# Jigsaw puzzle involving perimeter of semi-circles

• Jan 12th 2011, 11:47 AM
Mukilab
Jigsaw puzzle involving perimeter of semi-circles
Intermediate Mathematical Challenge 2010

Puzzle number 9

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

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

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

• Jan 12th 2011, 11:54 AM
Ackbeet
I don't think you were adding properly. I get, for the perimeter of the un-fitted pieces,

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

$\displaystyle 22+2\pi.$

The difference is then $\displaystyle 8+4\pi,$ or answer C.
• Jan 12th 2011, 12:03 PM
Mukilab
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?
• Jan 12th 2011, 12:21 PM
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.
• Jan 12th 2011, 12:41 PM
Quote:

Originally Posted by Mukilab
Intermediate Mathematical Challenge 2010

Puzzle number 9

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

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

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) $\displaystyle 1+1+1+1+3+3+{\pi}+{\pi}$ 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.
• Jan 12th 2011, 12:42 PM
Mukilab
Quote:

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
• Jan 12th 2011, 12:45 PM
Ackbeet
All good now, then?