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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

Suppose that you have a screw with square head of side

that you want to undo. Unfortunately, the only tool you have around is a key with a hexagonal opening of side . What relations must and satisfy for you to succed in undoing the screw?Offline

**jk22****Member**- Registered: 2010-06-14
- Posts: 33

Hi heirot, nice to meet you,

I think I found a case with a=2*sin(60)*sin(45)/sin(75)*b ?

cya.

*Last edited by jk22 (2010-08-09 21:49:36)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

hi Heirot and jk22,

Firstly, to jk22, any chance of a diagram to go with this. You have 'equality' so, I'm assuming this is a limiting case. But which?

And now to my own attempt at your question.

Whenever, mathematicians try to model a 'real life' problem from engineering (or science ...), they should consider and state their assumptions.

Here are mine.

(i) I've assumed the centre of the square is the same as the centre of the hexagon.

(ii) I've assumed the two polygons are perfectly regular (all sides and angles equal for each) and that the polygons don't have rounded vertices.

(iii) See first diagram. I'm assuming that if the hexagon is rotated anticlockwise the point marked x will cause the square to rotate too (no slippage).

Diagram two shows the upper limit for the hexagon. The distance to x from the centre to the square is a maximum and this is the largest hexagon I can make that 'just touches'. So if 'b' is any bigger there will be no contact at all.

Use Pythagoras to calculate the distances. (If you want help with this, let me know).

Square: centre to x is a.root2/2

Hexagon: centre to x is b.root3/2

Thus b.root3/2 < a.root2/2 ... => b < a.root(2/3)

Diagram three is, I think, the lower limit for the size of the hexagon. The top and bottom sides of the hex are parallel to the top and bottom sides of the square. Any smaller and parts of the hexagon overlap the space of the square.

Angle POX = 60 - 45 = 15

b.cos15 > a.root2/2 (using trigonometry, again ask if you need help with this)

so b > aroot2/(2.cos15)

Together 0.732a < b < 0.816a

hope this helps,

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**jk22****Member**- Registered: 2010-06-14
- Posts: 33

Hi, bob,

i thought there was a domain, but couldn't figure it out with geometry. Maybe it's the first case i fell upon, I used the sine theorem.

d/sin(60)=b/sin(180-60-45) [hope you can see the pics]

Do you think other possibilities with 3 touching spots are not possible ? I mean this would imply non concentric centers, and I don't know if it's possible to (un)screw in this eccentric case ? ( I could manage to upload pictures) :

I'll look at the equation and tell you if i manage to understand these.

cya.

*Last edited by jk22 (2010-08-11 03:14:39)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

Sorry jk22,

I cannot see these pictures at all. What I do is (i) screen shot of the pic (ii) paste into 'Paint' and edit (iii) save as a jpg file (iv) upload using the forum upload facility.

I can give more details if any of these steps is unclear to you.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

Here's a non concentric, 3-point contact.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**jk22****Member**- Registered: 2010-06-14
- Posts: 33

Hi Bob,

yeah it's dumb, it's a link to google docs I have, but only me can see them.

Does there exists a 4-point eccentric case ? I cannot figure it out.

bye.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

Hi

4 points? Very much doubt it ... all to do with 60 and 90 degree symmetries. I haven't got a rigorous proof though.

Bob

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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

Hi, Bob,

I agree with you on the concentric case. What's your conclusion on the non concentric case? I think, due to symmetry, the non concentric case must also fall within the range of the inequality you gave.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,020

hi Heirot,

My diagram software (Sketchpad) is a vector geometry program so it's fairly accurate for playing about with this problem.

My three points post (again below), has b = 2.021 whereas the previous concentric post had b = 1.934 or less.

I think the best solution is to buy an open ended spanner (wrench).

Bob

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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

Bob, you said it all! If a wrench can do it, so can a hex key. Now, is your three points solution unique?

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**jk22****Member**- Registered: 2010-06-14
- Posts: 33

Hi bob and heirot,

what about the reverse case : a hexagon in a square ? Only eccentric case exist ? I suppose it's possible to unscrew.

*Last edited by jk22 (2010-08-15 04:15:00)*

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