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#1 2015-12-20 04:26:15

helpmeplz
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Registered: 2015-12-20
Posts: 7

Geometry Problem

The solid block S shown on the right below is made from a unit cube of wood shown on the left below by cutting away a triangular block, where P, Q, R are the midpoints of sides AB, CD, EF respectively.

I dont have a pic, but the first unit cube has its verticies labeled like this, the top square has A in the left hand corner closest to us, with B, C, and D rotating clockwise around. Directly, below D, there is E, and the farthest away corner from us that we can see that is not labeled yet is F. Sorry about such a bad description, it is the pest that i can do.

A fly, an ant and a termite are all placed at point A of S, and they all wish to get to C, where there is food. Flies can walk on wood and fly. Termites can walk on wood or tunnel through it but can’t fly. Poor ants can only walk; they can’t tunnel or fly. For each insect, determine with proof the shortest path from A to C and the length of that path. In proving that no other path for that insect is shorter, you may use without proof that the shortest path of all between any two points is the straight line segment between them.

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#2 2015-12-20 06:21:54

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

I, personally, cannot tell just how the triangular block has been cut yet, and I like word problems. If you cannot give a link, can you email a file? If not, perhaps you could describe the layout of the triangular cut in more detail, specifically the lines connecting A to C? Sorry that I cannot see it just yet (:

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#3 2015-12-20 06:59:08

helpmeplz
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Registered: 2015-12-20
Posts: 7

Re: Geometry Problem

download.php?id=YXR0YWNobWVudHMvNC9iL2Q2YzExYmRjN2ZmY2FiODhiN2ExNGU0NjI2ZWJkODRjNzdjYzRkLnBuZw==&rn=U2NyZWVuc2hvdCAyMDE1LTEyLTIwIGF0IDEwLjIyLjM2IEFNLnBuZw==

I think this is the link to the image

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#4 2015-12-20 14:57:09

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

Yes, that is it! Excellent smile
The only distance that is not given is AE = PR, but I'm sure it is reasonably well-known that 1-1-√2 is a Pythagorean triple since the hypotenuse is √(1^2+1^2) = √2.
Is there any restriction on the flight of the flies? Because if not, they can just fly from A directly to C. That distance has to be √2 as well, or about 1.4142, since it is completely symmetric to the distance AE. Because there is so much information provided, you can prove this with any rule you like (drawing AC on the left gives two triangles, each with 90, 45, 45 degree angles and two sides of length 1).

The ant, since that is the simpler, will prefer to walk AP + PC. The length PC is √5 / 2, since that is √(1^2+(1/2)^2), which is about 1.1180. Therefore AP + PC = 1/2 + √5 / 2 which is about 1.6180.
You can prove this simply by asserting that PC is the shortest distance from P to C and that PC + AP must be shorter than the sum of AE or AR and any available route (in fact, it is shorter than even AE + ER or AR + anything else (AR is 1.5)).

The termite is more complicated. The problem is to optimise the distance AX + XC where X is some point on PR. It's possible that this cannot be less than 1/2 + √5 / 2, and the termite will go the same way as the ant, but I haven't proved it yet.
AX = √(2x^2 + 1/4) where x is the fraction of the distance PR traveled from P (in other words, x = PX/PR).

Last edited by Relentless (2015-12-20 17:12:46)

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#5 2015-12-20 20:57:14

Bob
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Registered: 2010-06-20
Posts: 10,052

Re: Geometry Problem

hi helpmeplz

Welcome to the forum.

The best way to calculate the shortest surface walk is to draw the net of the solid.

RCBLEyn.gif

The dotted line shows the shortest route.

For the fly, I agree with Relentless.

I think, for the termite, 'he' should surface walk like the ant until reaching the line PR.  Then tunnelling will shorten the last stage.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2015-12-21 05:40:51

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

I will get back to this problem within another day, and play around with general rules for the distance from C to any point on PQ, C to any point on PR, and between points on PQ and PR. Notice also that I didn't yet irrefutably prove that the ant will prefer AP + PC to moving to PR and then climbing to PQ; I strongly suspect it is true though.

It looks like the termite will obey the path of bob's line, but determining that distance will take more work. But time for sleep first! (:

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#7 2015-12-21 15:36:49

helpmeplz
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Registered: 2015-12-20
Posts: 7

Re: Geometry Problem

Thanks for everything already, it has been really helpful, I will go ponder the problem of the termite also now.

