DejMar
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Posts posted by DejMar
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You are missing some parameters in your problem. What do you mean by "how can this watch be designed to work?" Is it not already designed to work? Are you asking how an LCD device works?
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Let two squares share a side. (These two squares will share an infinite number of points on the line segment that represents the side of the two squares.)
Let a third square be placed such that one of its corners will form an isoceles triangle with its base about 1/3 of the length of the line segment in the middle of the line segment.
(At this juncture, the three squares share only two points.)
Let a fourth square mirror the third along this line segment.
(The four squares now share the same two points.
Without loss of generality, let the distance between the two points be an interger distance. It is not nessecary that these points be integral distance, but only to assist in demonstrating an infinite number of squares can share these two points.
Place the fifth square such that its corner forms a Pythagorean triple with the two points.
A sixth square can mirror this square along the line segment, as well.
Repeating the former two steps, using a different pythagorean triple (scaled as necessary), an infinite number of pythagorean triples, you can be placed, and each will only share the two points. Thus, the answer is that there does exist a common point for 6 (and 7, and 8, and 9, etc.).
*Remove any one of the first two squares, as these two shared more than one point, and were placed only for personal construction. -
Another question, as the problem sounds like a trick. Though the five squares share exactly two points, will it require some of the squares to intersect more than two points in total with all squares? Or are the shared two points only referring to the points shared between any two squares?
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Oops. I did make an error. I incorrectly read "It was a draw" for when all players used the teams, and for the subset of three members. K-man got it right. -
Multiplying both sides of the equation, x
3 + 1/x3 = q, by x3, and rearranging the equation in standard polynomial form, results in a sextic (aka hexic) polynomial:
x6 - qx3 + 1 = 0
The sextic polynomial has six roots, some of which may be multiple roots and degenerate roots. As the polynomial is with even degree, it has both a global maximum and global minimum in addition to local maximum(s) and local minimum(s).
By substituting a = x3, the discrimanant of the quadratic equation for the roots of a can be used to denote the inflection points of the polynomial. (One may also find these inflection points by taking the derivative of the function). These occur at -2 and 2.
If x = 0, we see the equation is undefined (1 = 0).
As x : x < -2 → -∞, q → -∞,
as x : x > 2 → +∞, q → +∞,
as x : x < -2 → 0-, q → -∞, and
as x : x > 2 → 0+, q → +∞.Restated,
x3 + 1/x3 = q is discontinuous where x=0 (thus, not satisfied for that real x),
For a < q < b : q = 0,
a = lim [x→0-] x3+1/x3 = -∞
b = lim [x→0+] x3+1/x3 = +∞ -
@bonanova, I had already demonstrated that in post #9. A difference is that I began at the board's heart instead of a corner.
Oh, and yes, kudos to k-man. -
x
3 + 1/x3 = q
multiplying both sides by x^3, and restating
x6 - qx3 + 1 = 0
let a = x3, then
a2 - qa + 1 = 0
using the quadratic formula, we find the discriminant as (q2 - 4)
The discriminant shows a,i.e., x3, is complex and not real if q2 < 4, i.e.,
-2 < q < 2. -
BMAD, It seems you left point 1 of gavinksong's request for clarifications unclarified. Your answer is ambiguous in that regard. And, to add a question of my own, are the squares to be considered residing in the same Euclidean plane?
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An assumption is made that the strength of a player's position in the tug-o-war chain has no bearing other than which side of the rope the player is on, and the strength of a team is the sum of the individual players' tug-o-war power.
Another assumption is that the two new mini-teams is composed of the two members lost to the original teams, and Carol+Douglas+Eric = Harry+Ian+Jessica.
Let the initial letter of the player's name represent the player's tug-o-war power. Then the following should be true:
A+B = F+G
A+C+D+E+F < B+G+H+I+J
Other than when forming teams of unbalanced numbers, I can not ascertain any other valid claims. -
Given the radius of the circle as 10, the circumference of the circle is 20π. The distance between A and B is 6π.
Each point is 108° apart from a point labeled with the alphabetic character that is adjacent in the alphabetic sequence, with a and j also 108° degrees apart, and each point being 2π units apart, with a complete circuit being thrice around the circle (three rings ... so send in the clowns.) -
A definition is a statement that clarifies the meaning of a word or the extent of distinctness of an abstract or real object, image, or sound.
In this case, the clarifying statement given here is for the meaning of the word definition.-
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1) Power is Money [Money is Power]
[Costs an arm and a leg + pulling your leg]
2) empty vessels make the most noise (loudest sound)
3) The pen is mightier than the sword.
4) Honesty is the best policy.
7) Time and tide waits for no one.
8) Knowledge is power
9) A barking dog never bites.
10) An apple a day keeps the doctor away.
11) There is strength in numbers
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@bonanova
Im curious how you found a move length of 13.3. Wouldn't that take you off the playing field? A chessboard is 8x8, but as the initial starting square is not counted*, the longest number of squares moved (as opposed to distance*) is 7. -
...then the highest number of moves may be 9 (10, if a zero distance move is permitted).
One solution is
(0) C1 = 0
(1) C2 = 1
(2) D1 = √2 (≈1.414)
(3) F1 = 2 (en passant: E1)
(4) D3 = 2√2 (≈2.828) (en passant: E2)
(5) G3 = 3 (en passant: E3 and F3)
(6) G8 = 5 (en passant: G4, G5, G6, and G7)
(7) A8 = 6 (en passant: F8, E8, D8, C8, and B8)
(8) A1 = 7 (en passant: A7, A6, A5, A4, A3, and A2)
(9) F6 = 5√2 (≈7.071) (en passant: B2, C3, D4, E5)
Under the interpretation that the total squares 'visited' are to be counted, and not simply the number of moves, the number of squares visited (starting, passing through, and ending in) is much higher than the count of 9 (or 10, the number of moves). Rather, including the beginning square, the 9 squares where the move ended, and the 23 squares passed through, the count would be 33 (or 32, if the starting square is not counted as being visited.)
