A triangle \(ABC\) has sides \(BC\), \(CA\), and \(AB\) of lengths \(a\), \(b\) and \(c\) respectively, and angles at \(A\), \(B\) and \(C\) are \(\alpha\), \(\beta\) and \(\gamma\) where \(0 \leq \alpha ,\beta ,\gamma \leq \frac{1}{2}\pi\).

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The points \(P\), \(Q\) and \(R\) are respectively the feet of the perpendiculars from \(A\) to \(BC\), \(B\) to \(CA\), and \(C\) to \(AB\) as shown.

Prove that \[\text{Area of $PQR$} = (1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma) \times (\text{Area of $ABC$}).\]

For what triangles \(ABC\), with angles \(\alpha\), \(\beta\), \(\gamma\) as above, does the equation \[\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1\] hold?

Use this applet to explore the behaviour for different triangles \(ABC\).

The blue triangle \(PQR\) is called a “pedal triangle” because it is formed from the *feet* of perpendiculars.

Remember that in the question, the angles at \(A\), \(B\) and \(C\) cannot be obtuse.

When does the blue triangle disappear?