Challenge Math Challenge by Charles Link #1

[Charles Link]

Sorry, my Latex is not good enough... that's why I waited till another correct solution was posted. But I think my description should make it pretty straightforward.

From what was described, I think what @maline is referring to is perhaps $V=8 \int_{0}^{\frac{1}{2}} x^2 \, dz$ where $(x-\frac{1}{2})^2+(z-\frac{1}{2})^2=\frac{1}{4}$. I will need to give it a try. $\\$ Editing. It's a little more complicated than that because of the hole in the z-direction. Something like that might work, but I think it's still going to be somewhat complex. $\\$ Further editing: I evaluated what I just wrote out, and it is a correct solution for the case of two holes. Perhaps some further refinement will make it work for 3 holes, but I think the integrals will not be nearly as simple.

 

[maline]

No, my solution is much simpler than that & takes all three holes into account. Should I write it out by hand and post it?

 

[Greg Bernhardt]

Should I write it out by hand and post it?

Yes. Use latex if possible here.

 

[Nidum]

@maline : Is this what you meant with your 'small cube' geometry description ?

With the corner cube gone the remaining wing shapes are all just elongated square section blocks cut away by cylindrical surfaces ?

 

[maline]

Ok, I think I can do it using a couple of diagrams.

I am assuming R=1 rather than 1/2, for simplicity. Therefore we will calculate the volume of only one of the 8 corner pieces, the one where x,y,z>0.

Consider the point x=y=z= √½. It is clear that this point is on the boundary of all 3 cylinders, so that the cubical region √½ ≤ x,y,z ≤ 1 is fully within our solid. Its volume is

(1-√½)³=1-3√½+3⋅½-(½)^3/2)=5/2-7/2⋅√½

Now we subtract the cube from our solid. Note that each of the three edges of this cube, x=y=√½, y=z=√½, & x=z=√½, lies on the boundary of one of the cylinders, parallel to its axis. So after subtracting the cube, we are left with three disconnected regions that I called "prongs", each one lying along an edge of the large cube. You can see the prongs clearly in Nidum's post #2. Since the three are identical, we will consider only one of them, the one along the edge x=y=1. It is bounded on the "outside" by the planes x=1 & y=1, and "on top" by the plane z=√½, which is a square face that it shares with the small cube. Its "inner" boundaries are the curves surfaces of the two "horizontal" cylinders. It touches the "vertical" cylinder only at the the corner point x=y=z=√½. The point of the prong extends "downward" from z=√½ to z=0 where it narrows to a point.

Here is a diagram of a cross section parallel to the xy plane, with 0 < z < √½:

https://drive.google.com/file/d/0B3BoyrPDqVLwc2FRX25fVnc0UkU/view?usp=sharing

The circle is the cross-section of the "vertical" cylinder. The lines in the x and y directions are the edges of the "horizontal" cylinders. The shaded square in the upper right-hand corner is the cross-section of our prong. It should be clear that the cross-section is indeed a square, with side 1-√(1-z²). Therefore its area is

(1-√(1-z²)²=2-z²-2√(1-z²)

To get the volume of the prong, we just integrate from z=0 to √½. The integral of "2" gives 2⋅√½=√2; the integral of "-z²" gives -⅓⋅(½)^3/2=-1/6⋅√½; and we are left with -2 times the integral of √(1-z²). But that integral can be solved geometrically:
https://drive.google.com/file/d/0B3BoyrPDqVLwb2NlMVl5MlVaQzg/view?usp=sharing

The shaded wedge is 1/8 of a unit circle, with area π/8, and the triangle is half a square of side √½, so its area is 1/4. Altogether, the volume of the prong is

√2-1/6⋅√½-2⋅(π/8+1/4)=-½+11/6⋅√½-π/4

Multiplying by 3 for three prongs, & adding the small cube:

3⋅(-½+11/6⋅√½-π/4)+5/2-7/2⋅√½=
=1+√2-¾π

 

[maline]

Oh, now that I finished my post I saw's Nidum's new beautiful diagrams. Yes, that's what I mean by "prongs". Too bad it's hard to see that their cross-section is square.
Thank you!

 

[Nidum]

The cross section of anyone wing at any point along the length is square and the function defining the edge length is a simple geometric one ?

 

[maline]

The cross section of anyone wing at any point along the length is square and the function defining the edge length is a simple geometric one ?

That's right.

 

[Nidum]

 

[Charles Link]

@maline A very good solution! I will ask @Greg Bernhardt to give you credit along with @MAGNIBORO for also having a successful solution. And thank you @Nidum for some very helpful diagrams! :) :)

 

[MAGNIBORO]

@maline if you latex is bad you can use: https://www.codecogs.com/latex/eqneditor.php?lang=en-us

@Nidum What is the program you use to make the diagrams?

