Mystic cevians 2

Geometry Level 5

Cevians \(AD, BE, CF\) are constructed in a triangle \(ABC\). Let the feet of perpendiculars on \(AD\) from \(B\) and \(C\) respectively be \(B_{a}\) and \(C_{a}\), the feet of perpendiculars on \(BE\) from \(C\) and \(A\) respectively be \(C_{b}\) and \(A_{b}\) and the feet of perpendiculars on \(CF\) from \(A\) and \(B\) respectively be \(A_{c}\) and \(B_{c}\).

Suppose:

\(\large{\frac{BB_{a}}{CC_{a}}=\frac{CC_{b}}{AA_{b}}=\frac{AA_{c}}{BB_{c}}=\kappa}\)

It may be observed then, that the cevians shall be concurrent if and only if \(\kappa = 1\). Suppose \(\kappa\) differs from \(1\) by \(\delta \%\). Then the cevians shall meet pairwise at three different points \(P_{1},P_{2}\) and \(P_{3}\).

Determine for what value of \(\delta\) will the triangle \(P_{1}P_{2}P_{3}\) have an area \(\frac{\Delta}{2014}\) where \(\Delta\) is the area of triangle \(ABC\).

\(\large{Note:}\) Report the rounded off value for \(\delta\), i.e. the integer closest to \(\delta\). Inspired by a popular problem, otherwise, original!

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