Fun math problem

[flyinlow]

God, don't let the bean counters at work see that.

[green69rt]

GreenRT,

Here is what you suggested in case it is a pain.

Now we need to find some more problems.  

I just wonder, and don't know the answer, whether the size of the block has to be the extra space and the missing space between each line, or whether it can be done with just one?  Maybe that should be the next question.

No you can't take those four pieces and make a proper right triangle, just won't happen unless you bend space  

[XH29N0G]

So I finally figured out my question about how the space adds up. 

Looking at the line for the large triangle there is one side at 13 and another at 5 (13:5 aspect ratio).  There is a gap that is barely visible in the top diagram and an overlap that is barely visible in the lower diagram.  This is because the 5:2 and the 8:3 triangles are close to 13:5, but not exactly 13:5 aspect ratio.

Thinking about the areas:

The red triangle is 8 x 3 with an area of 8x3x1/2 = 12.
The green triangle is 5x2x1/2 =5.

An 8 side 13:5 aspect ratio would have a vertical side of 3.08, giving it an area of 8x3.08x1/2 = 12.308 boxes.
A 5 side 13:5 aspect ratio would have a vertical side of 1.92, giving it an area of 5x1.92x1/2=4.808 boxes. 

The gap in the other is the difference in area 12.308-12 = 0.308 and 5-4.808=.192 which added together make up 1/2 of a box.
The other 1/2 box is made up by the overlap, which is calculated the same way.  So both the overlap and the space need to be considered to explain the empty box in the lower diagram.  I find this very strange, and don't mind being corrected if someone sees an error.

[XH29N0G]

So I have got a problem (to pay back into the pot). 

Say we have a 20 black dots and 20 white dots and we randomly split them into groups of two.  We will get some groups with two black dots, some groups with two white dots, and some groups with one of each.

If we use all 40 dots to make all possibilities in a fully randomized proportions, How many will we get of
    2 black dots:
    2 white dots:
    1 of each:

Why?

[ws23rt]

So I finally figured out my question about how the space adds up. 

Looking at the line for the large triangle there is one side at 13 and another at 5 (13:5 aspect ratio).  There is a gap that is barely visible in the top diagram and an overlap that is barely visible in the lower diagram.  This is because the 5:2 and the 8:3 triangles are close to 13:5, but not exactly 13:5 aspect ratio.

Thinking about the areas:

The red triangle is 8 x 3 with an area of 8x3x1/2 = 12.
The green triangle is 5x2x1/2 =5.

An 8 side 13:5 aspect ratio would have a vertical side of 3.08, giving it an area of 8x3.08x1/2 = 12.308 boxes.
A 5 side 13:5 aspect ratio would have a vertical side of 1.92, giving it an area of 5x1.92x1/2=4.808 boxes. 

The gap in the other is the difference in area 12.308-12 = 0.308 and 5-4.808=.192 which added together make up 1/2 of a box.
The other 1/2 box is made up by the overlap, which is calculated the same way.  So both the overlap and the space need to be considered to explain the empty box in the lower diagram.  I find this very strange, and don't mind being corrected if someone sees an error.

I like your effort to do the math.  I did the same thing when I first saw this puzzel.
The next thing I did was look at as if it was presented to me to fabricate.

That is when I realized that important information detail was missing.
The missing stuff is assumed by the picture but would be spelled out on a fabrication drawing.

I guess that should be a part of---explaning the missing square in a concise way?

[flyinlow]

5 groups both black, 5 groups both white. 10 groups mixed


because I said so.....er ,wait, law of probability

[ws23rt]

5 groups both black, 5 groups both white. 10 groups mixed


because I said so.....er ,wait, law of probability

Me too--

[green69rt]

While you guys think about your balls, here's  a little design/logic problem that I had to solve in the Navy electrician "A" school.

You have three single pole switches, three light bulbs and a battery.

Design a circuit that will do the following.

1. All switches open and then close switch #1 and only light #1 comes on.
2. All switches open and then close switch #2 and only light #2 comes on.
3. All switches open and then close switch #3 and lights #1, #2 and #3 come on at the same time.
4. You are not allowed to use any other devices in building the circuit.

The picture shows how to represent the items.

[XH29N0G]

I think I got it, but don't know much about wiring.  I'll try to draw it up, but will wait a little before posting so as not to spoil the fun.

Regarding the probability question, what I always find difficult to explain is why the one each has twice as much as the double black or double white.  I had thought it was just that you counted each side separately, so there were 5 black white combos and 5 white black combo in addition to the 5 white white and 5 black black.

I brought it up because I have been wrestling with a quantum mechanics problem someone else has been explaining to me where this does not appear to be the case. 

[XH29N0G]

Almost a day,  This is what I thought might work.  Don't know what rules I broke.

[green69rt]

I think that works and it is the same as my answer but I've had a few years to make it look pretty.

[flyinlow]

That would work. With switch 3 closed , lights 1,2&3 would be in series and very dim. The original question said nothing about the light  bulbs brightness....so congratulations 

I wanted to use diodes, but the question prohibits additional devices.

[funknut]

Yours is much cleaner than mine, 

but here's what I came up with.

[ws23rt]

Here is a problem/puzzle that is an old favorite.  I think I posted this one a couple of years back.

This is a 3D object that can be fabricated.  
This view of it is a top view.   It also is a front view.
What it the third (side) view?

BTW this is a true drawn view as it would be on a a fabrication drawing.---The only thing left out here are dimentions. Also material is a non issue with this.

[green69rt]

Yours is much cleaner than mine, 

but here's what I came up with.

I could get picky on yours.  It looks like all the switches would light all the lights unless the lights had some kind of way to block reverse flow of the current.  As a hint, the lights do not have three connections. 

[XH29N0G]

WS,  I think it has to be the same from the side but might be wrong.  I'll keep thinking.  I am assuming that the circle is some part of a cylinder (or a shape made by intersecting a cylinder - in this case a sphere) and that the square is some type of rectangular prism which ends up being a cube on intersection.

[flyinlow]

WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)

[funknut]

I could get picky on yours.  It looks like all the switches would light all the lights unless the lights had some kind of way to block reverse flow of the current.  As a hint, the lights do not have three connections. 

Interesting!  Thanks for the feedback. I see now that it is sloppily drawn, sorry.  I intended for the switches to connect to the same terminal on all the lights, the battery on the other.

I will rethink this...

[XH29N0G]

WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)

Didn't think about it that way. 

I was thinking shape in shape.  I also realized I was too limited for the internal shape of a shape in shape option because the intersection of two cylinders could be anything from a the limit of a + to a square.  A sphere would be but one option.

[green69rt]

Maybe this is the side elevation.  The dotted lines represent the hole drilled through the wedge.

[ws23rt]

WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)

If a side view had dshed lines to show a hole those lines would also show in the other views.

[green69rt]

WS , the side would be a square.....you could add dashed ghost lines to show the hidden 90* hole (like a 90* plumbing elbow)

If a side view had dshed lines to show a hole those lines would also show in the other views.

Not in the top view but yes they would be there in the side view, so back to the drawing board

Edit: oops, I see we have two conversations going but my dashed line comment still applies.

[flyinlow]

ok,  the side view is a square

[ws23rt]

ok,  the side view is a square

If the side view is a square - -where are items shown in the top and front views---?  This can be fabricated to include what is seen from the top and front.
This a 3D object.

A cube with circles drawn on two sides is not it.

[XH29N0G]

Now I feel completely lost. I suppose it could be something like this, but I do not know how the third dimension is represented. 

PS.  This should be my #1970 post...