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TA: Bitcoin Consolidates Gain: What Could Trigger Fresh Rally

Bitcoin is showing positive signs above the $61,200 support against the US Dollar. BTC could start a fresh rally if it clears the $62,200 resistance zone.

  • Bitcoin is trading in a positive zone above the $60,000 and $61,200 levels.
  • The price is now trading above $61,200 and the 100 hourly simple moving average.
  • There is a major bullish trend line forming with support near $61,000 on the hourly chart of the BTC/USD pair (data feed from Kraken).
  • The pair could start a fresh rally and it might rise towards the $63,500 level in the near term.

Bitcoin Price Remains Supported

Bitcoin price remained strong above the $60,000 zone and traded in a positive zone. BTC jumped above the $62,000 level before starting a downside correction.

A high was formed near the $62,885 before there was a minor decline. The price declined below the $62,000 and $61,200 levels. However, the price remained supported near the $59,200 level and the 100 hourly simple moving average.

Recently, the price climbed higher and climbed above $61,200. However, the price is struggling above the $62,200 level. A high was formed near $62,672 before there was another decline. It traded as low as $59,900 and recently recovered above $61,000.

Bitcoin surpassed the 50% Fib retracement level of the downside correction from the $62,672 swing high to $59,900 low. There is also a major bullish trend line forming with support near $61,000 on the hourly chart of the BTC/USD pair.

The pair is now testing the 76.4% Fib retracement level of the downside correction from the $62,672 swing high to $59,900 low. On the upside, an initial resistance is near the $62,000 level. The first key resistance is near the $62,200 level.

Source: BTCUSD on TradingView.com

A clear break above the $62,200 resistance could open the doors for more upsides. The next major resistance sits near the $63,500 level, above which the price might rise towards the $65,000 level.

Dips Supported In BTC?

If bitcoin fails to clear the $62,200 resistance zone, it could start a downside correction. An immediate support on the downside is near the $61,200 level.

The first major support is now forming near the $61,000 level, the trend line, and the 100 hourly SMA. A downside break below the $61,000 level could push the price towards $60,000.

Technical indicators:

Hourly MACD – The MACD is now losing pace in the bullish zone.

Hourly RSI (Relative Strength Index) – The RSI for BTC/USD is now above the 50 level.

Major Support Levels – $61,200, followed by $61,000.

Major Resistance Levels – $62,200, $62,850 and $63,500.

Can a series of attached tetrahedrons be used for cryptography?

(based on a prior question with triangle) Given a regular tetrahedron T1 (with his 4 vertices A,B,C,D lying in a finite Field $mathbb F_N^3 $ with $N$ a prime) another attached regular tetrahedron can be constructed by reflection 1 vertex at the plane defined by the 3 other vertices. E.g. vertex $A$ can be reflected::Listen

(based on a prior question with triangle)

Given a regular tetrahedron T1 (with his 4 vertices A,B,C,D lying in a finite Field $mathbb F_N^3 $ with $N$ a prime) another attached regular tetrahedron can be constructed by reflection 1 vertex at the plane defined by the 3 other vertices. E.g. vertex $A$ can be reflected to $A’$ with:
$$A’ equiv (B+C+D)cdot 2 cdot(3)^{-1} – A mod N$$
This can be repeated multiple times (with alternating vertices).
The edge length in between all vertices of a tetrahedron remains constant during this (square euclidean norm e.g. $const = |A-B|^2 mod N$ ).


Question:
Given just two random tetrahedron T1 and T2 (and $mathbb F_N^3 $) can an adversary compute how to construct T2 out T1?

(Assuming there is a way and modulo $N$ chosen high enough that testing all tetrahedron would take to long)
Or can he do it (much) faster than just applying it step by step?
(like for example at EC a generator $g$ don’t need to be applied $m$ times if we want to compute $g^m$. I’m looking for a more-than-one-dimensional technique which can not be reduced to a single dimension problem and need to be computed (to some amount) step-by-step.)


Some tests:
In euclidean space a regular tetrahedron can not fill up all space. If the attachment progress is in one direction it has an equal structure as the Boerdijk–Coxeter Helix with vertices coordinates $(rcos ntheta,rsin ntheta,n h)$. Two vertices can not have the same $x$ and $y$ coordinates there.
Doing the same inside a finite Field $mathbb F_N^3 $ they will match up again at some point.

In tests this took up to $N^2+N$ steps.

If we attach the next tetrahedrons around an edge it won’t fill up in euclidean space (left side) but in tests it did match up again with the starting tetrahedron after $N-1$ steps in $mathbb F_N^3 $(right side)

euclidean around edge around edge in finite field

If they just share one common point it also won’t work in euclidean space (left side) but it does in $mathbb F_N^3 $ :
euclidean around vertex around vertex in finite field
The number needed for $mathbb F_N^3 $ depends on $N$. Except for $N=11$ ($66$ there) the max number seems to be $3N pm 3$
(could not find any rule yet)
some examples:

N8389971011031071091131271311371397192017100374001350021
number42264241021025433048384664084202154504301083078150060

The prior related question with triangle instead of tetrahedrons could be reduces to a much simpler problem because a large subset sharing the same orientation.
In tests so far tetrahedrons have more than $N^2$ orientations if they also need $N^2+N$ steps in one direction to match up again (for $N>11$). It looks like they even get bigger with larger $N$ ($11: 110$, $13: 182$, $17: 408$, $31:2480$).

Given a tetrahedron with a certain edge length more than $N^5$ different tetrahedron could be produced (for $N$ where they also need $N^2+N$ steps in one direction to match up again (for $N>11$)). It also looks like the amount gets even bigger with larger $N$ ($11: 11^4cdot 10$, $13: 13^4cdot 14$, $17: 17^4cdot 21$).


Finding $1$ tetrahedron in $N^5$ would be very hard even $1$ in $N^2$ would be hard. Does the problem scale linear with the number of orientations or can it simplified to an easier problem? Or is it even harder than the number of orientations?

TA: Bitcoin Consolidates Gain: What Could Trigger Fresh Rally

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