Monday 17 December 2012

New Scientist Enigma 1728

This week's Enigma puzzle was interesting enough to publish an analysis.

This diagram shows the dynamics of the situation. $V_j$ is Jack's speed, $V_k$ is Ken's speed. $g$ is their initial goal line separation.



Defining $D$ as the distance between Joe and Ken, we will try to find the minimum of $D^2$ and hence the minimum of $D$

$D^2 = (g-V_j t cos\theta)^2 + t^2(V_k - V_j sin\theta)^2 $

$D^2 = g^2 - 2 g V_j t cos\theta +(V_j^2 + V_k^2)t^2 - 2 V_j V_k t^2 sin\theta$           ..............(1)

The minimum occurs when:

                    $\frac{\partial{D^2}}{\partial \theta}$ $= 2gt V_j sin\theta - 2 V_j V_k t^2cos\theta =0$                               .................(2)

                    $\frac{ \partial{D^2}}{\partial{t}}$ $= -2g V_j cos\theta +2t(V_j^2 + V_k^2) - 4t V_j V_k sin\theta =0$ ...............(3)

From (2), 
$t=$ $\frac{g}{V_k}$ $tan\theta$                                                  ..............................(4)

Substituting (4) into (3):
$ -2g V_j cos\theta +2\frac{g}{V_k} (V_j^2 + V_k^2)\frac{sin\theta}{cos\theta} - 4g V_j\frac{sin^2\theta}{cos\theta} =0$           .....................(5)

Defining $s=sin\theta$ and multiplying (5) through by $\frac{- cos\theta}{2gV_j}$ :

$ (1-s^2)-(\frac{V_j}{V_k}+\frac{V_k}{V_j})s+2s^2= (s-\frac{V_j}{V_k})(s-\frac{V_k}{V_j}) = 0$ 


If $V_j < V_k$, this quadratic has one solution satisfying $s\lt1$, namely $s=\frac{V_j}{V_k}$

so 

$sin\theta=$$\frac{V_j}{V_k}$ and $t=$ $\frac{g}{\sqrt{1-V_j^2/V_k^2}}$

Substituting these into (1)  gives:

$D^2 = $$\frac{g^2}{V_k^2}$$(V_k^2-V_j^2)$

so

$D = g\sqrt{1 - V_j^2/V_k^2}$        ............................... (6)

Substituting the values $g=25ft, V_j=12mph, V_k=12.5mph$ into (6) gives the answer to the puzzle.

The notable aspect of this analysis is that to minimise the distance between them when $V_j \lt V_k$, Joe runs at an angle $\theta$ where $sin\theta=\frac{V_j}{V_k}$. When $V_j \gt V_k$, Joe would run at an angle $\theta$ where $sin\theta=\frac{V_k}{V_j}$ in order to intercept Ken.



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