Henry Ernest Dudeney/Modern Puzzles/163 - The Russian Motor-Cyclists/Solution
Jump to navigation
Jump to search
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.
- 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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $163$. -- The Russian Motor-Cyclists
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $281$. The Russian Motorcyclists