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Put a cork in it [original thread]

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A while ago we asked for a shape that fits snugly through three different holes:

Actually, the answer is not unique. So now we ask:

What is the volume of the smallest shape that fits snugly through the holes?

The largest?

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Posted · Report post

Since you say "volume" and not "area" I get 0 for smallest and no limit for largest.

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Since you say "volume" and not "area" I get 0 for smallest and no limit for largest.

Zero volume [a point] will not fit any of the holes snugly.

A cube 2 inches on a side has finite volume [8 in3] but is too large to fit through the circular or triangular holes.

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Have to assume you're starting with the same 2" x 2" x 2" cube and by snuggly you mean the shape that, at a maximum, fits thru the hole in two dimensions while being the smallest volume? Dont know if that made any sense but it's late and I've just returned home after a night out.

the three dimensional shape described in your previous post would have a volume of pi in2 as would the two inch long volume of cross section equal to the incircle of the described triangle both of which may be the smallest depending on how you define "snuggly". A two inch long volume with an equilateral triangular cross section of side length sqrt(3) would have a volume of 6 sq. inches which would be my guess at the greatest. Still seems too many variables have been left open by the OP. Maybe bn has also just returned home after a night out? More likely I'm missing something...

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Posted · Report post

Have to assume you're starting with the same 2" x 2" x 2" cube and by snuggly you mean the shape that, at a maximum, fits thru the hole in two dimensions while being the smallest volume? Dont know if that made any sense but it's late and I've just returned home after a night out.

the three dimensional shape described in your previous post would have a volume of pi in2 as would the two inch long volume of cross section equal to the incircle of the described triangle both of which may be the smallest depending on how you define "snuggly". A two inch long volume with an equilateral triangular cross section of side length sqrt(3) would have a volume of 6 sq. inches which would be my guess at the greatest. Still seems too many variables have been left open by the OP. Maybe bn has also just returned home after a night out? More likely I'm missing something...

Snugly means completely fills each of the hole cross sections at some moment as it passes through.

Of all the 3-dimensional shapes that do this, one has a maximum and another has a minimum volume.

The question is what are these volumes?

Been watching TV all evening, so you have the advantage on me. B))

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Posted · Report post

A while ago we asked for a shape that fits snugly through three different holes:

Actually, the answer is not unique. So now we ask:

What is the volume of the smallest shape that fits snugly through the holes?

The largest?

If I understood it correctly, for the largest shapes, you could choose rods with the same cross-sectional area as the holes to be filled. So, for ex, for a triangular hole, you could have a rod with a triangular cross-section. The volume would then be the area of the cross-section multiplied by the depth of the hole.

For smallest shapes, you could have very thin rods or even thin cardboard or paper fitting snugly into the holes. The volumes would be very small.

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Oops, I just realised it was rather easy... so I decided to make sure what i understood was correct and that led me to the link to the previous question. So, now that I have realised that what I had previously understood was way off, pls ignore my previous post.

Although, for the shape with the smallest volume, I would still stick to the previous answer! Although the material would be wood and the shape could still be made by chopping and chiseling it to a very small breadth while keeping the length as 2".

Edited by DeeGee
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One option for the shape could be as below. Having triangular and circular shaped protrusions on either side of the cube. Naturally, the length of the cube is reduced by the amount of length of protrusions on either side.

post-17784-12523110083857.jpg

In order to have maximum volume, the length of the square cross-section should be max and length of circle and triangle cross-sections should be small. That is the length of protrusions on either side must be smallest.

Let x be the length of circular cross-section part then volume here = pi.12.x = pi.x

Let y be the length of triangular cross-section then volume here = root(3)/4.4.y = root(3).y

Then volume of square crossection = 4.(2 - x - y)

Then volume of total shape = 8 - 2.27y - 0.86x

Now we need x and y to be smallest possible such that they can also fit snugly and restrict movement of the shape when inverted.

Lets say the minimum length of x and y must be 1/4", then volume of the shape is 7.2 cubic inches.

If the minimum length of x and y must be 1/8", then the volume would be 7.6 cubic inches.

So, the volume of the shape depends on how long (minimum) must be the protrusions on either side of the reduced cube.

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The shape must pass completely through each of the holes.

