Henry Ernest Dudeney/Puzzles and Curious Problems/61 - Right and Left/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $61$

Right and Left

At what time between three and four o'clock will the minute hand be as far from $12$ on the left side of the dial plate as the hour hand is from $12$ on the right side of the dial plate?

Solution

$41 \tfrac 7 {13}$ past $3$.

Proof

Let $T$ be the time in question.

Let $m$ be the number of minutes past $3:00$ 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 $3:00$ and $4:00$ we have that $90 < \phi < 120$.

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

$12 \paren {\phi - 90} = \theta$

Then we are told that:

$\theta = 360 - \phi$

Hence:

\(\text {(1)}: \quad\)

\(\ds 360 - \theta\)

\(=\)

\(\ds \phi\)

\(\text {(2)}: \quad\)

\(\ds 12 \paren {\phi - 90}\)

\(=\)

\(\ds \theta\)

\(\ds \leadsto \ \ \)

\(\ds 12 \paren {360 - \theta - 90}\)

\(=\)

\(\ds \theta\)

substituting for $\phi$ in $(2)$ from $(1)$

\(\ds \leadsto \ \ \)

\(\ds 13 \theta\)

\(=\)

\(\ds 12 \times 270\)

\(\ds \leadsto \ \ \)

\(\ds \theta\)

\(=\)

\(\ds \dfrac {12 \times 270} {13}\)

\(\ds \leadsto \ \ \)

\(\ds m\)

\(=\)

\(\ds \dfrac {12 \times 270} {13} \times \dfrac 1 6\)

Speed of Minute Hand

\(\ds \)

\(=\)

\(\ds 41 \tfrac 7 {13}\)

calculating

Hence the result.

$\blacksquare$

Sources

• 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $61$. -- Right and Left
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $53$. Right and Left