You have an equilateral triangle you want to wrap with a square piece of paper. If the triangle has unit sides, what is the smallest square piece of paper that can be folded around the triangle such that both the front and back sides of the triangle are completely covered by the paper?

Initial Placement:
Place the triangle such that an axis of symmetry overlays the center line of the square and the apex touches the top of the square.

| s/2 |
+---------|---------+--
| /|\ |
| / | \ |
| / | \ |
| / | \1 | s
| / |h \ |
| / | \ |
| / 60 | \ |
| /_______|_______\ |
| |d |
+---------|---------+--

h = height of the triangle (sqrt(3) / 2 or cos60)
d = distance from the base of the triangle to side
s = side of the square, and h + d

Fold either flap (doesn't matter as this is symmetric):

|
+---------|
| /|\
| / | \
| / | \
| / | \1
| /_ | \
| /C '._| \
| / |'._ \
| /_______|___'.__\
| A | B '.\
+---------|---------+

B = point where the edge of the square crosses the base of the triangle.
C = point where the corner of the square touches the side of the triangle.
AB = exposed base of the triangle
AC = exposed side of the triangle

Lets look at the triangle ABC closely:

/
C/
/'._
/90 '._
/ '._ |
/ '._ |
/ '.|E \
/ |'._ \
/ y| '._ \
/ 60 | 30'._ \
/______________________|_________'.___________\
A |D x B
|
d|
|
-----------------------------------------------------

Note:

1. AC has length of 1 - s/2
2. AB has length of 2*AC or 2 - s
3. AD has length of 0.5
4. DB is AB - AD or (2-s) - 0.5 or 1.5 - s
5. x equals DB, thus is 1.5 - s
6. EDB is a 30-60-90 triangle, with angle EBC being 30.
7. The ratio of opp/adj is 1:sqrt(3).
8. y is x/sqrt(3) or (1.5 - s)/sqrt(3)

Now, the goal here is to make d and y the same. Well, if d = s - sqrt(3)/2 and y = (1.5 - s)/sqrt(3), the only way they can equal is if there is an s that satisfies both, thus, set both equations equal to each other and solve.

1. s - sqrt(3)/2 = (3/2 - s)/sqrt(3)
2. s * sqrt(3) - 3/2 = 3/2 - s
3. s + s * sqrt(3) = 3
4. s * (sqrt(3) + 1) = 3
5. s = 3 / (sqrt(3) + 1)
= 3 * (sqrt(3) - 1) / 2
= 1.09807621135331594029116951225881....

Thus, a square with sides 1.0980762... will cover the triangle. Fold each exposed flap over the triangle and presto, a covered triangle. As far as it being the best, I have no idea.

Posted by Michael on 2005-10-13 03:22:36