All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Origamic (Posted on 2003-09-23) Difficulty: 3 of 5
A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle.

What is the ratio of the longer side of the rectangle to the shorter?

See The Solution Submitted by DJ    
Rating: 4.4167 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution No Subject | Comment 2 of 11 |
Let the rectangle's corners be A, B, C, and D. Assume the fold unites corners A and C. The fold creates a line EF. Let E be on AB and F be on CD. From the folding, AF and FC are congruent. Let the center of the rectangle be G. AC is perpendictular to EF and the lines intersect at G. G also bisects AC and EF.

Since AD is the shorter side, let AD=1 and AB=x (and EF=x). Let DF=a. Then FC=x-a (and AF=x-a), FG=x/2, and AG=(sqrt(1+x^2))/2.

ADF is a right triangle, then AD^2+DF^2=AF^2: 1 + a^2 = (x-a)^2. AFG is also a right triangle, then AF^2=FG^2+AG^2: (x-a)^2 = (1+x^2)/4 + x^2/4

Solving the pair of equations yeilds x=1.27202, hence the ratio of the sides is 1.27202
  Posted by Brian Smith on 2003-09-23 15:13:33
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information