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## #1 Re: Help Me ! » no. of triangle » 2012-01-17 10:22:36

I see that zero factorial is defined as 1. I figured tough, I need zero factorial to be zero so I will use it that way. But that isn't the problem. The problem is that my factorial statement is wrong. For n points there are not (n-1)! unique combinations. For 3 points the possible line segments are 12, 13, 23. That is 3 possibilities, not 2  [(3-1)!].
The possible line segments from n points = (n-1) + (n-2) + (n-3)...
not (n-1) x (n-2) x (n-3)...
Once more multiplying when I should have been adding. And that addition sequence is represented by the symbols you know but I don't.
Thanks for hanging in there with my misguided thinking. Need to brush up on that binomial theory and symbology.

## #2 Re: Help Me ! » no. of triangle » 2012-01-17 09:55:54

The terms for the 2 other sides to connect a line segment should be added, not multiplied. Each of the line segments on side l can connect with m+n points, not mn points. It should read:

lmn + (l-1)! x (m+n) + (m-1)! x (l+n) + (n-1)! x (l+m)

So for 1,1,1 the factorials become zero and you get one.
For 3,4,5 you get

60 + (2 x 9) + (6 x 8) + (24 x 7) = 294

Which isn't what you got. Hmmmm. The logic seems right. Have to think more on this one.

## #3 Re: Help Me ! » no. of triangle » 2012-01-17 08:59:28

Sorry Bobby, I didn't realize that is what those high parenthesis meant. I am not familiar with the notation, though I probably have learned it in the past. Does that equate to what I came up with on second try?

Thanks

## #4 Re: Help Me ! » no. of triangle » 2012-01-17 08:55:41

Thinking about it more I see that a triangle does not have to have a point from each side. It can have 2 points on one side. On side m, (m-1)! line segments can be made. Each of those line segments can join with any point from the other 2 sides so I would now say number of possible triangles =

lmn + (l-1)! x mn + (m-1)! x ln + (n-1)! x lm.

## #5 Re: Help Me ! » no. of triangle » 2012-01-17 08:42:54

It seems to me that all of the points are "on the side" meaning they are between the endpoints A, B, and C. So a triangle will need to contain one point from each of the 3 groups. Any set of one point from each group will make a triangle, so I think there are l x m x n possibilities.

Bobby, take another look at your example. 4/3 -1/3 -1 = 3? Though I agree that if one side has no points there are no triangles possible I think it is because if l or m or n =0 then lmn = 0.

## #6 Re: Help Me ! » Cards probability!!! » 2011-12-21 08:32:14

Do you mean on the first draw or do they work their way all through the decks?

## #7 Re: Help Me ! » compactness and non- compactness of objects » 2011-12-21 04:12:05

I believe the cocept of compactness is related to what in science we would call density. In those 2 figures if you assume that the area enclosed by the 2 shapes is the same then the first manages to enclose the same amount of area in a smaller enclosure. In 2 dimensions the circle encloses the most area in the smallest length of perimeter. In 3 dimensions the sphere encloses the most volume with smallest area of surface. If there is a quantitative measure of compactness I would think that the first figure would have a value of 1 and the second would be less than 1.
Suppose you had a bunch of items of different sizes and shapes you had to send through the Post Office. It would make sense to figure out how to arrange them to fit in the smallest box possible, since the Post Office charges more for a bigger box. You would be trying to maximize compactness.
Have you seen the puzzles where a 3 X 3 cube is cut into 27 smaller cubes, but some of the smaller cubes are stuck together to make differently shaped pieces? The trick is to figure out how to put them back together as a 3 X 3 cube. That would be the most "compact" solution. If you don't get it right the pieces add up to something bigger than the 3 X 3 cube. I think the original puzzle was called Soma, but now it is copied all over.

## #8 Re: Help Me ! » boundary in complex plane » 2011-12-20 11:15:46

What part are you having trouble with? I assume by the complex plane they mean the horizontal axis is real numbers and the vertical axis is imaginary numbers, which are a real number multiplied by SQRT(-1). That is the 2 dimensional "plane" on which you would plot complex numbers. Do you understand what an ellipse is? If not a search will get you plenty of information.

## #9 Re: Help Me ! » Divide edges into equal distance points » 2011-10-31 11:19:46

Model,
I am new here and not yet profficient with the Latex language or posting charts. I can ask you to think of the 2 points you want to connect. There is a horizontal (X) and vertical (Y) distance between them. Like if you had to go from one to the other but you can only go straight up or down and straight sideways. You break each of those distances into 10 parts and take one step at a time. Each step takes you 1/10 of the way to the next point. The 10th step would be
X1 + 10(dx/10) = X1 + dX  which by definition is X2.
Suppose your 2 points are at the origin (0,0) and at (10,10). Can you see how your in between points would be (1,1), (2,2), (3,3)... and how that fits into the equation I gave?
One of the more experienced people here hopefully can make you a graph.

