The top line must be shorter than 9cm and longer than 5 cm. Every length within that range gives the same total perimeter, 30. This is a shape where you can find the perimeter, but you can't determine the width except that 5<w<9.
Woot! I saw a version of this a tuthree weeks ago and sent it off to my 13 y.o. grandaughter. The solution there was "42" and there were several comments about know-nothings having an ironic Hitchhiker's punt at the answer and getting the win.
Your explanatory diagram makes the algebraic solution pretty clear, but I found it helpful to consider the width of the isthmus between the top and bottom blocks. This can only be between 0 and 4 and those extremes give the same perimeter; it's then clear that the two unknown horizontal line segments must be trading off against each other. What's interesting is that any ancient Greek would go at the problem geometrically (scratching lines in the dust of the agora) but we now teach kids in school the abstracts of algebra and divorce such problems from the real dusty world. [rant OFF]
34: look at the 5cm line and the two vertical lines connected to it left A and right B. Imagine sliding the 5cm line up 1mm so that A becomes 1mm shorter, and B 1mm longer. The operation doesn't change their total length. So slide the 5cm line all the way up until it meets the top line. The shape is now an L shape with a double line off one corner. The double line is 10cm. Remove it. The L shape is a rectangle with a smaller rectangle removed from one corner. If you un-remove it, the final rectangle has the same perimeter, which is 6x4=24cm. Plus the 10cm double gives an answer of 34.
(Similar approach) We can't know the lengths of the three lines in between the 5 and 4. If the puzzle's to be solvable, then we should be able to make them whatever convenient thing we want. I like zero. Now it's a 6 by 9 rectangle, with perimeter 6+9+6+9.
5 cm line = 6 cm line (measure it.) It the rules of space/time have been suspended the perimeter could be anything!
ReplyDeleteSomebody goofed. But consider the 6 to be a 5 and continue solving.
DeleteThe top line must be shorter than 9cm and longer than 5 cm. Every length within that range gives the same total perimeter, 30. This is a shape where you can find the perimeter, but you can't determine the width except that 5<w<9.
ReplyDeleteWoot! I saw a version of this a tuthree weeks ago and sent it off to my 13 y.o. grandaughter. The solution there was "42" and there were several comments about know-nothings having an ironic Hitchhiker's punt at the answer and getting the win.
ReplyDeleteYour explanatory diagram makes the algebraic solution pretty clear, but I found it helpful to consider the width of the isthmus between the top and bottom blocks. This can only be between 0 and 4 and those extremes give the same perimeter; it's then clear that the two unknown horizontal line segments must be trading off against each other. What's interesting is that any ancient Greek would go at the problem geometrically (scratching lines in the dust of the agora) but we now teach kids in school the abstracts of algebra and divorce such problems from the real dusty world. [rant OFF]
That is a circle with a diameter of 9.55 units.
ReplyDelete34: look at the 5cm line and the two vertical lines connected to it left A and right B. Imagine sliding the 5cm line up 1mm so that A becomes 1mm shorter, and B 1mm longer. The operation doesn't change their total length. So slide the 5cm line all the way up until it meets the top line. The shape is now an L shape with a double line off one corner. The double line is 10cm. Remove it. The L shape is a rectangle with a smaller rectangle removed from one corner. If you un-remove it, the final rectangle has the same perimeter, which is 6x4=24cm. Plus the 10cm double gives an answer of 34.
ReplyDeleteCorrection: 30. The final rectangle has a perimeter of 2x(6+4)=20cm.
Delete(Similar approach)
ReplyDeleteWe can't know the lengths of the three lines in between the 5 and 4. If the puzzle's to be solvable, then we should be able to make them whatever convenient thing we want. I like zero. Now it's a 6 by 9 rectangle, with perimeter 6+9+6+9.