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

 Validity Vindication (Posted on 2014-06-18)
Given a square piece of paper, prove using paper folding alone this trigonometric identity:

arctan(1/3) + arctan(1) = arctan(2)

*** No geometry instrument aids like straightedge, compass etc are allowed.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 a rough method | Comment 1 of 2

Note that arctan(2)-arctan(1/3) = arctan(1).

'Unfold' means unfold the last folding action, not earlier folds.

Fold the paper in half horizontally once.

Fold the paper in half vertically four times. Unfold.

Fold the paper along the last vertical crease.

Fold this rectangle diagonally. Unfold.

Count 8 vertical creases away from the larger - angled diagonal crease. Fold the paper backwards along the 8th crease.

This will create a square. Fold it diagonally so that the diagonal meets the larger - angled diagonal crease at a corner. Unfold.

The original diagonal crease is arctan(2) (precisely, arctan (15/8))

The diagonal of the square is arctan(1).

The angle between them is arctan(1/3) (precisely, arctan (7/23))

The governing formula is u^2-v^2= uv+1; with {u=5,v=3} this gives {16,30,34} for which{8,15,17} is a solution. Larger solutions are more accurate but require a lot of folding; e.g. arctan(67104/33553) = 1.99994 - arctan(1) =arctan(33551/100657) =0.33332... etc.

However, this doesn't affect the underlying principle.

Edited on June 18, 2014, 12:07 pm
 Posted by broll on 2014-06-18 11:53:25

 Search: Search body:
Forums (0)