The following trigonometric expression arises in Malfiatti's problem
[math]\;\;\; \sin² λ = \sin² α + \sin² β+ 2 \sinα \sinβ \cosλ [/math]
What is not obvious, is this: if the equality holds, then [i]α+β = λ[/i]. And we may draw λ as the external angle of a triangle, with opposite internal angles α, β.
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[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)
