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Posts posted by BMAD


I have a negative value for x as my min and a different x max
36 minutes ago, Nightfox__ said:Max of x is 1/2+(1/√2) and min 0??
This is not the answer but as an example of this possibility:

Suppose we have the following system
x^2+y^2=r; x+y=r, such that the line crosses the circle at exactly two places.
Obviously with two equations and three variables, we have a solution set of answers that can satisfy the given conditions. What I want to know is of the given solutions that satisfies this problem, what is the smallest and largest values x can possibly be?

I think you are on the right track rocdocmac


I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?

On 4/17/2019 at 6:02 AM, rocdocmac said:
116 seconds (or possibly even 106 seconds) presuming first two pieces are already in place.
Toasting alone takes 90 seconds for slices numbered 1, 2 and 3 with sides A and B.
Stages: (1) 1A and 2A under grill (30 sec), followed by (2) 1B and 3A under grill (30 sec); finally (3) 2B and 3B under grill (30 seconds)
Normal additional time for single movements:
End of 1st stage: Slice 1 flipped from 1A to 1B (3 sec), 2A removed (5 sec), 3A inserted (5 sec)
End of 2nd stage: Slice 1 removed (5 sec.), slice 3 flipped from 3A to 3B (3 sec), slice 2 (2B) reinserted with flip side up (5 sec)
End of 3rd stage: Done
Normal additional time = 26 seconds
Total time = 116 seconds.(1:56)
Exchange in one go:
If one is allowed to do simultaneous movements (exchanging in one switch of 5 sec), 10 seconds can be deducted, i.e. a total time of 106 seconds (1:46)
14 hours ago, Pirate8 said:Left
side
Right
side
15
A1 in
610
B1 in
635
A1 toast
1140
B1 toast
3638
A flip
4145
B out
3968
A2 toast
4650
C in
6974
A out
5180
C1 Toast
7579
B2 in
8183
C flip
80109
B2 toast
84113
C2 toast
110114
B out
114116
C out
Manual operations must not overlap.

Three slices of bread are to be toasted under a grill. The grill can hold two slices at once but only one side is toasted at a time. It takes 30 seconds to toast one side of a piece of bread, 5 seconds to put a piece in or take a piece out and 3 seconds to turn a piece over. What is the shortest time in which the three slices can be toasted?


I wish to share 30 identical individual sausages equally amongst 18 people. What is the minimum number of cuts I need to make? What is the minimum number of pieces I need to create?

Assume that only four buttons work on your calculator: 5, 7, enter, and plus. What whole numbers can you not use your calculator to make?

20 hours ago, EventHorizon said:
I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points.
How to find the arc length of a function:
If the function was a straight line, you could use the Pythagorean theorem to find it.
sqrt( (y2y1)^2 + (x2x1)^2 )
If you assume the arc length was a straight line, and slowly bring the points on the function closer together, eventually you get the change in y approaching the slope of the line times the difference in x's.
The arc length for that infinitesimally small section is then sqrt( ( f'(x)dx )^2 + dx^2 ) = sqrt( dx^2 * (f'(x)^2 + 1 ) = sqrt( f'(x)^2 + 1 ) dx. Where f'(x) is the derivative of f(x).
Integrate that between two points to get the arc length between those two points.
Example: f(x)=2x
f'(x) = 2
integrate( sqrt( 2^2 + 1 ) dx ) = integrate( sqrt(5) dx) = x*sqrt(5) + c
If you evaluate the integral between 0 and 1 you end up with arc length = sqrt(5). Which is the same as if you just used the Pythagorean theorem. Yeah, I'm too lazy right now to find a function that ends up with a nice arc length function that isn't just a straight line.
Restatement of problem using previous spoiler:
One answer is simple, the other is less so.
yes, this is what I mean.

James and Mike are running in a race. They both walked and ran for part of the rate. They each walked and ran at the same speed. James ran for half the distance and walked for half the distance. Mike ran for half of his time and walked for half of his time. Who finished first?

On 12/15/2018 at 4:16 PM, CaptainEd said:
I think not: please check my counting
In what follows, the rubber band’s path goes on the outside of the nails marked “xo o o o o o o
o o o o o o o o x
o o o o o o o o
o o o o o o o
o o o o o o
o o o o o x
N = 43
P = 21 + sqrt(5)
S = 62: (5x5:1, 4x4:4, 3x3:10, 2x2:18, 1x1:29)Compare with
o o o o o o o
o o o o o o o o x
o o o o o o o o
o o o o o o o
o o o o o o
o o o o o x
N = 43
P = 21 + sqrt(5)
S = 56: (4x4:2, 3x3:8, 2x2:17, 1x1:29)All you showed is the bound is wider than one thought.

On 12/12/2018 at 4:52 PM, CaptainEd said:
I question whether the information given about a figure is sufficient to uniquely determine the number of included squares:
If I understand right, these two figures have the same N and P, but don’t have the same number of squares.o o o o
o o o o o
o o o
Nails = 12, perimeter = 7+sqrt2 + sqrt5
Squares =5
o o o o
o o o o o
o o o
nails=12, perimeter = 7 + sqrt2 + sqrt5
Squares = 6So then if we know N and P we should be able to bound squares x by a two values. Are those values always consecutive?

On 12/11/2018 at 9:00 PM, CaptainEd said:
While this doesn’t answer the question with randomly placed 1s, I think it does give an upper bound to the number of columns that can be removed.
What is the smallest number of remaining columns that would allow (a) the 30 rows to be distinct, (b) each row to have 5 1bits, and (c) all remaining columns to be distinct?
7 columns: 7!/2!5! = 21, too few distinct rows
8 columns: 8!/3!5! = 56, that’s enough.The first 12 columns in each of the 30 rows should be one of these patterns:
11000 00000 00
01100 00000 00
00110 00000 00
00011 00000 00
00001 10000 0000000 11000 00
00000 01100 00
00000 00110 00
00000 00011 00
00000 00001 1000000 00000 11
10000 00000 01The last 8 columns should be chosen from the C(8,5) possible patterns.
The resulting matrix satisfies the requirements because:
(a) the last eight columns differentiate the 30 rows
(b) the first twelve columns are distinct by design
(c) the last eight columns are distinct because of the distinct patterns chosen from the C(8,5)
(d) all row sums are 7 before column removal, and 5 afterwardsI agree

Find a function where the arc lenth and area between any two randomly defined points is the same. There are two.

Say we have the function:
y=x^x^x^x^x.....
Find an x value for which the derivative of this function converges.
If you are really clever you'll find the interval that converges.


What is (1/2)! x (1/2)!

Write the complex form (a + bi) for:
Sqrt ( i )

Find the Limit as n goes to infinity for:
(1^n + 2^n + 3^n + 4^n....+ n^n)
over
(n^1+ n^2 + n^3 + n^4 ... + n^n)

On 11/12/2018 at 9:12 AM, Donald Cartmill said:
The area would be that of the original triangle,however that area would vary as determined by the length of the 3rd side
I agree that it will vary but we can bound the area.

Suppose you have a triangle that has 21 inch lengths. Divide this triangle into half by drawing a line from vertex between the two identical sides, choose one of the sides randomly and shade it. The nonshaded side is cut in half again. Choose one of these sides randomly and cut it in half again shading one random piece. If this pattern of cut, shade, cut, cut, shade, cut, cut, shade cut, cut,.... was to be continued forever, what would be the area of the shaded region?

Easy question about a system (or is it)
in New Logic/Math Puzzles
Posted
On the right track but see if you can find what x is approaching.