## #10 Re: Help Me ! » Divide edges into equal distance points » 2011-10-31 11:09:27

Bobby,
Sorry to jump in on you, I just had a feeling you 2 were not quite on the same page.

## #11 Re: Help Me ! » Divide edges into equal distance points » 2011-10-31 11:02:48

I thought you might since you suggested the distance formula.
You are going to add 9 in between points between each consecutive pair of points.
dX = X2 - X1
dY = Y2 - Y1
X coordinates of in between points = X1 + (dX/10); X1 + 2(dX/10); X1 + 3(dX/10)....
Same for Y coordinates.

## #12 Re: Help Me ! » Divide edges into equal distance points » 2011-10-31 10:41:24

Model,
Do you have the X and Y coordinates of the red points?

## #13 Re: Help Me ! » Combinations vs Permutations » 2011-10-31 10:37:24

Since the only difference in the paths is the order of the steps isn't that by definition a permutation?

## #14 Re: Help Me ! » all triplets » 2011-10-20 11:18:45

Since the terms are sort of symmetrical if you are doing trials you can leave a few out. For example if you try (2,2,3) then you don't have to try (2,3,2) or (3,2,2), since every combination of addition and multiplication is used and in addition and multipliction order does not matter (they are commutative).
I guess I am saying that you only need to try different combinations, not permutations. Order does not matter.
At least that's how I see it.
If like me you sometimes have trouble remembering which is which between combinations and permutations, just remember this four word sentence:

A combination lock isn't.

I know, you have to let it sink in for a minute...

## #15 Re: Help Me ! » Tough task » 2011-10-03 06:55:55

I got logic slightly different from Bob, though I don't have another answer either:

2^x ends in 2, 4, 6 or 8

5^y ends in 5

c^2 ends in 0, 1, 4, 5, 6 or 9

2^x + 5^y ends in 1, 3, 7 or 9

c^2 can't end in 3 or 7 so c^2 ends in 1 or 9

So c ends in 1, 3, 7 or 9.

## #16 Re: Help Me ! » Higher Mathematics » 2011-09-14 04:11:22

Lightning,
I was also tought these triangles in high school geometry. We were  even made to put that chart on a 3x5 index card and keep it handy. Knowing these 2 triangles, the 45-45-90  and the 30-60-90, helps you understand basic trigonometry and the Pythagorean theorem. These triangles are called "right" triangles because they contain a 90 degree angle, which is called a "right" angle. Your handwritten page contains a table of the values and pictures of the 2, labelled "right triangle values". It does contain an error, however. In the 30-60-90 triangle the right angle is between the 1 and sqrt3 sides, not between the 1 and 2 sides. 2 is the hypotenuse (that is the name of the longest side of a right triangle). The right angle is indicated by putting the little square on it. See how you have it on the wrong angle? That is why Au101 said that based on that picture the hypotenuse would be sqrt5 instead of sqrt3.
For the values of the lengths of the sides, by convention the shortest side is given a value of one. This is for consistency.
The term "exact value" I believe refers to the fact that they use the symbols for sqrt2 and sqrt3. This is the only way to exactly describe these values. You could put 1.414 and 1.73 for those values but that is not exact. No matter how many decimal places you use you can never give an exact value for those values except by writing sqrt2 and sqrt3.
Sorry I haven't learned to use the Latex language and print the radical (square root) sign yet.

## #17 Re: Help Me ! » scientific calculators (casio) » 2011-08-25 07:49:48

Diamond,
The Casio website has the 83/85GT Plus manual available online for free. I tried to do a cut and paste of the polar to rectangular coordinate section but it would not work. I know I am not allowed to put links in these messages, but if you go to Casio and look for calculator manuals you should be able to find it. They give an example of converting rectangular (sqrt 2, sqrt 2) to polar (answer r = 2, theta = 45).
Let me see if I can cheat the system:
Aych Tee Tee Pee colon forward slash forward slash
support dot casio dot com forward slash pdf forward slash 004 forward slash fx-83_85GT_PLUS_E.pdf

Maybeb that will make it through. You want pages E-18 and E-20.

## #18 Re: Help Me ! » Question about the Girl, Boy n Dog puzzle. » 2011-08-25 03:38:52

Anandbrar,
I don't know that it would be possible to predict where the dog would be. Think of when the three just start to walk. About a microsecond later the humans are a miniscule distance apart and the dog is bouncing  back and forth a million times a second or so. We are talking infinitesimals here.

It is not hard to visualize the opposite of the process. Imagine a very bouncy ball. You throw it straight down hard and it starts bouncing up and down between two boards. Now the top board starts moving down toward the bottom board. The closer the boards get the faster the ball changes direction. Its frequency increases. Theoretically that frequency will reach infinity just before the boards come together and squash the  oscillation. That is not hard to imagine. But your puzzle is the same thing in reverse. The boards are the people, the dog is the ball. We are asked to decide at what point the distance between the people goes from zero to, say, one over almost infinity. Does the dog go forward or backward first? What is the smallest distance that is not zero? In the real world the first measurable distance is when the signal is first able to be separated from the noise.