Your cube will not pass through either the circular or triangular holes.

Consider them to be hole in a wall separating two rooms.

The shape must pass from one room to the other, in turn, through

all three holes, fitting each hole snugly as it passes.

There are actually an infinity of 3-dimensional solids that will do this.

One has a largest and another has a smallest volume.

What are these volumes?

Refer to the figure in the for dimensions of the holes.

The circle has a 2" diameter; the square and triangle have 2" sides.

Since this puzzle asks a new question, it's fair game to look at the answers given there.

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Posted (edited) · Report post

the OP's above mentioned walls to be infinitesimally thin and the cut outs made with an infinitesimally thin blade, reserve the cut outs and glue the square perpendicular to the circle, cut the triangle in half with the same blade and glue it perpendicular to both, you have a volume that approaches zero as the smallest possible volume (8pi x thickness of the shapes in3). The largest volume is as described in the referenced post and is pi in3.

Edit: inches squared/inches cubed, what really is the difference?

Edited by plainglazed
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Posted · Report post

having multi-dimensional issues this a.m. - above should be (4 + 2 + pi) x thickness and not 8pi

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Since you say "volume" and not "area" I get 0 for smallest and no limit for largest.

I agree with the assessment of the smallest area being zero. Take three pieces of paper and cut one of them into a circle, one into a square, and one into a triangle. Then cut some slits into them so you can fit them together in three orthogonal planes like so; the circle standing up and facing you like a mirror, the triangle flying forward like an airplane, and the square coming at you like an axe about to chop you into two symmetrical halves. Total volume for three planes is zero.

post-15489-12524675377728.jpg

The max volume question might be more interesting though.

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the smallest hole is the triangle with an area of (2 x square root 3).. so the maximum volume must be smaller than 2x2x1.732(square root of 3)

post-22187-12525940342872.jpg

i cant think of a shape larger than this that shall pass all the holes SNUGGLY

just don't know how to calculate its volume

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I agree with various answers posted but to summarize

Min - although infinitely small is theoretical, the orig prob says you have a wood block, so plainglazed was close but the shapes can be outlines instead of solids. So if we define CA and the min cross section the wood can be cut down to without breaking, then it is CA*(perimeter of square + perimeter of triangle + perimeter of circle or:

CA * (8 + 2*(1+sqrt(5)) + 2*pi)

Max - after much calc I believe I was able to show that the wedges cut off = the wedge left and therefore plainglazed was right 1/2*circle area*length = 1/2 * pi * 2 = pi

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To revive this thread by means of a duplicate post was an oversight.

Nevertheless it contains the most discussion [the duplicate thread has none.]

One condition for the solid is that it be convex.

This rules out the zero-volume shapes.

The volume that has been put forward as the largest is convex,

but in fact it pertains to the convex solid with the smallest volume!

If it helps, the shape of the largest solid is the simplest, and has

a volume that's larger by almost 1 in3.

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...

post-52566-0-42755000-1344346974_thumb.p

Finding its volume was a fairly straightforward calculus problem. Letting the circular base lie in the x-y plane and the linear ridge lie along the y-axis:

post-52566-0-98273900-1344347396_thumb.p.

Letting R=2 (as in OP) yields post-52566-0-16118900-1344349535_thumb.p, or about 7.83 in3.

(The integrand of the inside-integral is the area of a circle that's had left- and right-edges sliced off, the limits are the amount to slice as given by the slope of the triangular cross-section.)

Of course, I had to go back to my CRC to look up some of the integrals, so I wouldn't be surprised if there were a small transcription error in there. Especially as my 3D-modeling program claims the volume is

3.62 in3.

Anyone else find the volume of a solid that looks like this?

[edited to catch factor-of-2 error]

Edited by austinm
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Depending on whether the 2" altitude of the triangle is parallel or perpendicular to the axis of the circular cylinder, I get two possible convex solids. The one which austinm shows has a volume of 3.616 in3, the other has a volume of 3.142 in3.

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Depending on whether the 2" altitude of the triangle is parallel or perpendicular to the axis of the circular cylinder, I get two possible convex solids. The one which austinm shows has a volume of 3.616 in3, the other has a volume of 3.142 in3.

