

Thanks. By the way the formula was:

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

## #8 Re: This is Cool » Julianthemath's number 3 » 2013-04-14 20:10:24

How about this sequence:

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

Guess the pattern.

## #9 This is Cool » Berliner's aperiodic tiling » 2013-04-14 09:24:37

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.

## #10 Re: This is Cool » Julianthemath's number » 2013-04-12 23:31:45

Sorry becuase your sequence is in the OEIS. It is never nice when you find something that someone else has already discovered.

## #11 Re: This is Cool » Julianthemath's number » 2013-04-12 21:50:36

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

## #12 Re: This is Cool » Julianthemath's number » 2013-04-12 19:57:14

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.

## #13 Re: Introductions » Hello :) » 2013-04-09 05:18:37

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.

## #14 Re: Introductions » Hello :) » 2013-04-09 03:50:43

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

## #15 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-08 20:25:31

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.

## #16 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-08 04:39:23

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 = 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°. ∠AMR = ∠HRP because they have in common the line r and PR. Triangles MAR and PHR are similar, because ∠AMR = 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.

## #17 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-06 08:21:48

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.

## #18 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-06 02:20:50

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.

## #19 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-04 22:25:55

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

## #20 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-04 09:17:18

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.

## #21 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-04 08:09:04

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.

## #22 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-04 07:11:36

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.

## #23 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-04 06:00:19

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.

## #24 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-03 07:25:12

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.

## #25 Re: This is Cool » Theorem in Euclidian geometry » 2013-04-03 03:00:14

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.

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