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#8 2015-12-21 22:18:27

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

I am back to it.
Notice that bob's net is properly to scale, magnified to 4x. It makes the mathematics easy to visualise smile

The equation for the distance CY where Y is a point on PQ is:
CY = sqrt(y^2 + 1/4) where y = YQ/PQ

You can use the cosine rule to create a formula for any XY!

So XY = √(x^2 - √2*x(1-y) + (1-y)^2) where, again, x = PX/PR and X is any point on PR.

To prove the ant's path, once and for all, we just have to minimise AX + XY + CY. Which is this formula: √(2x^2 + 1/4) + √(x^2 - √2*x(1-y) + (1-y)^2) + √(y^2 + 1/4)

It looks like I was wrong about the ant. There is a solution to that equation which gives a marginally smaller distance than AP + PC ~ 1.6180.

I don't have the exact answers, but the distance is approximately 1.60759 if the ant walks from A to a point roughly 0.0939912 from P along PR, and then from there to a point roughly 0.128874 from P along PQ, and from there to C. That is, x is about 0.0664618 and y is about 0.871126 (where these are fractions along the lines PR from P and PQ from Q respectively).
This doesn't look to be quite the line drawn in bob's net to me (it looks like the climb up the wall is a little bit steeper), but all the equations are given to analyse it and why it is the way it is.



The termite proof will use a right triangle with small sides CY and XY and hypotenuse CX, where x = 1 - y (that equation ensures that the points referred to are in the same vertical line). I will continue this at some point, but that is simple enough now; it is not even a three-dimensional problem anymore smile

Last edited by Relentless (2015-12-22 01:41:15)

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#9 2015-12-22 01:54:18

Bob
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Registered: 2010-06-20
Posts: 10,052

Re: Geometry Problem

My method (straight line along the net) gives 1.603.... for the ant.

The question only asks for the shortest routes; it doesn't say that ants prefer not to climb up.  From what I've noticed about ants, they are quite happy to do this.  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#10 2015-12-22 02:11:45

helpmeplz
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Registered: 2015-12-20
Posts: 7

Re: Geometry Problem

I think for the termite, you can create a net too, it would be similar to bob's net, but it would have just 2 planes, APR and a cross section of PRC, forming a triangle, that way you can tell which tunnel route is shortest.

I need help finding the dimensions of that plane

Last edited by helpmeplz (2015-12-22 02:12:10)

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#11 2015-12-22 02:18:57

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

I tried to solve for the straight line along the net while looking at the net, and came up with 1.5584 :S I need another break or a good explanation lol

What distances do you need? We already know the diagonals are √2

Last edited by Relentless (2015-12-22 02:25:02)

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#12 2015-12-22 04:33:53

Bob
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Registered: 2010-06-20
Posts: 10,052

Re: Geometry Problem

sLyzFeL.gif

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#13 2015-12-22 04:41:29

helpmeplz
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Registered: 2015-12-20
Posts: 7

Re: Geometry Problem

the distances from P to C and P to R

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#14 2015-12-22 04:54:06

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

They were √5 / 2 and √2.
1.118033988749894848204586834365638117720309179805762862135448...
and
1.414213562373095048801688724209698078569671875376948073176679...

Last edited by Relentless (2015-12-22 04:56:06)

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#15 2015-12-22 05:05:50

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

Using my equations I obtained a minimum length for the termite, but because Bob disagrees with my result of 1.60759 for the ant I am not entirely sure of them.

I obtained a length of about 1.58451 for the termite. He walks to a point about 12.3929% along PR from P.


To sum up, these are the results I got.
Fly: √2 ~ 1.4142
Termite: ~1.58451
Ant: ~1.60759

If these results are not correct it is because I made a mistake on one of the following equations:

1. The distance from C to any point on PQ: CY = √(y^2 + 1/4) where y = YQ/PQ

2. The distance from A to any point on PR: AX = √(2x^2 + 1/4) where x = PX/PR

3. The distance from any point on PR to any point on PQ (using the Law of Cosines): XY = √(x^2 - √2*x(1-y) + (1-y)^2)

And finally,
4. The distance from C to any point on PR: CX = √(CY^2 + XY^2) where x = 1 - y
Or the long version: CX = √(√(x^2 - √2*x(1-y) + (1-y)^2) + y^2 + 1/4)


So my results are based on the minimisation of these systems of equations.
For the ant: AX + XY + CY
For the termite: AX + CX
If the equations are correct, I consider this minimisation to be the final proof.

