##
Five Triangles in a Square

Cut a unit square into 5 triangles such that the difference of the areas
of the smallest one and the largest one is minimal.
Source: Reader Philippe Fondanaiche.

Solutions were received from
Al Zimmermann, Jimmy Chng Gim hong, Dane Brooke, Joseph DeVincentis,
David Green, R.E. Saunders Jr, Berinda Radu, Jozef Hanenberg, Ajay Singh,
Claudio Baiocchi.
From Joseph DeVincentis (same as David Green, Al Zimmermann):

My best idea so far is cutting the triangles in this configuration:
A-------B----+
| /| |
| / \ |
| / | |
| / \ |
| / \ | |
| / \ \ |
|/ \_ | |
C--_ \ \ |
| --_ \ ||
| --_ `\|
| --_\
+------------+

If the distance AB=AC is 3/5, then the smallest triangle is ABC, and the
largest are the two in the middle, at areas of 18/100 and 21/100
respectively.

If you increase AB=AC to (sqrt(5)-1)/2 then the areas of triangle ABC and
the triangles at the far right and bottom become the same, (3-sqrt(5))/4,
and the middle triangles have area (3sqrt(5)-5)/8 for a difference of about
0.0225.

Minimum Area = 0.190983

Maximum Area = 0.213525

Difference = 0.022542

From Dane Brooke, the same result with a different configuration. The bottom
triangle in both figures has the same dimensions:
+------------+
|\ | /|
| \ | | |
| | B / B / |
| \ | | |
| | | / |
| \ / / |
| | | / |
C |/ / A |
| A \|/ __-f
| '_--- |
|_--- A |
+----f-------+
f = (3-sqrt(5))/2.
A = f/2 = (3-sqrt(5))/4 = 0.190983
B = (1-3A)/2 = 0.213525
difference = 0.022542

From Berinde Radu, again the same result with a different configuration.
The bottom triangle has the same dimensions as the previous figures.
+---f--------+
|\ B __-|
| \ __- |
| \_-- |
y A |-_ A |
| / \_ |
| | - |
| / \_ |
| | B \|
| / __-f
|| __--- |
|_--- A |
+------------+
f = (3-sqrt(5))/2.
y = (1 - x^2) / 2 = (3*sqrt(5)-5)/4.
A = f/2 = (3-sqrt(5))/4 = 0.190983
B = (1-3A)/2 = 0.213525
difference = 0.022542

Mail to Ken