Source: Reader Philippe Fondanaiche.
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.
+------------+ |\ | /| | \ | | | | | 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
+---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