Last edited by Relentless (2015-12-22 05:37:09)

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#16 2015-12-22 11:31:28

Bob
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Registered: 2010-06-20
Posts: 10,052

Re: Geometry Problem

Relentless wrote:

3. The distance from any point on PR to any point on PQ (using the Law of Cosines): XY = √(x^2 - √2*x(1-y) + (1-y)^2)

Shouldn't that be

XY = √(2x^2 - √2*x(1-y) + (1-y)^2)

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#17 2015-12-22 12:19:47

Bob
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Registered: 2010-06-20
Posts: 10,052

Re: Geometry Problem

hi

I was just about to go to bed when I realised there is another root 2 missing:

XY = √(2x^2 - 2*x(1-y) + (1-y)^2)

Now my calculations work out with yours.

Try x = 0.0798  and y = 0.792893219

Bob

Good night from the UK sleep


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#18 2015-12-22 12:21:53

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

Yes!! You're right! I neglected the fact that PX = x*PR, not just x. Thank you!
Time for another crack at it...

Last edited by Relentless (2015-12-22 12:29:27)

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#19 2015-12-22 12:54:41

Relentless
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Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

I believe the problem is now about to be solved.
I will lay out the formulas.

1. The distance from C to any point on PQ: CY = √(y^2 + 1/4) where y = YQ/PQ

2. The distance from A to any point on PR: AX = √(2x^2 + 1/4) where x = PX/PR

3. The distance from any point on PR to any point on PQ (using the Law of Cosines): XY = √(2x^2 - 2x(1-y) + (1-y)^2)

4. The distance from C to any point on PR: CX = √(CY^2 + XY^2) where x = 1 - y
CX = √(2x^2 - 2x(1-y) + y^2 + (1-y)^2 + 1/4)

For the fly: AC = √(1^2+1^2) = √2 ~ 1.414213562

For the termite: The minimum possible AX + CX where x=1-y, i.e.
Minimise: (√(8x^2 + 1) + √(8x^2 - 8x + 5)) / 2 (y=1-x)

The minimum is precisely 1/2 * √(3-√3) * (1+√3) at
x = (√3 - 1) / 4 and
y = (5 - √3) / 4

AX = 1/2 * √(3-√3)
CX = 1/2 * √(9-3√3)

In approximate terms:
The termite travels 1.538189001 to a point 18.30127019% along PR from P, and tunnels to C. He travels 0.5630162503 to get to that point (36.603% of the journey), and 0.9751727510 to complete the route (63.397%).





For the ant: The smallest possible AX + XY + CY, i.e.
Minimise: √(2x^2 + 1/4)+√(2x^2 - 2x(1-y) + (1-y)^2)+√(y^2 + 1/4) where 0<=x,y<=1

The minimum is precisely 1/2 * √(6+3√2) at
x = 3/(14√2) - 1/14 and
y = (3 - √2) / 2

AX = 1/7 * √(15 - 3/√2)
XY = 1/14 * √(102 - 69√2)
CY = √(3 - 3/√2)

In approximate terms:
The ant travels a total of 1.600206290 by walking to a point 8.009431025% along PR from P, then up to a point 20.71067812% along PQ from P, then to C. He travels 0.5126696764 to the line PQ (32.038% of the journey), 0.1501574717 up to PQ (9.3836%), and 0.9373791423 to C (58.579%)

Last edited by Relentless (2015-12-22 13:48:20)

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#20 2015-12-22 13:53:36

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Geometry Problem

Final results in one place:
Fly: √2
Termite: 1/2 * √(3-√3) * (1+√3)
Ant: 1/2 * √(6+3√2)

Fly: 1.414213562373095048801688724209698078569671875376948073176679...
Termite: 1.538189001320851548483012931968361209659295428862529812720317...
Ant: 1.600206290382531050560655069826726875794131986410577501483748...


Whoever guessed a simple unit cube could yield such a difficult problem? Thanks for the puzzle!

Last edited by Relentless (2015-12-22 13:55:49)

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#21 2015-12-22 14:22:48

helpmeplz
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Registered: 2015-12-20
Posts: 7

Re: Geometry Problem

You're welcome

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