Thread: Jigsaw puzzle involving perimeter of semi-circles

1. Jigsaw puzzle involving perimeter of semi-circles

Intermediate Mathematical Challenge 2010

Puzzle number 9

I'm not satisfied with the explanation because I did $26+4\pi$ - $22+2\pi$ to get $4+2\pi$

With the 26+4 pi being what I got from them being separate pieces $(3+3+3+3+3+3+1+1+1+1+1+1+4(\frac{2\pi}{2}))$

And the second part being the same way without using two 1s and four of 2pi over 2

2. I don't think you were adding properly. I get, for the perimeter of the un-fitted pieces,

$30+6\pi,$ and for the fitted pieces, I get

$22+2\pi.$

The difference is then $8+4\pi,$ or answer C.

3. How can you get 6 pi if there are only 4 sides with semi circle sticking out/indented?

If anything now I seem to be getting 8 pi for the first perimeter.

Mind explaining how you got that answer?

4. I don't mind at all! There are three pieces. Each piece has two semi-circular regions of perimeter pi. Ergo, three pieces times two regions times pi gives you six pi.

5. Originally Posted by Mukilab
Intermediate Mathematical Challenge 2010

Puzzle number 9

I'm not satisfied with the explanation because I did $26+4\pi$ - $22+2\pi$ to get $4+2\pi$

With the 26+4 pi being what I got from them being separate pieces $(3+3+3+3+3+3+1+1+1+1+1+1+4(\frac{2\pi}{2}))$

And the second part being the same way without using two 1s and four of 2pi over 2

The perimeters of all pieces summed are

(1) $1+1+1+1+3+3+{\pi}+{\pi}$ for the piece on the left

(2) same for the centre piece

(3) same for the piece on the right.

You correctly calculated the perimeter for the assembled jigsaw.

6. Originally Posted by Ackbeet
I don't mind at all! There are three pieces. Each piece has two semi-circular regions of perimeter pi. Ergo, three pieces times two regions times pi gives you six pi.
sorry it seems I ignored the middle one

Thanks for pointing this out

7. All good now, then?