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... then the count may be 13 (14, if a zero move is permitted).
(Note: It has been inferred that only a "Queen's move" is permissible, i.e. a diagonal move must be parallel to the diagonal of the square playing field, and otherwise, only orthogonal moves are performed.)
(0) E5 = 0
(1) D5 = 1
(2) E4 = √2 (≈1.414)
(3) C4 = 2
(4) E2 = 2√2 (≈2.828)
(5) H2 = 3
(6) D2 = 4
(7) A5 = 3√2 (≈4.243)
(8) F5 = 5
(9) B1 = 4√2 (≈5.659)
(10) B7 = 6
(11) G2 = 5√2 (≈7.071)
(12) A8 = 6√2 (≈8.485)
(13) H1 = 7√2 (≈9.899)-
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...then the highest number of moves may be 9 (10, if a zero distance move is permitted).
One solution is
(0) C1 = 0
(1) C2 = 1
(2) D1 = √2 (≈1.414)
(3) F1 = 2 (en passant: E1)
(4) D3 = 2√2 (≈2.828) (en passant: E2)
(5) G3 = 3 (en passant: E3 and F3)
(6) G8 = 5 (en passant: G4, G5, G6, and G7)
(7) A8 = 6 (en passant: F8, E8, D8, C8, and B8)
(8) A1 = 7 (en passant: A7, A6, A5, A4, A3, and A2)
(9) F6 = 5√2 (≈7.071) (en passant: B2, C3, D4, E5) -
Are we allowed to use negative numbers? And if we are, how does primality work?
Primality only applies to positive integers (i.e., natural numbers) by definition. Yet, in a non-formal definition of a prime number, the negative of a positive prime number can be associated as being the same prime number. Thus, with the second definition permitting negative values, 2 and -2, for example, are the same prime, while by the first definition negative numbers are not prime.
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There is a published paper that demonstrates that there is a unique solution to the order-3 magic hexagon.
http://www.yau-awards.org/English/N/N92-Research%20into%20the%20Order%203%20Magic%20Hexagon.pdf
------ 3--17--18------ Condition 1 is satisfied with (4,8) in the fourth horizontal row,
----19-- 7-- 1--11---- and (2,4) in the second L-to-R-diagonal row.
--16-- 2-- 5-- 6-- 9-- Condition 2 is sastisfied {3,9,10,15,16,18} only 3 is prime.
----12-- 4-- 8--14---- Condition 3 fails {1,2,4,6,7,8} both 2 and 7 are prime.
------10--13--15------Condition 4 is satisfied - all fifteen rows sum to 38. -
...that it does not follow that every region of diameter 1 will fit inside a circle of diamter 1. The diameter is defined as the greatest distance between any two points on the boundary of a closed figure. For a polygon, it can be defined as the the greatest distance between any two vertices. Let this figure be an equilateral triangle. The diameter then is congruent and equal to any of the sides of the triangle. The altitude of this equilateral triangle is √3/2 (= cos(30°) ≈ 0.866), which is greater than the radius of the circle (r =
½), thus would extend outside the circle given that the triangle's base would need be placed congruent to the circle's diameter for the base to fit within the circle. -
Your three reals {2, 2pi, 4-pi} did show that the proposition that all terms will be a multiple of the irrational value if an irrational value did exists is not necessarily true. (Hence a proposition and not an assertion). What you did demonstrate is that not only scaling, the multiplicative aspect, is an operational factor to be considered, but that the additive aspect employed is a factor, as well.
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Perhaps check it again,
It was my suspicion that you did not expect the factorial character to be used in the manner of a subfactorial. I also guessed you were not aware of the multifactorial notation a!(b), sometimes written a!(b), a!b, or a!b such that a ∉ ℤ- where a mod b = 0, and b ∈ ℤ+, but more often recognized with b (i.e., multiple) factorial characters following the number. Of course it would be difficult to find a solution for the multifactorial given the limitation of three 8's and the other notation symbols.
There indeed is a special character (i.e., a single code point) for the double factorial character -- just as I had used in my post for the double factorial -- hence my question about concatenation with the symbol. Of course, two factorial characters are often used in ASCII notation, as the code point is extended out of the normal ASCII range. -
8×√(!√(8+8)) = 24
[Here the factorial sign preceding the number is the common notation for the subfactorial function. The subfactorial (derangement) of 4 is 9.] -
A note regarding the rule modification. I see now the word that caused confusion was the word 'unless'.
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Yes, division by zero does create a situation where all values are correct (and thus undefined). In addition to this error, I see where I had an error while calculating.
There may be less steps, but here goes...
(1) Y=2, X=4 - Π....new Y=Π - 2
(2) Y=2Π, X=4 - Π....new Y=3Π - 4
(3) Y=3Π - 4, X=Π - 2...new Y=2Π - 2
(4) Y=2Π - 2, X=Π - 2...new Y=Π(5) Y=Π - 2, X =4 - Π...new Y=2Π - 6
(6) Y=Π, X=4 - Π...new Y=2Π - 4
(7) Y=2Π - 4, X=2Π - 6...new Y=2
(8) Y=2, X=4 - Π...new Y = Π - 2
(9) Y=Π - 2, X =4 - Π...new Y = 2Π - 6
(10) Y=2Π - 6, X=2Π - 6...new Y = 0
all triangles are equilateral!
in New Logic/Math Puzzles
Posted · Edited by DejMar