If the cylinder's axis is perpendicular to the triangle's 2" altitude, are you sure it would still fit snugly through a square hole?

To illustrate

post-15489-0-32182800-1344468051_thumb.j

If this is what you mean (taking a cylinder and cutting off the excess material so the cylindrical object can fit through the triangular hole), the figure on the left will look like a square if you imagine yourself looking at it from either the left or the right side after it goes through the triangle, but I think the object on the right will not look like a square if you look at it from above. Before being cut into a triangular shape, the cylinder on the right will look like a square from above because of material at it's fattest part if you go halfway up vertically, and some of that material will be cut by the triangular shape at that altitude.

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If the cylinder's axis is perpendicular to the triangle's 2" altitude, are you sure it would still fit snugly through a square hole?

To illustrate

post-15489-0-32182800-1344468051_thumb.j

If this is what you mean (taking a cylinder and cutting off the excess material so the cylindrical object can fit through the triangular hole), the figure on the left will look like a square if you imagine yourself looking at it from either the left or the right side after it goes through the triangle, but I think the object on the right will not look like a square if you look at it from above. Before being cut into a triangular shape, the cylinder on the right will look like a square from above because of material at it's fattest part if you go halfway up vertically, and some of that material will be cut by the triangular shape at that altitude.

I think you're correct. In that case, I think there is only one convex solid which works -- the minimum volume is the same as the maximum volume.

I'm assuming here that the faces which present the three shapes (circle, square, and triangle) are mutually orthogonal. I have not considered non-orthogonal cases yet.

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...

The volume that has been put forward as the largest is convex,

but in fact it pertains to the convex solid with the smallest volume!

If it helps, the shape of the largest solid is the simplest, and has

a volume that's larger by almost 1 in3.

I've got a convex solid with a smaller volume than the one posted so far...

After a while of wondering how another solid could work I found that if I had a clay model of the solid already found, that I could simply use a plane and slice a small piece off along the curve connecting the base of the triangle to the top of the square.

So how much could I shave off? The requirement of being convex would actually give me the answer. First, take the required orthogonal cross sections and simply connect each point in the cross sections to each other (repeat if necessary) to get the convex closure / convex hull. Easy to visualize now, huh? (kidding of course)

Still a bit confused about how connecting all those points would look, I found that the triangle is superfluous given the other two (already have the two base vertices from the circle and the top from the square, and the convex closure given these points includes the triangle cross section). That simplified things greatly. So I simply connect the circular base to every point on the top edge of the square. You can think of this as skewing (to keep the cross section parallel to the base a circle) the top of a cone from one top vertex of the square to the other and including all the volume it passes through. Another visualization would be to have two intersecting skewed cones terminating at the top two vertices of the square and connect the two skewed cones with planes.

Now for the volume of this solid. Everything I need to solve this is in the visualization of two skewed cones. Each cross section parallel to the circular base would be a rectangle with semicircles on either side. The width of the rectangle would be 2 - h, which is easy to see from the cones. This makes the radius of the semicircles (2-h)/2. The length of the rectangle would be 2 - 2*radius of semicircles = 2 - 2*((2-h)/2) = 2-2+h = h.

The area for the cross section at any particular height would then be

f(h) = area = (2-h)*h + pi * ((2-h)/2)^2

area = 2h-h^2 + .25 * pi * (2-h)^2

area = 2h-h^2 + .25 * pi * (4-4h+h^2)

area = 2h-h^2 + pi - pi*h + .25 * pi * h^2

Integrating with respect to h gives

F(h) = h^2 - (h^3)/3 + h*pi - .5*pi*h^2 + (1/12) * pi * h^3

Evaluating from 0 to 2 gives

F(2)-F(0) = 2^2 - (2^3)/3 + 2*pi - .5*pi*2^2 + (1/12) * pi * 2^3 - (0 - 0 + 0 - 0 + 0)

F(2)-F(0) = 4 - 8/3 + 2*pi - 2*pi + (8/12) * pi

F(2)-F(0) = 4/3 + (2/3) * pi

F(2)-F(0) = (4 + 2*pi)/3, which is about 3.4277 in3

As of yet, I'm still at a loss for the solid with the maximum volume.

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