## Task 36/82

It is New Year's Eve, a few minutes before midnight. On a clock, the following observations can be made:

1. The hour hand and the line connecting the tips of the hands are the legs of a right triangle, whose hypotenuse is the minute hand.
2. The side lengths of this triangle (measured in cm) are integers and pairwise coprime.
3. For the perimeter $U$ and the area $A$ of this triangle, it holds that $A: U = A: U = 1 \mathrm{~cm}$.
4. The shortest side of the triangle is the line connecting the tips of the hands.

How much longer will the old year last?
Let $a$ be the length of the hour hand, $b$ the distance between the tips of the hands, and $c$ the length of the minute hand in $\mathrm{cm}$. Then we have:

1. $a^{2}+b^{2}=c^{2}$, thus $c=\sqrt{a^{2}+b^{2}}$
2. $0.5 a b:(a+b+c)=1$, thus $a b=2(a+b+c)$.

By substitution, we get

$$
a b=2\left(a+b+\sqrt{a^{2}+b^{2}}\right)
$$

Solving this equation for $a$ (since $b \neq 0$): $a=4+\frac{8}{b-4}$.

Since $a \in \mathbb{N}$, we have $5 \leq b \leq 12$ and the solutions are only for

$$
\begin{array}{cccc}
b_{1}=5 & b_{2}=6 & b_{3}=8 & b_{4}=12 \\
a_{1}=12 & a_{2}=8 & a_{3}=6 & a_{4}=6 \\
c_{1}=13 & c_{2}=10 & c_{3}=10 & c_{4}=10
\end{array}
$$

The requirement for pairwise coprimality excludes the solutions $\left(a_{2} ; b_{2} ; c_{2}\right)$ and $\left(a_{3} ; b_{3} ; c_{3}\right)$. Furthermore, $\left(a_{4} ; b_{4} ; c_{4}\right)$ is excluded due to condition 4. Thus, only the solution $a_{1}=12 \mathrm{~cm} ; b_{2}=5 \mathrm{~cm}$; $c_{2}=13 \mathrm{~cm}$ remains.

If we denote the angle that the minute hand has to traverse until 24:00 with $\varphi_{1}$ and the corresponding angle of the hour hand with $\varphi_{2}$, then we have

$$
\tan \left(\varphi_{1}-\varphi_{2}\right)=\frac{5}{12} ; \varphi_{1}=\arctan \frac{5}{12}+\varphi_{2} \quad \text { (1) } \quad \varphi_{1}=12 \varphi_{2} ; \varphi_{2}=\frac{1}{12} \varphi_{1}
$$

thus

$$
\varphi_{1}=\arctan \frac{5}{12}+\frac{1}{12} \varphi_{1} \quad \text { becomes } \quad \varphi_{1}=\frac{12}{11} \arctan \frac{5}{12}
$$

Since the angular velocity $\omega_{1}$ of the minute hand is given by

$$
\omega_{1}=2 \pi h^{-1}=\frac{2 \pi}{3600} s^{-1}=\frac{\Delta \varphi}{\Delta t}
$$

the time follows as

$$
\Delta t=\frac{\Delta \varphi}{\omega_{1}}=\frac{\varphi_{1}}{\omega_{1}} \approx 246.8 \mathrm{~s}=4 \min 6.8 \mathrm{~s}
$$

# 2.23 Problems and Solutions 1983 

#### Abstract