Henry Ernest Dudeney/Modern Puzzles/44 - Find the Distance/Solution

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Modern Puzzles by Henry Ernest Dudeney: $44$

Find the Distance
A man named Jones set out to walk from $A$ to $B$,
and on the road he met his friend Kenward, $10$ miles from $A$, who had left $B$ at exactly the same time.
Jones executed his commission at $B$ and, without delay, set out on his return journey,
while Kenward as promptly returned from $A$ to $B$.
They met $12$ miles from $B$.
Of course, each walked at a uniform rate throughout.
Now, how far is $A$ from $B$?


Solution

$18$ miles.


The "absurdly simple" rule being referred to by Dudeney:

$3$ times the distance from $A$ that they first cross minus the distance from $B$ when they cross for the second time.


Proof

Let $d$ miles be the distance between $A$ and $B$.

Let $V_J$ and $V_K$ miles per hour be the speeds of Jones and Kenward respectively.

Then we have:

\(\ds \dfrac {10} {V_J}\) \(=\) \(\ds \dfrac {d - 10} {V_K}\) on the road he met his friend Kenward, $10$ miles from $A$
\(\ds \dfrac {d + 12} {V_J}\) \(=\) \(\ds \dfrac {2 d - 12} {V_K}\) They met $12$ miles from $B$
\(\ds \leadsto \ \ \) \(\ds V_J\) \(=\) \(\ds \dfrac {10 V_K} {d - 10}\)
\(\ds V_K \dfrac {d + 12} {2 d - 12}\) \(=\) \(\ds V_J\)
\(\ds \) \(=\) \(\ds \dfrac {10 V_K} {d - 10}\)
\(\ds \leadsto \ \ \) \(\ds \paren {d + 12} \paren {d - 10}\) \(=\) \(\ds 10 \paren {2 d - 12}\)
\(\ds \leadsto \ \ \) \(\ds d^2 + 2 d - 120\) \(=\) \(\ds 20 d - 120\)
\(\ds \leadsto \ \ \) \(\ds d^2 - 18 d\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds d \paren {d - 18}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds 18\) as $d = 0$ is not a valid answer

$\Box$


For the general case, let $d_1$ and $d_2$ be the distances from $A$ and $B$ that they crossed respectively.

We have:

\(\ds \dfrac {d_1} {V_J}\) \(=\) \(\ds \dfrac {d - d_1} {V_K}\)
\(\ds \dfrac {d + d_2} {V_J}\) \(=\) \(\ds \dfrac {2 d - d_2} {V_K}\)
\(\ds \leadsto \ \ \) \(\ds V_J\) \(=\) \(\ds \dfrac {d_1 V_K} {d - d_1}\)
\(\ds V_K \dfrac {d + d_2} {2 d - d_2}\) \(=\) \(\ds V_J\)
\(\ds \) \(=\) \(\ds \dfrac {d_1 V_K} {d - d_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {d + d_2} \paren {d - d_1}\) \(=\) \(\ds d_1 \paren {2 d - d_2}\)
\(\ds \leadsto \ \ \) \(\ds d^2 + \paren {d_2 - d_1} d - d_1 d_2\) \(=\) \(\ds 2 d_1 d - d_1 d_2\)
\(\ds \leadsto \ \ \) \(\ds d^2 - \paren {3 d_1 - d_2} d\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds d \paren {d - \paren {3 d_1 - d_2} }\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds 3 d_1 - d_2\) as $d = 0$ is not a valid answer

So the general rule is:

$d = 3 d_1 - d_2$

$\blacksquare$


Sources

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