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PostPosted: Wed Feb 06, 2008 1:55 pm 
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Joined: Fri Sep 07, 2007 1:12 pm
Posts: 151
Please could someone explain the method for the following question:-

The diagram shows a circle with circumference of 1 cm being rolled around an equilateral triangle with sides of length 1 cm.

How many complete turns does the circle make as it rolls around the triangle (without slipping) to its original starting position?

The diagram shows a triangle with a circle balanced just above halfway on the right hand side.

The answer given is 4. I don't understand as I thought it would be 3.

Thanks for your help.


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PostPosted: Wed Feb 06, 2008 2:15 pm 
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Joined: Mon Jun 18, 2007 2:32 pm
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Location: East Kent
:? I would have said 3 too :?

rolling pennies round teh table even as we speak!!

I'd love to find out oo. This will puzzle me now :shock:


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 Post subject: triangles
PostPosted: Wed Feb 06, 2008 3:42 pm 
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Joined: Wed May 09, 2007 2:09 pm
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Location: Solihull, West Midlands
The question has been discussed here too:

http://forum.skyatnightmagazine.com/tm.asp?m=54289

I've been trying to visualise it too, but if you imagine a line from the centre of the circle to the point where they initially touch - think of the circle balanced on top of the triangle to start with, so the line (radius) is vertical

Then imagine rolling the circle down the LH side. The line rotates anti-clockwise but by the time the circle reaches the next apex the line will have gone past the vertical again - it will infact have gone an extra 1/3 of a turn

Wish I could find a diagram or animation!


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PostPosted: Wed Feb 06, 2008 6:16 pm 
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Location: Watford, Herts
Imagine rolling the circle along one side of the triangle until it reaches the corner. Now think about the radius from the corner of the triangle to the centre of the circle. (Drawing a picture will help a lot here.) That radius is perpendicular to the side of the triangle. Now we have to rotate the circle around the corner until that radius is perpendicular to the next side. What angle to we have to rotate it through? That is, what is the angle between the original and final positions of this radius we're focussing on? The full circle is 360, the angle between the sides is 60 degrees and we have these two perpendiculars, so the angle of rotation is 360 - (90+60+90) = 120 degrees. (Check your picture.) We have to do that 3 times, so there's your extra rotation.


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PostPosted: Wed Feb 06, 2008 8:58 pm 
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WoW - Any child that got that right deserves a GCSE immediately!!


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PostPosted: Wed Feb 06, 2008 10:19 pm 
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Joined: Mon Jun 18, 2007 2:32 pm
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Location: East Kent
GCSE???

Degree!


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PostPosted: Wed Feb 06, 2008 10:30 pm 
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Location: Berkshire
WoW! WP, I'm seriously impressed!! :D


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PostPosted: Thu Feb 07, 2008 1:10 am 
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There's another way that doesn't involve calculation. At each corner, draw the perpendiculars to the sides, so there is a wedge on the outside of each corner. Then
  1. argue that the angles of each wedge is the angle that the circle must be rotated on that corner, and
  2. prove that the angles of the wedges sum to a full circle.
Thus there is one more rotation to go the whole way round. This argument works for any convex polygon.

This is a ridiculously tricky question for an 11+.


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PostPosted: Fri Feb 15, 2008 2:38 pm 
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Joined: Wed Oct 31, 2007 3:15 pm
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Hi all,
I recognise this question from one of the UKMT Junior Maths Challenges. These are aimed at bright year 7 or year 8 pupils so the question is particularly tough if aimed at 11+ ~presumably to be taken in the first few months of year 6.

I can't improve on WP's answer. The circle completes one full rotation along each of the three sides of the triangle. At each corner it has to rotate 1/3 of a turm (120 degrees), so with three corners that makes the additional rotation.


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