Perimeter of a folded square

Geometry Level 4

From a square, ABCD of side length 1, the vertex A is folded down to meet the midpoint of the side CD. The perimeter of the created shape can be written in the form a b + c ϕ \frac{a}{b} + c\phi , where a a , b b and c c are positive integers, with a , b a,b coprime, and ϕ \phi denotes the golden ratio .

What is the value a b c abc ?


The answer is 66.

Alex Burgess by Alex Burgess , 26, United Kingdom

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1 solution

Alex Burgess
Aug 2, 2018

The new shape formed is a hexagon, CDEFBG.

DA = 1 2 =\frac{1}{2} , ED = x =x , EA = 1 − x =1-x .

By Pythagoras x = 3 8 x=\frac{3}{8} .

AC = 1 2 =\frac{1}{2} , and by similar triangles, CG = 2 3 =\frac{2}{3} , AG = 5 6 =\frac{5}{6} .

GB = 1 6 =\frac{1}{6} , and by similar triangles, BF = 1 8 =\frac{1}{8} , FG = 5 24 =\frac{5}{24} .

FH = = (CG + + FG) − - ED = 1 2 =\frac{1}{2} , EH = = 1.

So EF = 5 2 =\frac{\sqrt{5}}{2} .

Hence perimeter CDEFBG = 7 3 + 5 2 = 11 6 + ϕ =\frac{7}{3}+\frac{\sqrt{5}}{2} = \frac{11}{6} + \phi .

So a b c = 66 abc = 66

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