The following trigonometric expression arises in Malfiatti's problem [math]\;\;\; \sin² λ = \sin² α + \sin² β+ 2 \sinα \sinβ \cosλ [/math] But this is an equality I know, in a [i]very[/i] frumpy disguise. Let me find it.
That is, this expression says [i]α+β = λ.[/i] But [i]An exterior angle of a triangle is equal to the sum of its two opposite, internal angles.[/i] I may draw λ as the external angle of a triangle, with opposite internal angles α, β. And the knot at the heart of Malfatti's problem is undone. ______________ [b]Malfatti's Problem:[/b] 1. Describe the constraints: [url]http://www.geogebratube.org/material/show/id/32079[/url] 2. Solution [url]http://www.geogebratube.org/material/show/id/32233[/url] [b]→3. Trig Supplement[/b] This is problem # 30 in Heinrich Dorrie's [i]100 Great Problems of Elementary Mathematics.[/i] More: (http://www.geogebratube.org/material/show/id/73813)