Solutions to the ARML Power Question 2006: The Power of Origami

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1 Solutions to the L ower uestion 006: The ower of Origami 1. a) The crease is the perpendicular bisector of, no justification is necessary. b) The crease is the bisector of two of the angles formed by the two lines; no justification is necessary. c) old onto line! at! such that the crease passes through. Now = ' by (1a) applied to and ' so ' is on the circle with radius centered at. '. a) 6 does not require that! 1 and! be distinct; in the case where! 1 =! =!, the fold simply places both points onto line!. Therefore, given two lines! 1 and!, mark points and arbitrarily on! 1 ; then apply 6 to place both points on! and that folds! 1 onto!. b) One situation is where and are on the same side of line!, but is less than the distance from to the line!. Then the circle described in (1c) fails to intersect line!, as shown at the right. 3. a) Use to fold onto point ; by (1a), the crease is the perpendicular bisector of and is, therefore, the perpendicular bisector of. roof: call the crease s intersections with and points S 1 and S, respectively; then S 1 S is a quadrilateral with three right angles, hence a rectangle, and S = S 1 equals = 1 =. S 1 T 1 T S Use to fold onto ; similar logic shows that the crease is the perpendicular bisector of and, intersecting at T 1 and T respectively. inally, let be the intersection of the creases; S 1 T 1 is a quadrilateral with three right angles, hence a rectangle, and all sides are of length 1/, so it is a square of area 1/4. b) To locate the center of the square use either the construction in (3a) or use either or 1 twice; first on diagonally opposite points &, and then again on points &. The creases form the diagonals, hence intersect in the center. If 1 is used, the diagonal formed is the one joining the two points employed; if is used, the other diagonal is formed. With the center of square located, use four times to fold each corner to the center of the square.

2 Without locating the center of the square, use as in (3a) to locate the midpoint of each side of the square and then use 1 on each pair of adjacent midpoints such as and. Note: In folding a crease through and, you fold each corner into the center. ' 4. The eno construction applied to a square of side length s and area s produces a s square of side length and area s. The uarter construction produces a square of side length s and area s 4. In particular, doing a eno construction twice is H G equivalent to doing one uarter construction as shown by folding into GH in the diagram. So at 1 each step, the side length can be reduced by a factor of or 1, and the area by a factor of 1/ or 1/4. y induction, then, the possible areas formed for the smallest square are of area 1 n, for n a nonnegative integer. However, smaller squares can be combined into larger ones by folding back strips along two adjacent edges, as shown in the diagram at the right: 9/16 Therefore, the possible areas are of the form k n, where k! n/. inally, notice that the construction at the right is not allowed in this problem since it uses creases other than those formed by eno and uarter constructions. 10/16 5. a) string such as will mean a eno construction, followed by a uarter construction, followed by two eno constructions, followed by another uarter construction. n Ways to fold a unit square into a square of area 1 n 1 1, 3,, 3 4,,,, 5 5,, 8,,, 6,,,,,,,,,,, 13 (n)

3 b) Yes, it s the ibonacci sequence, namely (n) = f n+1 where f n+ = f n+1 + f n with f 1 = f = 1. The basic idea can be seen in rows 3 and 4. The squares in row 3 have area 1/8. rom those squares we obtain a square in row 5 of area 1/3 by a uarter construction. So, to the end of any string of letters in row 3 we will add a and that new string will be entered in row 5. The squares in row 4 have an area of 1/16 and from those squares we can obtain a square of area 1/3 by using a eno construction. So, to the end of any string of letters in row 4 we will append an and enter the new string in row 5. Thus, row 5 will have as many members as the sum of rows 3 and 4. ore formally, we have the following proof: roof: Let W n be the set of all ways to form a square of area 1 n as represented by the strings described above. Thus, (n) = W n. Writing "x" for the concatenation of string x and the letter representing a uarter construction, we can write W n+1 = x : x! W n"1 { } # { y : y! W n }. That is, a square of area 1 n+1 can be formed by applying a uarter construction to a square of area 1 n!1 or it could be formed by { } # { y : y! W n } = $ applying a eno construction to a square of area 1 n. Note that x : x! W n"1 since the strings in the first set end in while the strings in the second set end in. Thus we have { } + { y : y! W n } = W n!1 + W n, giving (n + 1) = (n 1) + (n), and W n+1 = x : x! W n"1 since (0) = (1) = 1, the assertion is proved. 6. a) There are many possible constructions, but perhaps the simplest is below: Step 1: Using, fold onto, labeling the midpoint. Step : Without unfolding, use to fold onto. N N T 1 T ' Step 3: Unfold. ecalling (3a), let N and be the midpoints of and respectively. Label the crease passing through N as line segment N. Now use 5 to fold point up to N with the new crease passing through. Step 4: Now let T 1 and T be the places where the creases through points and pass through!. Justification: (3a) shows that the creases are all perpendicular to side, so the creases are parallel, and parallel lines that cut equal segments on one transversal (here ) cut equal segments on any transversal (here! ). b) To split into k congruent segments, pick n so that n > k and then repeatedly bisect a total of n times, producing n! 1 parallel creases. Then unfold the side, and using 5, fold vertex up to the k th crease (counting from ), so that the new crease passes through. Then the intersections of the other horizontal creases pass through divide into k equal segments, since the horizontal creases are equally spaced. lternately, it s not necessary to bisect as long as the creases formed in steps 1 3 are parallel and equally spaced. This can be achieved by picking an arbitrary point 1 on near and folding a perpendicular to through 1 using 4. Now denote the image of by. learly, 1 = 1 by 4 3 1

