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Tuesday, 3 January 2012

My answer

Following the totally overwhelming response to yesterday's little puzzle, here is my solution. Please feel free to kick holes in it.
Start with a line, any line. Lets call the left end A and the right end B

Adjust the compass us to the length of the line and draw 2 circles, centred on A and B 

Draw a line that passes through A and through  one of the points of intersection of the 2 circles.  Do the same from B to the other point of intersection. These 2 new red lines are parallel (by rotational symmetry if you insist).  When that is done we can rub out the 2 blue circles because they just get in the way.

Choose a point between A and less than half way along the red line.  The exact distance doesn't matter but somewhere around half way will give the most accurate results. Draw a circle from this point that passes through A.  Now repeat exercise on the other red line with a circle of exactly the same size that passes through B.  If you were smart you would realise that you need to keep the compass set at the same width and you can find the centre of the second circle by scribing an arc centred on B passing through the other red line, and then drawing a circle centred at the point of intersection.

We now draw 2 more lines (purple), each one being drawn from the centre of these new circles to the point where that circle intersects the other red line (and not the black line AB) 

Rub out the green circles and we can see that we have a construction with  similar triangles on both sides of the black line.  We know the distances between adjacent points along the red lines are equal to the radius of the green circles, and by rotational symmetry we know that the purple lines are parallel (just like the red lines are parallel to each other), so it is simple to figure out that the triangles are all similar and that the distance from A to the first point on the black line is half the distance from A to the second point, and that the distance from the first point to B is twice the distance from the second point to B, from which it follows that the line AB has been divided into 3 equal sections...   Neat, huh?

2 comments:

A K Haart said...

Very neat, but difficult to work out. I'm not surprised you didn't get an overwhelming response (:

Alex said...

Just testing the readership. I have a *very* long and technical post coming up, relevant to 2012.