Henry Ernest Dudeney/Modern Puzzles/164 - Those Russian Cyclists Again/Solution

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Modern Puzzles by Henry Ernest Dudeney: $164$

Those Russian Cyclists Again
In the section from a map given in the diagram we are shown three long straight roads, forming a right-angled triangle.
Dudeney-Modern-Puzzles-164.png
The General asked the two men how far it was from $A$ to $B$.
Pyotr replied that all he knew was that riding round the triangle, from $A$ to $B$,
from there to $C$ and home to $A$, his odometer showed exactly $60$ miles,
while Sergei could only say that he happened to know that $C$ was exactly $12$ miles from the road $A$ to $B$ --
that is, to the point $D$, as shown by the dotted line.
Whereupon the General made a very simple calculation in his head and declared that the distance from $A$ to $B$ must be ...
Can the reader discover so easily how far it was?


Solution

$25$ miles.


Proof

Let $BC = a$, $AC = b$ and $AB = c$.

Let $CD = d$.

Let $AD = c_1$ and $BD = c_2$.

We have:

\(\text {(1)}: \quad\) \(\ds a + b + c\) \(=\) \(\ds 60\) riding round the triangle, from $A$ to $B$ ... exactly $60$ miles
\(\text {(2)}: \quad\) \(\ds d\) \(=\) \(\ds 12\) $C$ was exactly $12$ miles from the road $A$ to $B$
\(\text {(3)}: \quad\) \(\ds a^2 + b^c\) \(=\) \(\ds c^2\)
\(\text {(4)}: \quad\) \(\ds a^2\) \(=\) \(\ds d^2 + c_2^2\)
\(\text {(5)}: \quad\) \(\ds b^2\) \(=\) \(\ds d^2 + c_1^2\)
\(\ds \leadsto \ \ \) \(\ds c_1^2\) \(=\) \(\ds b^2 - d^2\) rearranging $(5)$
\(\ds c_2^2\) \(=\) \(\ds a^2 - d^2\) rearranging $(4)$
\(\ds \leadsto \ \ \) \(\ds c_1^2 c_2^2\) \(=\) \(\ds \paren {a^2 - d^2} \paren {b^2 - d^2}\)
\(\ds \leadsto \ \ \) \(\ds d^4\) \(=\) \(\ds a^2 b^2 - d^2 \paren {a^2 + b^2} + d^4\) $d^2 = c_1 c_2$ from some theorem I can't find
\(\ds \leadsto \ \ \) \(\ds a b\) \(=\) \(\ds c d\) $a^2 + b^2 = c^2$ and simplification
\(\ds \leadsto \ \ \) \(\ds a b\) \(=\) \(\ds 12 c\) from $(2)$


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Then we have:

\(\ds a + b\) \(=\) \(\ds \paren {a + b + c} - c\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \paren {a + b}^2\) \(=\) \(\ds \paren {\paren {a + b + c} - c}^2\)
\(\ds \leadsto \ \ \) \(\ds a^2 + 2 a b + b^2\) \(=\) \(\ds c^2 - 120 c + \paren {a + b + c}^2\)
\(\ds \leadsto \ \ \) \(\ds c^2 + 2 c d\) \(=\) \(\ds c^2 - 2 \paren {a + b + c} c + \paren {a + b + c}^2\) from $a^2 + b^2 = c^2$ and $a b = c d$
\(\ds \leadsto \ \ \) \(\ds 2 c d\) \(=\) \(\ds -2 \paren {a + b + c} + \paren {a + b + c}^2\)
\(\ds \leadsto \ \ \) \(\ds 2 c \paren {\paren {a + b + c} + d}\) \(=\) \(\ds \paren {a + b + c}^2\) simplification
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds \dfrac {\paren {a + b + c}^2} {2 \paren {\paren {a + b + c} + d} }\) simplification

Setting $a + b + c = 60$ and $d = 12$ gives the result.


Hence Dudeney's General simply squared $60$ and divided by $2 \times \paren {60 + 12}$:

$c = \dfrac {60^2} {2 \times \paren {12 + 60} } = \dfrac {3600} {144} = 25$

$\blacksquare$


Sources

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