Programming Enigma Puzzles
From New Scientist #2871, 30th June 2012 [link] [link]
Joe drew a right-angled triangle on an A6 file card. He asked Penny to cut it out and then cut it in two to make two right-angled triangles. Then he asked her to cut each triangle in two to make four right-angled triangles in total. Now Joe’s triangle was very special. Penny found that the lengths of all the sides of all the triangles were a whole number of millimetres.
What was the area of the smallest triangle?
Note: An A6 card has dimensions of 105 mm × 148 mm.
[enigma1704]
 | Frits on Enigma 773: Duodecimal |
 | Jim Randell on Enigma 773: Duodecimal |
 | Brian Gladman on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Jim Randell on Enigma 773: Duodecimal |
Enigmatic Code 
The following Python program runs in 50ms.
from enigma import irange, is_square, printf # find possible right-angled triangles that fit on an A6 card triangles = [] for a in irange(1, 105): for b in irange(a + 1, 148): c = is_square(a * a + b * b) if c is None: continue triangles.append((a, b, c)) # split a triangle into two def split(t): (a, b, c) = t # find triangles (a1, b1, c1), (a2, b2, c2) such that # a = c1, b = c2, c = a1 + b2, b1 = a2 for (a1, b1, c1) in triangles: if a != c1: continue (a2, b2, c2) = (b1, c - a1, b) if (a2, b2, c2) not in triangles: continue yield ((a1, b1, c1), (a2, b2, c2)) # t1 is the first triangle for t1 in triangles: # split it into two for (t2, t3) in split(t1): # and split each of these into two for (t4, t5) in split(t2): for (t6, t7) in split(t3): printf("triangles = {t1} {t2} {t3} {t4} {t5} {t6} {t7}") # and find the smallest triangle (by area) printf("min area = {m}", m=min(a * b // 2 for (a, b, c) in (t4, t5, t6, t7)))Solution: The area of the smallest triangle is 486 sq mm.
Here’s a diagram showing the solution:
I’m not sure if the intention is to use the whole of the card in some way, but if not there are multiple solutions.
No there aren’t. I missed one of the equations from the code:
squares = {n*n:n for n in range(3,149)} for c in range(1,148): for a in range(3,c): if c*c-a*a in squares: b=squares[c*c-a*a] for g in range(3,100): if a*a +g*g in squares: f=squares[a*a +g*g] for d in range(3,100): if b*b+d*d in squares: e=squares[b*b+d*d] if (g+b)**2 + (a+d)**2 == (f+e)**2 and f**2 +c**2==(g+b)**2: print a,b,c,d,e,f,gMore understandable code, and a picture:
squares = {n*n:n for n in range(3,149)} pythags = [(x,y,squares[x*x+y*y]) for x in range(3,100) for y in range(3,100) if x*x+y*y in squares] for (a,b,c) in pythags: for (g,f) in [(y,z) for (x,y,z) in pythags if x==a]: for(d,e) in [(y,z) for (x,y,z) in pythags if x==b]: if (g+b)**2 + (a+d)**2 == (f+e)**2 and f**2 +c**2==(g+b)**2 and e**2 +c**2==(a+d)**2: print a,b,c,d,e,f,gHere is my solution:
# All the triangles are scaled up copies of the same primitive triangle # (u, v, w) with u <= v and w^2 = u^2 + v^2. The linear scale factors # for the three different internal triangles are u^2, u * v and v^2 and # that for the original triangle is (u^2 + v^2). The A6 card limits v # to less than 6. fs = 'Area = {0:d}, sides = {1:s}' sq = { x ** 2: x for x in range(1, 11)} for u in range(1, 6): for v in range(u, 6): t = u ** 2 + v ** 2 if t in sq: prim = (u, v, sq[t]) scale = (u * u, u * v, v * v, u * u + v * v) sides = tuple(set(x * y for x in prim for y in scale)) print(fs.format((u * u) ** 2 * (u * v // 2), sorted(sides)))Brian’s observation that all the triangles are similar is the key to a simple solution for non-coders. This diagrams illustrates the case of a 3,4,5 triangle scaled up. https://www.dropbox.com/s/ip38qog0gy9xpne/Enigma%201704%20a.PNG
The bottom side is 125n/12, so n must be 12.
I’m guessing that a similar exercise on the next triangle, 5,12,13 will exceed the limits of A6.
But the answer is a whole number of millimetres. And A6 is 105mmx148mm.
There is definitely room for it I think.
Hi Claire-Louise,
Yes, but it also has to have the four integer sided right angled angles triangles inside it and this is not possible