Henry Ernest Dudeney/Modern Puzzles/163 - The Russian Motor-Cyclists/Solution

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Modern Puzzles by Henry Ernest Dudeney: $163$

The Russian Motor-Cyclists
Two Russian Army motor-cyclists, on the road $A$, wish to go to $B$.
Now Pyotr said: "I shall go to $D$, which is $6$ miles, and then take the straight road to $B$, another $15$ miles.
But Sergei thought he would try the upper road by way of $C$.
Curiously enough, they found on reference to their odometers that the distance either way was exactly the same.
Dudeney-Modern-Puzzles-163.png
This being so, they ought to have been able easily to answer the General's simple question,
"How far is it from $A$ to $C$?"
it can be done in the head in a few moments, if you only know how.
Can the reader state correctly the distance?


Solution

$3 \tfrac 1 3$ miles.


The "pretty little rule" as Dudeney so patronisingly puts it is applied thus:

... divide $15$ by $6$ and add $2$, which gives us $4 \tfrac 1 2$.
Now divide $15$ by $4 \tfrac 1 2$, and the result $3 \tfrac 1 3$ miles) is the required distance between the two points.


Thus if $a + x$ and $b$ are the legs of the right triangle:

$x = \dfrac b {\dfrac b a + 2}$

which is actually a pretty unwieldy little rule considering how rarely one would end up using it.


Proof

Let $x$ miles be the distance from $A$ to $C$.

Let $y$ miles be the distance from $C$ to $B$.

We have:

\(\text {(1)}: \quad\) \(\ds \paren {x + 6}^2 + 15^2\) \(=\) \(\ds y^2\) Pythagoras's Theorem
\(\text {(2)}: \quad\) \(\ds x + y\) \(=\) \(\ds 6 + 15\) ... the distance either way was exactly the same.
\(\ds \leadsto \ \ \) \(\ds x^2 + 12 x + 36 + 225\) \(=\) \(\ds \paren {21 - x}^2\) multiplying out and substituting for $y$ from $(2)$
\(\ds \leadsto \ \ \) \(\ds 54 x\) \(=\) \(\ds 180\) more simplifying
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {10} 3\) more simplifying


The general case can be explored as:

\(\text {(1)}: \quad\) \(\ds \paren {a + x}^2 + b^2\) \(=\) \(\ds \paren {\paren {a + b} - x}^2\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds a^2 + 2 a x + x^2 + b^2\) \(=\) \(\ds \paren {a + b}^2 - 2 \paren {a + b} x + x^2\) multiplying out
\(\ds \leadsto \ \ \) \(\ds a^2 + 2 a x + x^2 + b^2\) \(=\) \(\ds a^2 + 2 a b + b^2 - 2 a x - 2 b x + x^2\) more multiplying out
\(\ds \leadsto \ \ \) \(\ds \paren {2 a + b} x\) \(=\) \(\ds a b\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {a b} {2 a + b}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac b {2 + b / a}\)

as we were to show.

$\blacksquare$


Sources

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