Help with a geometry problem from the Australian national mathematics competition?

A while ago i participated in the national Australian mathematics competition there was a question which had me well and truly stumped and in extra attempts outside the paper continued to stump me any help would be appreciated, thanks.

Details:

AB = 11cm

BE : EG : GH = 1 : 2 : 3 (ratio)

CF : FH = 4: 5 (ratio) (independent of the previous ratio)

Find the length of AD

https://docs.google.com/drawings/d/17Buz7xNnO6imtEeNApGdpl_QX7QSGrLBZwu6YqPCPcs/edit?usp=sharing

Note by Felix Haydock
4 years, 6 months ago

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Suppose that \(BE = x\), \(EG = 2x\), \(GH = 3x\), \(CF = 4y\), \(FH = 5y\). Using Menelaus' Theorem: \[ \begin{array}{rcl} AC \times BE \times FH & = & AB \times EH \times CF \\ (BC+11)\times x \times 5y & = & 11 \times 5x \times 4y \\ 5xy(BC+11) & = & 220xy \end{array} \] and hence \(BC+11=44\), so that \(BC=33\). Applying Menelaus' Theorem again: \[ \begin{array}{rcl} BD \times CF \times GH & = & CD \times FH \times BG \\ (BC + CD) \times 4y \times 3x & = & CD \times 5y \times 3x \\ 12xy(BC + CD) & = & 15xy CD \end{array} \] so that \(5CD \,=\, 4(BC+CD)\), and hence \(CD = 4BC = 132\). Thus \(AD = 11+33+132=176\) cm. There are several more ways in which we could invoke Menelaus' Theorem to work out all the other lengths.

Mark Hennings - 4 years, 6 months ago

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