This is solved by using similar triangles leading to a quartic
equation which has to be solved by numerical methods. Sue Liu of
Madras College used similar triangles and then the Newton Raphson
to give a numerical solution.
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Let $AB = a$, $BC = b$, $CD = c$, $AC = 20$, BD = 30$, EF = 8$. By
similar triangles:$${x \over b}= {8\over c}, \quad {b -
x\over b} = {8 \over a}.$$ Adding these two expressions: $${8\over
a} + {8\over c} = 1.\quad (1)$$ We can write $c$ in terms of $a$
using Pythagoras Theorem which gives: $$b^2 = 900 - c^2
= 400 - a^2. \quad (2)$$ and hence $$c^2 = 500 + a^2.$$ So (1)
becomes $${8\over a} + {8\over \sqrt (500 + a^2)} = 1
\quad (3)$$ and we have to solve equation (3) to find $a$ and then
we can find $b$ from equation (2). Rearranging (3) and squaring we
get $${64\over (a^2 + 500)} = {(a - 8)^2\over a^2}.$$ Rearranging
this equation gives: $$a^4 - 16a^3 + 500a^2 - 8000a + 32000 = 0.$$
Using the Newton Raphson method, or alternatively 'interval
halving' gives the solution: $a = 11.712$ because the other roots
of the quartic equation are less than 8 or complex. From this we
get $b=16.212$ and $c=25.242$ so the distance between the bases of
the trees is 16.212 metres (to 3 decimal places). Another approach
is to set up a spreadsheet to give the height of the intersection
of the ladders $h$ in terms of the distance $b$ between the trees
at ground level using the equivalent of equation (1), namely:
$${1\over \sqrt(400 - b^2)} + {1\over \sqrt (900 - b^2)} = {1\over
h}.$$ Use the spreadsheet calculation, and interpolation, until $h
= 8$ as nearly as required. This method will then give the
corresponding value of $b$.