4 construction. Unfold and fold again through a perpendicular to. Let the images of and 1 be 4 and 3 respectively. Now 1 = 1 = 3 = 3 4. ontinue in this fashion until k creases are formed where k < n!1. Then proceed as in steps 3 and 4 of part (6a). The only difficulty is in selecting a point 1 such that 1! 1 k, otherwise there won't be enough room to make all k folds. 7. Now the sides can be folded into any fraction of the form m n or m n So the area can be folded into any fraction of the form m n or m n., where m and n are integers, n! m. roof: use the construction in (6b) to mark points 1/n, /n, etc. of the way from to. Then choose the point that is m/n of the way from to. Use 3 to find a corresponding point on. t both and use 4 to create a crease perpendicular to each side of. The creases meet at point forming square that has an area of ( m n) = m n. The eno construction can be used on to halve its area. 8. a) Given collinear points,,, with = 1 and = r, first fold a crease! through with!! using 4. Unfold the paper and fold onto using ; let O be the midpoint of determined by the crease. Unfold the paper and fold onto! with a crease passing through O using 5. Let! be the image of on crease!. Now O = by and O = O! by construction, so O! is a radius of the circle with diameter. Since! " by construction, then! is the geometric mean of and, i.e.! ( ) = = 1 r = r!! = r. b) Start with square and point on with = r. old point onto using and mark the midpoint of =. Now = r + 1! r r + 1 =, so is a radius of a circle centered at with diameter r + 1. Using 4 fold a crease through perpendicular to ; call the crease!. Using 5 fold onto! with the crease passing through. Let! be the image of on! ; note that! = = So! is the intersection of the circle centered at with the perpendicular through. Thus (! ) = (r + 1 ) = r (r + 1 r) = r. Therefore,! = r. ' 1 r O r ' (1 + r) r Using the constructions in (6) and (7), any side length of the form k/n, k n, can be constructed on the square. k Using (8b), any side length of the form can be constructed. Therefore, any area of the form k/n can be n constructed. ut in fact, we can do even better: by square-rooting twice, we can construct a side length of the form k 4 and thus an area of the form k n n. That is, it s possible to construct a square of area = 1 using these constructions, and certainly any area of the form ( k n) 1/m by repeated applications of (8b).

5 3 10. a) The altitude of a unit equilateral triangle is, and yields an area of 3/4. One can be constructed by folding point onto point using and labeling the midpoint, then folding onto the crease through with the new crease passing through using 5. That's folds. Let the image of be!, then! = 1, ' = 1/, and! = 3. Now fold a perpendicular to! through! using 4 and denote the intersection of that crease and by point. Using 3 fold onto marking the image of on as point!. That makes 4 folds. Using 4, fold a perpendicular to at!. That's 5 folds and,, H, and! determine the vertices of a square whose area is 3/4. ' ' H b) One quite efficient construction is to use the method of (6b) to divide into five equal parts. This requires making 3 folds to subdivide into 8 equal parts and then folding up to the fifth crease allowing one to subdivide into the five '!! equal parts. Using 3, those can also be marked off on using one fold, marking such that = 1/5 and such that = /5. That's five folds. The sixth is to fold over using 1 to produce the crease and point!. y the ythagorean theorem, = We need to construct a perpendicular to through but that can t be constructed using 4 since is an endpoint of. However, we can still manage if we fold using 3 so that lands on! with crease through, creating angle. Let m! = m! " = # and 5 5 ' ', making it the side of a square of area 1/5. '! " " m! " = m! = #. Since! + " = 180, then! + " = m# = 90. Using 3, fold onto! with crease S. That allows us to construct two sides of the square. That makes 8 folds. S ' Using 3, we fold S onto, producing crease W. Using 4 we fold a S crease through W that is perpendicular to. oints,, W, and form the vertices of a square with area 1/5. It took 10 folds. V W