# A Case for Determinants

### Solution

Denote $x+y=s\,$ and $xy=p.\,$ Then

\begin{align} c_2s&=(ax^2+by^2)(x+y)=ax^3+by^3+xy(ax+by)\\ &=c_3+c_1p \end{align}

Similarly, $c_3s=c_4+c_2p\,$ and $c_4s=c_5+c_3p.\,$ Thus, the linear system

$\left(\begin{array}{ccc}-c_2 & c_3 & c_1\\-c_3 & c_4 & c_2\\-c_4 & c_5 & c_3\end{array}\right)\left(\begin{array}{c}\,s\\1\\p\end{array}\right)=\mathbf{0}$

has a non-trivial solution, implying that the determinant of the matrix equals zero:

$D=\left|\begin{array}{ccc}-c_2 & c_3 & c_1\\-c_3 & c_4 & c_2\\-c_4 & c_5 & c_3\end{array}\right|=-\left|\begin{array}{ccc}c_2 & c_3 & c_1\\c_3 & c_4 & c_2\\c_4 & c_5 & c_3\end{array}\right|=0.$

Expanding by the last row we obtain

$c_4\left|\begin{array}{cc}\,c_3 & c_1\\c_4 & c_2\end{array}\right|+c_3\left|\begin{array}{cc}\,c_2 & c_3\\c_3 & c_4\end{array}\right|=c_5\left|\begin{array}{cc}\,c_2 & c_1\\c_3 & c_2\end{array}\right|.$

Further expansion of the $2\times 2\,$ determinant proves the statement and at the same time unveils the mystery of the identity

$\displaystyle c_5=\frac{2c_2c_3c_4-c_3^2-c_1c_4^2}{c_2^2-c_1c_3}.$

### Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc with a link to a post by Francisco Javier Gracía Capitán who had referred to Problem 15 at AIME 1990. His is a generalization of the latter.

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• Three Similar Triangles
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• Divisibility Of Prod-Dif And The Vandermonde Determinant
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