SERVING THE QUANTITATIVE FINANCE COMMUNITY

 
User avatar
wileysw
Topic Author
Posts: 593
Joined: December 9th, 2006, 6:13 pm

cut & assemble

August 15th, 2015, 5:40 pm

here is a cute puzzle i encountered recently (no obvious spoiler please):suppose you are given a wooden rectangular board with a point marked. your task is to cut the board into two pieces and when you put them back into a new rectangular board, the marked point is the center.follow-ups: how about other type of symmetric shapes? and can you generalize to an arbitrary shape with two arbitrary points to be swapped?
 
User avatar
winthroptsmith
Posts: 72
Joined: March 14th, 2008, 7:50 pm
Contact:

cut & assemble

August 15th, 2015, 8:18 pm

Call the rectangle R, the marked point P and the rectangle's center C. Draw a line segment PC to connect these points. Find the midpoint M between P and C. Now draw a rectangle Q that is entirely inside R, which has a pair of sides which are parallel to PC, has M as its center, and also contains P and C. This might need to be a very skinny rectangle if P is close to a corner (I assume P is an interior point). Q will have the same orientation as R only if P is directly above, below, left, or right of C.Now cut Q out of R. We now have two pieces: a smaller rectangle, and large rectangle with a hole. Pick up Q, rotate 180 degrees and place back in the hole. Now P and C have traded places. P is now in the center of the rectangular board.
 
User avatar
savr
Posts: 48
Joined: January 21st, 2013, 3:28 pm

cut & assemble

August 17th, 2015, 11:40 am

@OP, is that solution accepted?The fact that the cut does not start at the edges takes imagination - it needs a different tool than your standard scissors, unless you can fold the wooden rectangular board. On the other hand if the rectangle Q is not entirely inside R, then in general when placing it back in the hole after rotating 180 degrees, the funny shape obtained is not rectangular.
 
User avatar
winthroptsmith
Posts: 72
Joined: March 14th, 2008, 7:50 pm
Contact:

cut & assemble

August 17th, 2015, 5:19 pm

We're talking about a board, so why not a jigsaw? Also, it should always be possible to find a Q that is entirely inside R. Find the point S on the perimeter of R that is closest to P, and let T be the midpoint between S and P. Let T be one of the corners of the inner rectangle Q. The other corners of Q will satisfy the conditions above and these will all be no closer to the perimeter than T (remember that P is closer to the perimeter than C). So Q will fit entirely inside R. If you don't want a hole in one of the pieces, consider the sides of Q that are parallel to PC. In the vicinity of P, extend these sides to the perimeter. The perimeter will now contribute one or two sides to the revised Q' (two if a corner of R is captured), replacing an original side of Q that was orthogonal to PC. Q' is now a trapezoid or a pentagon with two parallel sides. We won't be able to rotate it and fit in perfectly back in to R. But we could change the opposite end of Q' (the part closer to C) so that Q' has rotational symmetry about M. Now Q' is a parallelogram or a hexagon with two parallel sides, and it can be rotated in placed perfectly back into the gap.
 
User avatar
winthroptsmith
Posts: 72
Joined: March 14th, 2008, 7:50 pm
Contact:

cut & assemble

August 18th, 2015, 12:23 pm

The simplest solution I think is to let Q share a corner of R as well as R's orientation. Let U be the corner of R that is closest to P, then let Q be the rectangle that has U as a corner, M as its center, and parts of two sides of R for two of its sides. Q can be separated with two simple cuts, leaving an L-shape for the other piece.
 
User avatar
winthroptsmith
Posts: 72
Joined: March 14th, 2008, 7:50 pm
Contact:

cut & assemble

August 18th, 2015, 12:23 pm

The simplest solution I think is to let Q share a corner of R as well as R's orientation. Let U be the corner of R that is closest to P, then let Q be the rectangle that has U as a corner, M as its center, and parts of two sides of R for two of its sides. Q can be separated with two simple cuts, leaving an L-shape for the other piece.
 
User avatar
Traden4Alpha
Posts: 23951
Joined: September 20th, 2002, 8:30 pm

cut & assemble

August 18th, 2015, 12:43 pm

The more general solution for an arbitrary convex shape S with points P1 and P2 (at least one of which lies in the interior of S) is to:1. Mark the midpoint, M, between P1 and P22. Duplicate shape S with the three points to create template S' with points P1', P2', and M'3. Rotate S' 180° about M' to place on S such as P1' lies on P2 and P2' lies on P1.4. Use the overlap shape SS' (the intersection of S and S') as the superset of piece to cut out.The cutout can be any shape that is mirror symmetric about point M, in SS', contains P1 & P2, and that does not cut S into three pieces.Cases in which both P1 or P2 lie on the edge of S may be doable but the preconditions for feasibility (adhering to the cut-into-two-pieces restriction) are quite complex
ABOUT WILMOTT

PW by JB

Wilmott.com has been "Serving the Quantitative Finance Community" since 2001. Continued...


Twitter LinkedIn Instagram

JOBS BOARD

JOBS BOARD

Looking for a quant job, risk, algo trading,...? Browse jobs here...


GZIP: On