Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $62$

At Right Angles
How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?


Solution

$10 \tfrac {10} {11}$ past $5$.
Dudeney-Puzzles-and-Curious-Problems-62-Solution.png


Proof $1$

Let $T$ be the time in question.

Let $m$ be the number of minutes past $5:00$ that $T$ is.

Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.

Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.

Between $5:00$ and $6:00$ we have that $150 < \phi < 180$.

Hence from Condition for Valid Time Indication, we have that:

$12 \paren {\phi - 150} = \theta$

Then we are told that:

$\theta = \phi - 90$

as (from how soon) we are interested in the earliest instance of the hands being at right angles.

Hence:

\(\text {(1)}: \quad\) \(\ds \theta + 90\) \(=\) \(\ds \phi\)
\(\text {(2)}: \quad\) \(\ds 12 \paren {\phi - 150}\) \(=\) \(\ds \theta\)
\(\ds \leadsto \ \ \) \(\ds 12 \paren {\theta + 90 - 150}\) \(=\) \(\ds \theta\) substituting for $\phi$ in $(2)$ from $(1)$
\(\ds \leadsto \ \ \) \(\ds 11 \theta\) \(=\) \(\ds 12 \times 60\)
\(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \dfrac {12 \times 60} {11}\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds \dfrac {12 \times 60} {11} \times \dfrac 1 6\) Speed of Minute Hand
\(\ds \) \(=\) \(\ds 10 \tfrac {10} {11}\) calculating

Hence the result.

$\blacksquare$


Proof $2$

From Time when Hour Hand and Minute Hand at Right Angle, the times when this is between $5$ and $6$ are:

$05:10:55$

and:

$05:43:38$

The first of these is the earlier of those two times.

$\blacksquare$


Sources

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