### Random Post

### Recent Posts

### Recent Comments

Jim Randell on Enigma 1691: Factory part… | |

Jim Randell on Puzzle 52: Football on the Isl… | |

geoffrounce on Enigma 1691: Factory part… | |

Hugh Casement on Enigma 1070: Time to work | |

Jim Randell on Enigma 1070: Time to work |

### Archives

### Categories

- article (11)
- enigma (1,157)
- misc (2)
- project euler (2)
- puzzle (40)
- site news (44)
- tantalizer (42)
- teaser (3)

### Site Stats

- 177,808 hits

Advertisements

This one doesn’t really need any programming. The solution falls out with a little analysis…

Consider the unfolded rectangle. It has an area of 24

a(whereais the length of the shorter side),and is composed of 4×

T1triangles (on the outsides) and 2×T2triangles (in the middle).Then consider the folded pentagon. It has an area 2/3 of the rectangle, so

16a, and iscomposed of 2×

T1triangles and aT1+T2triangle.So considering the areas:

4 × T1 + 2 × T2 = 24a3 × T1 + T2 = 16ahence:

T1andT2both have area4a.And they both have height

a, so they are identical triangles (T) with a base measuring 8 (=b).Which means the longest side of the pentagon is the hypotenuse of

T(=h), the other sides of the pentagon being the non-hypotenuse sides ofT(aandb).But when we make the fold the bases of two

Ttriangles lie along the hypotenuse of anotherTtriangle it follows thath = 2 × b, henceh = 16(anda = √192).Solution:The longest side of the pentagon is 16cm.