Sorry if it seems that I am being abastract and not specific, but my point is that I don't think you can be specific and predict this situation.

## #19 Re: Help Me ! » Tricky Coin Flipping Probability Problem » 2011-08-16 04:14:42

Bobby,
You mention not finding any literature on the subject of the drukard's walk. I remember reading about it years ago in the book One, Two, Three...Infinity by George Gamow. I don't know if he gives an equation for it but he does discuss the concept. It is also a great book for introducing many concepts of math and science. I recommend it for anyone interested in science, from high school age to adult. On Amazon the book got 38 reviews, and 35 of them are 5 stars (the max).

## #20 Re: Help Me ! » Rate of change (cone problem) » 2011-08-11 03:22:39

I got the same answer but there did not need to be any calculus involved. At 2m the level is rising at 20cm/min. So at that instant the amount of volume increase per minute is a cylinder with a radius of 2/3m and height of 0.2m. That volume is (pi)*(2/3)^2*0.2 and throw in a factor of 10^6 to turn m^3 into cm^3. Add the 10,000cc/min that is going out and that's it. Same answer.

This is an algebra/geometry problem all dressed up to look like a calculus problem. Watch out for these, especially when taking a timed test.

## #21 Re: Help Me ! » Golf Teams » 2011-08-08 14:51:10

I assumed that the team plays together all day, regardless of how many games are played in a day. Otherwise why state the 5 day limit. The first test I thought of was to consider golfer zero (actually I had called him golfer 1 but I like your system better). He has 5 days to play with the other 7 players twice. Each day he plays with 3 other players, so he has 15 slots available. Each of the other 7 players has to fill 2 of those slots, which requires 14 slots. This of course does not prove that it is possible, as Bobby said it is quite complex. But this first simple test did not prove it impossible.

## #22 Re: Help Me ! » Help with finding shaded area. » 2011-08-01 08:04:00

In the first question 4 is the height of the overlap, not the width.
Of course this assumes a bias toward the horizontal instead of vertical, but I think the writer of the question would see the circles as gAr drew them, but with less overlap.
I get 16 x (pi/3) - (8 x sqrt(3)).

## #23 Re: Help Me ! » Brightness values to db values conversion » 2011-07-27 03:34:32

Hi Freddy,
dB is by definition a ratio. The ratio has to be to a reference value. Since you are looking for negative values the reference will be the max value, 255. The equation for dB for a power quantity is

dB = 10 log P/P(ref)

10dB represents a ratio of 10:1, 20dB = 100:1, 30dB = 1000:1 ...

Your value of 65 gives a dB value of 10*log(65/255) = -5.94

The scale will not go down to -100 since that would require a ratio of 10^10 to 1. Your max ratio is 255:1. An intensity of 1 will give your lowest dB value of -24.1.

Hope this helps. Wikepedia also has a pretty good description.
Note the case of the letters, deci is a prefix that means 1/10, as always. A decibel is 1/10 of a Bel, an old unit that is no longer used. But since it was named after a person the B is upper case, but the d is not.

## #24 Re: Help Me ! » Data problem » 2011-07-21 16:41:58

Hi Bob and Winky,
I saw this post but wasn't able to reply because my computer was down. Sorry for this being so late. I think to just compute the upper and lower limits does not show you understand the concept of the absolute value. The absolute value can be used because a person's weight could be above or below the target value and still be within the limits.
My answer for 5'8" would be (and I still haven't learned the math shorthand):

ABS (x - 143) < 21

Inside the parenthesis could also be 143 - x.

See Bob's last post about whether to use (less than) or (less than or equal to). You can also change the 21 to 22.

Hope this helps and isn't too late.

Tom Morgan

## #25 Re: Help Me ! » Algebra Problem: Seeking The Why » 2011-04-13 16:00:12

Hello,
This is my first post here and I do not yet know how to make equations look good. I did however want to comment on this problem.
I believe there is a problem with the original premises. The equation
x = -y^2
and the statement
y is a function of x
are not, in my opinion, compatible. If you accept the definition of a function as a rule that assigns every element of one set (the domain) exactly one value of another set (the range), then in this equation x is a function of y. For every value of y (-1, 0 +1) there is exactly one value of x (-1, 0 -1). So x is a function of y. But for a value of x (-1) there are 2 values of y (-1, +1) so y is not a function of x.
As far as discussing what the correct values are for the domain and range it first must be agreed which variable (x or y) is a function of the other, which will determine which is the domain and which is the range. Also it must be clarified if the original equation
x = -y^2
is to be interpreted as
x = -(y^2) or
x = (-y)^2.
I think the way it is written implies the first, but the way bobbym solved for y, without putting a plus/minus before the square root sign, implies the second. I think either way y is still not a function of x.