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Big Fish

Stage: 4 Short Challenge Level: Challenge Level:1


Let the weights in kilograms of the tail, head and body of the fish be $t$, $h$ and $b$ respectively.

We are told that:

$t = 9$
$h = t + \frac{1}{3} b$
$b = h+t$

So:
$h = 9 + \frac{1}{3} b$
and $b = h+9$.

Therefore $b = 9 + \frac{1}{3} b + 9$,
$\frac{2}{3} b = 18$,
$\frac{1}{3} b = 9$,
$b = 27$.

Hence $h+t = 27$, and the whole fish weighed $54$kg.

Alternatively:
Since $b = h+t$
$2b = h+t+b$
$2b = 9 + \frac{1}{3} b + 9+ b$ 
$2b = \frac{4}{3} b + 18$
$\frac{2}{3} b = 18$,
$b = 27$
$h+t+b = 54$kg.

Jonah tackled this problem by setting up a sequence that converged on the solution.
Here is one way to write out Jonah's method:

We know that
$t=9$
$h=9+\frac{b}{3}$
$b=h+9$

Initially, let $b_0=9$
$h_n=9+\frac{b_n}{3}$
$b_{n+1}=h_n+9$

Substituting each term to get the next term in the sequence, you get the first few terms:
$h_0=9+\frac{9}{3}=12$
$b_1=12+9=21$
$h_1=9+\frac{21}{3}=16$
$b_2=16+9=25$
$h_2=9+\frac{25}{3}=17\frac{1}{3}$

If we generate the terms in a spreadsheet, we see that the sequences converge to $b=27$, $h=18$ Jonah's sequence 






This problem is taken from the UKMT Mathematical Challenges.