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# 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)

Note by Felix Haydock
3 years, 11 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.

- 3 years, 11 months ago