 

[Nidum]

Autodesk Fusion360

 

[JasonWuzHear]

Here's an imgur album of my work:

http://imgur.com/a/AttQ7

It's not at all rigorous and tidied up, but I believe it easily leads to a solution. I'll type out what is said in my notes and post a more rigorous solution when I'm away from job. I'm too lazy to integrate right now, but I'll leave that up to future Jason and all you people out there.

I decided to integrate the area of the holes from top-down to find the volume of the holes, then double it, then subtract from the volume of the cube. To integrate the area to get a volume of the holes, I just created a coordinate, z, which will parametrize moving from top to the middle of the cube. I will integrate areas of constant z times dz to get the volume.

To better understand what these areas look like, I imagined slices of these figures at constant z. At the z=0 slice, the area is a circle. At the z=r slice, the area is a square. Any slice in between looks like a circle + some curvy triangles, as shown in the images.

Analyzing the area of the circle + the curvy triangles is easy until the triangles begin to meet and overlap. At which point, you have to subtract the overlapping portions from the area. First I will talk about the area before the triangles overlap.

The area of the curvy triangles can be found by integrating some height away from the circle, h(x), times a small piece of dx from 0 to x(z). x is shown in the images, and grows with z. For example, x(z=0) = 0, and x(z=r) = r. x(z) is just the length of the curvy triangle. There area is then the area of a circle plus 8 curvy triangles before they start overlapping.

When the two curvy triangles meet, which is at x = r/(sqrt2), they begin to overlap with each other. The extra area is in the shape of an "L" type figure, which has an area of two rectangles minus a square. The images show how simple it is to find this area and find it using the variables l(z) and s, where l(z) is the length of the square and s is the width of the rectangle. The area then will be the circle, plus the eight curvy triangles when x=r/(sqrt2), plus the 8 rectangles, minus 4 little squares.

So to find the volume, we just integrate these two areas in their proper bounds times dz from z=0 to z=r. This gives us half the volume of the holes, which we then double to find the total volume of the holes. We then subtract this volume from the volume of the cube to find the volume of what's left of the cube.

Cheers y'all, hopefully you can follow it. Good luck with your solutions!EDIT: Found some errors in my work. Will update later when I get to my desktop at home. The procedure should hold true, though.

 

[PhysicsOfLearning]

How come it is not:

1'' = 2.54cm = diameter

2.54÷2 = Radius = 1.27cm

π(1.27)² = 5.067074... = a₁

(1 side of the squares circular, drill area, which can be substituted from the area of the face of the square):

area of sq.face = 2.54² = 6.4516cm = a₂

a₂ - a₁ = 6.4516 - 5.0670 = 1.3845

Cubing the answer:

1.3845³ = 2.654cm³

The volume leftover from the drill holes in the square is:

2.654÷2.54 = 1.044 cubic inches of cube left over.

Is that correct?

 

[PeroK]

The volume leftover from the drill holes in the square is:

2.654÷2.54 = 1.044 cubic inches of cube left over.

Is that correct?

Given that the original cube only has a volume of 1 cubic inch, probably not!

Cubing an area does not give a volume.

 

[PhysicsOfLearning]

Given that the original cube only has a volume of 1 cubic inch, probably not!

Cubing an area does not give a volume.

Oops!

Methodology wrong as well?

 

[PeroK]

Oops!

Methodology wrong as well?

Yes. Look at the graphic in post #2 to see what you are dealing with.

 

[JasonWuzHear]

okay I have my answer, the volume of what's left of the cube is...0.0580191... in^3here's an imgur album with my work:
http://imgur.com/a/Xzwjk

agh forgot to say that the 4 pi r^2 should be a pi r^2 obviously. When I evaluated the integral numerically in mathematica I fixed it. Seems I got the right answer from reading the past pages, but less efficiently.

 

[Charles Link]

okay I have my answer, the volume of what's left of the cube is...0.0580191... in^3here's an imgur album with my work:
http://imgur.com/a/Xzwjk

agh forgot to say that the 4 pi r^2 should be a pi r^2 obviously. When I evaluated the integral numerically in mathematica I fixed it. Seems I got the right answer from reading the past pages, but less efficiently.

The integrals actually can be computed in closed form to give $V=1+\sqrt{2}-\frac{3}{4} \pi$. It might interest you that you can also get a numerical answer with about 15 lines of computer code using 3 nested For-Next loops ("Do" loops) that divide the volume into 100x100x100 parts and using inequalities to test whether a point (tiny cube) lies inside or outside the 3 cylinders. If it lies outside the 3 cylinders, you count $N=N+1$. Once the "Do" Loops have completed all 1,000,000 points, you compute $V=N/1,000,000$. With 1,000,000 points (100 intervals on each Do Loop), you can get an answer accurate to about 3 or 4 decimal places or more.

 

[mfb]

You can speed up the code by a factor about 4 if you remove one cylinder from the loops. You don't have to add "0" 100 times.
Dividing 1/4 of the cube surface into a 1000x1000 grid and evaluating the length of solid material for one corner for each point should give an even better approximation with a similar computing time.